I spent the last week vacationing in South Carolina, where my parent's house seems to have Atlantic Monthly's and Harper's from the dawn of time. What luck, then, that one of the most interesting articles (at least statistically) was in an issue as recent as the July/August 2008 issue of the Atlantic. The article is called "American Murder Mystery" and it's by Hanna Rosin.
The article talks of the recent increase in violent crime in mid-sized cities. In many of these cities, government housing projects (called "Section 8" housing) have been torn down. In their place, the government has provided the poor with rent subsidies so that they can move to private housing. Rosin describes how Phyllis Betts and Richard Janikowski, of the University of Memphis, tie the increase in crime in these cities to the destruction of these projects. A striking quote in the article is from the Memphis police chief: '“It used to be the criminal element was more confined,” said Larry Godwin, the police chief. “Now it’s all spread out."'
The primary statistical evidence given in the article of an association between crime and former Section 8 residents, is a map that shows areas with high incidents of crime correspond to areas with a large number of people with Section 8 subsidies (i.e., former residents of housing projects). As convincing as this might sound, it has a fatal flaw: the map looks at total incidents rather than crime rate. This means that an area with 10,000 people and 100 crimes (and 100 Section 8 subsidy recipients) will look much worse than an area with 100 people and 1 crime (and 1 Section 8 subsidy recipient). However, both areas have the same rate of crime, and, presumably, the same odds of being a victim of crime (see my earlier blog about the safest place to live for some explanation of the use of rates in measuring crime). Yet in Betts and Janikowski's analysis, the area with 10,000 people has a higher number Section 8 subsidy recipients and higher crime, thus "proving" their theory of association.
Of course, there will be both a greater number of Section 8 subsidy recipients and a greater number of crimes in the area with 10,000 people than in the area with 100 people . Thus, while the map presented in the Atlantic article does indeed seem to indicate that there is higher crime in areas where there are more Section 8 subsidies, this differential might be entirely an artifact of population density, and, in fact, the crime rate may be completely unrelated to where Section 8 subsidy recipients reside. Without an adjustment for population density, the inferences made from the association are statistically meaningless.
Thursday, August 28, 2008
Wednesday, July 2, 2008
Statistics in Politics - Lies and Damn Lies
The nice thing about politicians and the newspaper columnists that write about them is that they lie a lot about statistics. That makes writing a blog that points out the errors easy to create. This weeks subject is David Brooks' latest New York Times column.
Brooks takes issue with Obama’s claim that his fundraising is from a broad base of small donors, and goes on to compare Obama money raised to McCain money raised by special interest group. I am not going to attack the actual dollar figures that Brooks gives. He cites no sources whatsoever, so that makes them hard to attack anyway. Instead, I am going to show how presenting raw numbers without proper context creates a biased picture.
Let’s take Brooks’ first claim: He says “lawyers account for the biggest chunk of Democratic donations” and have donated $18 million, as compared to $5 million for McCain. This sounds like 1) Obama is getting most (“biggest chunk”) of his donations from one big special interest group (lawyers) and 2) Obama is getting 3 times as much of his donations from this group as McCain.
Here’s the problem: Obama has out raised McCain by more than 2 to 1. According to CBS News, Obama’s total amount raised is $295.5 million compared to McCain’s $121.9 million. Thus, the $18 million raised from lawyers represents only 6% of the money raised. Still a lot of money, but it puts the “biggest chunk” in context. McCain’s $5 million raised from lawyers, on the other hand, represents 4% of the total money he raised. Thus, Obama is getting more as a percentage from lawyers but instead of 18 to 5, or 3 times as much, it’s 6% to 4%, or 50% more. Another issue is that there is a difference between individuals who are lawyers and public interest groups for lawyers. Brooks is trying to blur those lines by grouping all their donations together (to be fair, he does not say "special interest groups"). Sure, a lot of lawyers certainly support some of the public interest groups, but others do not. Also, these groups can be at odds with one another, so grouping all lawyers together gives you the bigger number but is inaccurate.
Brooks goes on to compare several other groups of professions. In each of these areas, Obama receives more money in absolute dollars. However, in terms of percentage of total donations, McCain is usually always receiving more: from financial securities workers, McCain gets 36% more as a percent of his total; from real estate workers, McCain gets 94% more; from bank workers, McCain gets 82% more; from hedge fund workers, McCain gets 29% more; from medical/health care workers, McCain gets 4% more.
There are two other areas (in addition to lawyers) where Obama is receiving more in percentage terms. The first is “communications and electronics”, where Obama is getting 106% more in percentage terms. The second is “Professors and other people who work in education.” In that area, Obama gets a whopping 4 times as much as McCain as a percentage of total funds raised. Brooks implies that these are "part of a spontaneous movement of small-money enthusiasts," but he doesn't support that with any evidence showing that these groups are anything more than an unorganized group of individuals--all the polls have indicated that more educated people lean toward Obama, so why wouldn't they give more?
The last thing that Brooks points out is that although, as Obama claims, 90% of his donors gave less than $200, only 45% of his donated money comes from such small donors. This is a good point, and Obama, who has been claiming this for awhile, should be called to the mat on it.
However, it would be more interesting to look at the percent of small donors and money from small donors in McCain’s campaign as a comparison. You can bet that it’s less than 45% of donated money and less than 90% of donors. Yet, a comment on Brooks’ article by a New Republic blogger puts it into context, pointing out that “31 percent of Bush's money in 2004 came from donations of $200 or less (compared to 16 percent in 2000). Kerry, meanwhile, raised 37 percent...” (the blog sites this article on 2004 donations (by Joseph Graf) as its source). Thus, 45% is a lot, but the number has been increasing for both parties, with the most obvious reason that the Internet has allowed candidates to easily reach out to everyone, rather than raising most of their money through $1,000 a plate dinners and the like (campaign finance reform, which limits individual contributions, also had a role in bringing up the percentage raised through smaller donors).
The lesson here is, of course: “don’t believe the numbers.” David Brooks is going to make them look good for McCain--he’s a columnist, not a reporter--just as other columnists are going to make them look good for Obama.
Brooks takes issue with Obama’s claim that his fundraising is from a broad base of small donors, and goes on to compare Obama money raised to McCain money raised by special interest group. I am not going to attack the actual dollar figures that Brooks gives. He cites no sources whatsoever, so that makes them hard to attack anyway. Instead, I am going to show how presenting raw numbers without proper context creates a biased picture.
Let’s take Brooks’ first claim: He says “lawyers account for the biggest chunk of Democratic donations” and have donated $18 million, as compared to $5 million for McCain. This sounds like 1) Obama is getting most (“biggest chunk”) of his donations from one big special interest group (lawyers) and 2) Obama is getting 3 times as much of his donations from this group as McCain.
Here’s the problem: Obama has out raised McCain by more than 2 to 1. According to CBS News, Obama’s total amount raised is $295.5 million compared to McCain’s $121.9 million. Thus, the $18 million raised from lawyers represents only 6% of the money raised. Still a lot of money, but it puts the “biggest chunk” in context. McCain’s $5 million raised from lawyers, on the other hand, represents 4% of the total money he raised. Thus, Obama is getting more as a percentage from lawyers but instead of 18 to 5, or 3 times as much, it’s 6% to 4%, or 50% more. Another issue is that there is a difference between individuals who are lawyers and public interest groups for lawyers. Brooks is trying to blur those lines by grouping all their donations together (to be fair, he does not say "special interest groups"). Sure, a lot of lawyers certainly support some of the public interest groups, but others do not. Also, these groups can be at odds with one another, so grouping all lawyers together gives you the bigger number but is inaccurate.
Brooks goes on to compare several other groups of professions. In each of these areas, Obama receives more money in absolute dollars. However, in terms of percentage of total donations, McCain is usually always receiving more: from financial securities workers, McCain gets 36% more as a percent of his total; from real estate workers, McCain gets 94% more; from bank workers, McCain gets 82% more; from hedge fund workers, McCain gets 29% more; from medical/health care workers, McCain gets 4% more.
There are two other areas (in addition to lawyers) where Obama is receiving more in percentage terms. The first is “communications and electronics”, where Obama is getting 106% more in percentage terms. The second is “Professors and other people who work in education.” In that area, Obama gets a whopping 4 times as much as McCain as a percentage of total funds raised. Brooks implies that these are "part of a spontaneous movement of small-money enthusiasts," but he doesn't support that with any evidence showing that these groups are anything more than an unorganized group of individuals--all the polls have indicated that more educated people lean toward Obama, so why wouldn't they give more?
The last thing that Brooks points out is that although, as Obama claims, 90% of his donors gave less than $200, only 45% of his donated money comes from such small donors. This is a good point, and Obama, who has been claiming this for awhile, should be called to the mat on it.
However, it would be more interesting to look at the percent of small donors and money from small donors in McCain’s campaign as a comparison. You can bet that it’s less than 45% of donated money and less than 90% of donors. Yet, a comment on Brooks’ article by a New Republic blogger puts it into context, pointing out that “31 percent of Bush's money in 2004 came from donations of $200 or less (compared to 16 percent in 2000). Kerry, meanwhile, raised 37 percent...” (the blog sites this article on 2004 donations (by Joseph Graf) as its source). Thus, 45% is a lot, but the number has been increasing for both parties, with the most obvious reason that the Internet has allowed candidates to easily reach out to everyone, rather than raising most of their money through $1,000 a plate dinners and the like (campaign finance reform, which limits individual contributions, also had a role in bringing up the percentage raised through smaller donors).
The lesson here is, of course: “don’t believe the numbers.” David Brooks is going to make them look good for McCain--he’s a columnist, not a reporter--just as other columnists are going to make them look good for Obama.
Thursday, June 12, 2008
Let’s make a deal problem
Monty Hall, the host, allows you to choose one of three curtains. Behind one of the curtains is a new car or another big prize, while behind the other two is a year’s supply of shampoo or the equivalent. You choose Curtain 1. Monty opens Curtain 2 and shows you it has a year’s supply of the shampoo. Then he gives you a choice:
a) stick with your original decision, or
b) switch to Curtain 3.
1) always shows a curtain with the shampoo behind it;
2) never reveals the curtain you chose; and
3) randomly decides which of the remaining two curtains to reveal if the curtain you chose contains the car,
then switching to Curtain 3 gives you a 2/3 chance of winning while sticking to Curtain 1 gives you a 1/3 chance of winning.
This problem, like many probability problems, is one of information. Initially, you have no information about any of the three curtains so each choice gives you a 1/3 chance of winning. By showing you the curtain with the shampoo, you have learned nothing new about the curtain you originally chose—because there was no way, whether your curtain had the car or the shampoo, that Monty was going to show you what was behind your curtain. Your curtain had, and still has (as far as you know), a 1/3 chance of containing the car. However, you did get information about Curtain 3: Monty did not choose to reveal it. This could mean one of two things:
A. Curtain 3 has the car, and therefore Monty had to show you Curtain 2, as he would never reveal the curtain with the car (1/3 chance, calculated by taking the 1/3 chance that Curtain 3 has the car and multiplying by the 100% chance that he reveals Curtain 2 when the car is behind Curtain 3); or
B. Curtain 1 has the car, and Monty chose to reveal Curtain 2 (1/6th chance, calculated by taking the 1/3 chance that curtain 1 has the car and multiplying by the ½ chance that Monty reveals curtain Number 2 when the car is behind Curtain 1).
These probabilities do not sum to 1, because we are excluding the outcomes, now impossible, where Monty reveals Curtain Number 3. In order to revise the probabilities to take into account what was revealed by Monty, we need to divide the probabilities in A (1/3) and B (1/6) above by the chances of the two possible remaining outcomes (1/3 plus 1/6 = 1/2). Thus, outcome A (car is behind Curtain 3) has a probability of (1/3) / (½)= 2/3, while outcome B (car is behind Curtain 1) has a probability of (1/6)/(1/2) = 1/3.
The intuition is as follows: Monty always reveals Curtain 2 or 3 when you choose 1, so you do not get any more information about whether it is behind 1 by this revelation, but you do gain information about 2 and 3 from this revelation, since he never reveals 3 if the car is behind it but does sometimes reveal 3 if the car is not behind it. Thus, the fact that Monty did not reveal Curtain 3 tells you something.
[Note: this problem has been around for awhile, but was made famous by Marilyn Vos Savant’s discussion of it and the subsequent outcry by those who insisted her answer, the correct one, was wrong. See, for example: http://www.letsmakeadeal.com/problem.htm]
Technical Explanation
There are a whole class of problems in probability that involve updating the chances based on new information. These problems are solved according to Bayes’ Rule, after a law in probability that specifies how to update probabilities with new information (for a full discussion, including discussion of whether the Reverend Bayes was actually the first to discover this theorem, see the Wikipedia entry: http://en.wikipedia.org/wiki/Bayes'_theorem).
To understand Bayes’ Rule, we need to first know the notation used for conditional probability. We use the vertical line ( | ) to denote a condition and, as in prior blogs, P(A) is the probability that event A occurs. Thus, P(A|R2) is the probability that A occurs, given that R2 already occurred. Bayes' Rule is:
P(A|R2) = P(R2|A)*P(A) / P(R2)
So let:
A=event that prize is under Curtain 3
R2= event that Monty reveals the curtain 2 contents
C=event that prize is under Curtain 1
Now we can figure out the right side of the Bayes’ Rule equation, in order to figure out P(A| R2).
We know P(R2|A) = 1, because Monty won’t reveal curtain 3 when it contains the prize and he won’t reveal curtain 1 because you chose curtain 1.
P(A) = P(C) = 1/3 ==> remember, this one is unconditional, so given three curtains, there’s a 1/3 chance of the prize being behind each.
To figure out P(R2), it is useful to note that for any events R2 and A, P(R2 and A) = P(A) * P(R2|A)
In our case, the P(R2) is the sum of the probabilities of 2 exclusive events:
1) prize is under curtain 3 (event A) and Monty reveals curtain 2 (event R2): 1/3 * 1=1/3
2) prize is under curtain 1 (event C) and Monty reveals curtain 2 (event R2) 1/3*1/2 = 1/6.
This sum, 1/3 plus 1/6 is ½=P(R2).
Thus, by Bayes Rule, P(A| R2) = (1*1/3) / ½ = 2/3
Just for fun, now you can compute P(C|R2) = P(prize is under Curtain 1 given that Curtain 2 is revealed) = 1/3 using Bayes’ Rule.
False Positives in Cancer Diagnoses
The outcome of Bayes’ Rule can be very confusing, and is important to keep in mind in more important problems than the Let’s Make a Deal problem. For example, suppose an MRI for breast cancer has a false negative rate of 1/100, meaning that the test will incorrectly indicate that you do not have cancer when you in fact do 1 in 100 times. Similarly, the test might also have a false positive rate of 1 in 100, meaning that the test will incorrectly indicate that you do have cancer when in fact you do not 1 in 100 times (false positive rates for MRIs over time can be much higher, because they are frequently done once or twice a year: see the recent article about a study of false positives in MRIs for breast cancer screening, which were around 25% over time.
Suppose your MRI result just came out positive for breast cancer. What are the chances you actually have breast cancer?
First, it’s useful to know that around 250,000 women a year get breast cancer (see this site) and there are about 60 million women above the age of 40 (see census site), when most cases occur. This represents an annual infection rate of nearly 1 in 200.
P(C) = Probability of breast cancer in a given year = 1/200 = 0.005
P(D| not C) = Probability that MRI diagnosed cancer given that you do NOT have cancer = false positive = 1/100 =0.01
P(N|C) = Probability of MRI did not diagnose given that you have cancer = false negative = 1/100= .01
P(D|C) = 1-P(N|C) = Probability that MRI diagnosed cancer given that you have cancer = 99/100 =0.99
We want P(C|D) = Probability of cancer, given a cancer diagnosis by MRI.
Before using Bayes’ Rule, we can first define P(D) as the sum of the probabilities of all exclusive events that include D. In English, the chance of diagnosis is the sum of 1) the chance that you have cancer and are diagnosed and 2) the chance that you do not have cancer and are diagnosed. Thus, P(D) = P(D|C)*P(C) + P(D|not C)* P(not C) = 0.99*0.005 + 0.01* .995 =.0149
Using Bayes’ Rule:
P(have cancer given the MRI result shows cancer) = P(C|D) = P(D|C)*P(C)/ P(D) = 0.99 * .005 / .0149 = .33 or about 1/3.
Thus, a very effective MRI test for cancer, which gives the wrong result only 1% of the time, is still suspect when it gives a result of cancer. In fact, an MRI diagnosis of cancer indicates only a 1/3 probability of actually having cancer (keep in mind while there are indications that false positives I used here for the MRI are made up, though they do appear to be at least in the 1% range).
It’s easy to understand what happens logically when you imagine that 200 women come in for screening. Only 1 will probably have cancer, since the cancer rate is about 1 in 200. The MRI will almost surely diagnose her (99% chance). For the other 199, the MRI will indicate no cancer for all but about 1%, which means it will indicate cancer for about 2 of them. Thus, of the 3 cases where the MRI indicates cancer, 2 of them will be false indications.
Thursday, May 8, 2008
Why are there too many boys in China?
For a long time now, the ratio of males to females in China has been increasing. In fact, one of the most recent articles I could find on it was from 2004, where the ratio stood at around 120 boys to every 100 girls (see the msnbc article).
It's clear to most that the combination of the one child law, preventing most chinese couples from having more than one child, and the preference in China for boys, is driving this (though there are other explanations, including the possibilities of different effects of some diseases: see this business week article).
There are two sinister mechanisms for ensuring that your only child is a boy: selective abortion or infanticide. Yet there is another option: just have another baby if the first is a girl, and don't tell the government. I think this third option is more likely, because I do not think most families can afford an abortion (illegal for sex selection) and very few mothers would kill their babies.
So how much does this non-reporting need to happen to change the ratio from the normal 106 to 100 male to female births to the abnormal 120 to 100?
The answer to this is the combination of three things: 1) percent of births that are girls (with no intervention), 2) percent of families that have another baby (hoping for a boy), given the first is a girl, and 3) the percent of families that do not tell the government about the first baby.
Lets call these percentages, Pg, P2, and Ps (for girl, 2nd child, and secret). Lets also call Pr the reported percent of girls, which is 100/220, or 45.45%. We'll assume also for simplicity that families quit trying when they have a boy or have 2 children, whichever comes first. Also, we'll assume families always report the first child if it is a boy or if they have no more children.
Pg is known at around 100/206=48.54%
P2 and Ps are unknown.
We want to figure out what P2 and Ps could lead to the Pr being 45.45% when Pg is 48.54%.
First, consider that, given the ground rules above, the following are the types of families that can exist (in birth order):
B (boy, one child only)
G (girl, one child only)
GB (girl boy, two children)
GG (girl girl, two children)
To figure out the percent of girls reported, we need the total girls reported divided by the total children reported. This is easy to figure out for each combination above:
B = 0 girls / 1 child
G = 1 Girl / 1 child
GB = 0 girls / 1 child Ps percent of the time and 1 girls / 2 children (1- Ps percent of the time)
GG = 1 girl / 1 child Ps percent of the time and 2 girls / 2 children (1-Ps percent of the time)
We are almost there. Now we just need to sum the numerators multiplied by their probabilities and the denominators multiplied by their probabilities. Here are the probabilities of each family combination:
B = 1-Pg
G = Pg*(1 - P2) ==> It's just Pg times the percent of families who do not have more children
GB = Pg*(P2)*(1-Pg) It's the chances of a girl, followed by the decision to have a 2nd, followed by having a boy.
GG = Pg*(P2)*Pg=Pg^2*P2
Thus the numerator (number of girls reported average is):
Num = (1-Pg)*0 +
Pg*(1-P2)*1 +
Pg*P2*(1-Pg)*Ps*0 +
Pg*P2*(1-Pg)*(1-Ps)*1 +
Pg^2*P2*Ps*1 +
Pg^2*P2*(1-Ps)*2
and the denominator (number of children reported on average):
Den = (1-Pg)*1 +
Pg*(1-P2)*1 +
Pg*P2*(1-Pg)*Ps*1 +
Pg*P2*(1-Pg)*(1-Ps)*2 +
Pg^2*P2*Ps*1 +
Pg^2*P2*(1-Ps)*2
We know that, in China, Pr= Num/Den = 45.45% and that, in general, Pg=48.54%. Thus, we can solve .4545=Num/Den in terms of Ps and P2.
Since we have 1 equations and 2 unknowns, there are an infinite number of solutions, but here are a few possibilities:
0% have a second child --impossible
10% have a second child -- impossible
15% have a second child and 85% of those keep the first a secret from the government
20% have a second child and 65% of those keep the first a secret from the government
30% have a second child and 45% of those keep the first a secret from the government
40% have a second child and 35% of those keep the first a secret from the government
50% have a second child and 30% of those keep the first a secret from the government
One thing to note (that is not necessarily obvious in these calculations) is that if everyone reports all the children they have (Ps=0), then the percent of girls will be exactly 48.54%, the same as if everyone had one child, as long as infanticide and selective abortion are not occurring.
But the main point here is that a small number (15%) of couples having second children and not reporting the first girl leads to the warped percentages of baby girls, if there is high under-reporting of these first children. You do not need to assume that infanticide or selective abortion plays a role at all.
It's clear to most that the combination of the one child law, preventing most chinese couples from having more than one child, and the preference in China for boys, is driving this (though there are other explanations, including the possibilities of different effects of some diseases: see this business week article).
There are two sinister mechanisms for ensuring that your only child is a boy: selective abortion or infanticide. Yet there is another option: just have another baby if the first is a girl, and don't tell the government. I think this third option is more likely, because I do not think most families can afford an abortion (illegal for sex selection) and very few mothers would kill their babies.
So how much does this non-reporting need to happen to change the ratio from the normal 106 to 100 male to female births to the abnormal 120 to 100?
The answer to this is the combination of three things: 1) percent of births that are girls (with no intervention), 2) percent of families that have another baby (hoping for a boy), given the first is a girl, and 3) the percent of families that do not tell the government about the first baby.
Lets call these percentages, Pg, P2, and Ps (for girl, 2nd child, and secret). Lets also call Pr the reported percent of girls, which is 100/220, or 45.45%. We'll assume also for simplicity that families quit trying when they have a boy or have 2 children, whichever comes first. Also, we'll assume families always report the first child if it is a boy or if they have no more children.
Pg is known at around 100/206=48.54%
P2 and Ps are unknown.
We want to figure out what P2 and Ps could lead to the Pr being 45.45% when Pg is 48.54%.
First, consider that, given the ground rules above, the following are the types of families that can exist (in birth order):
B (boy, one child only)
G (girl, one child only)
GB (girl boy, two children)
GG (girl girl, two children)
To figure out the percent of girls reported, we need the total girls reported divided by the total children reported. This is easy to figure out for each combination above:
B = 0 girls / 1 child
G = 1 Girl / 1 child
GB = 0 girls / 1 child Ps percent of the time and 1 girls / 2 children (1- Ps percent of the time)
GG = 1 girl / 1 child Ps percent of the time and 2 girls / 2 children (1-Ps percent of the time)
We are almost there. Now we just need to sum the numerators multiplied by their probabilities and the denominators multiplied by their probabilities. Here are the probabilities of each family combination:
B = 1-Pg
G = Pg*(1 - P2) ==> It's just Pg times the percent of families who do not have more children
GB = Pg*(P2)*(1-Pg) It's the chances of a girl, followed by the decision to have a 2nd, followed by having a boy.
GG = Pg*(P2)*Pg=Pg^2*P2
Thus the numerator (number of girls reported average is):
Num = (1-Pg)*0 +
Pg*(1-P2)*1 +
Pg*P2*(1-Pg)*Ps*0 +
Pg*P2*(1-Pg)*(1-Ps)*1 +
Pg^2*P2*Ps*1 +
Pg^2*P2*(1-Ps)*2
and the denominator (number of children reported on average):
Den = (1-Pg)*1 +
Pg*(1-P2)*1 +
Pg*P2*(1-Pg)*Ps*1 +
Pg*P2*(1-Pg)*(1-Ps)*2 +
Pg^2*P2*Ps*1 +
Pg^2*P2*(1-Ps)*2
We know that, in China, Pr= Num/Den = 45.45% and that, in general, Pg=48.54%. Thus, we can solve .4545=Num/Den in terms of Ps and P2.
Since we have 1 equations and 2 unknowns, there are an infinite number of solutions, but here are a few possibilities:
0% have a second child --impossible
10% have a second child -- impossible
15% have a second child and 85% of those keep the first a secret from the government
20% have a second child and 65% of those keep the first a secret from the government
30% have a second child and 45% of those keep the first a secret from the government
40% have a second child and 35% of those keep the first a secret from the government
50% have a second child and 30% of those keep the first a secret from the government
One thing to note (that is not necessarily obvious in these calculations) is that if everyone reports all the children they have (Ps=0), then the percent of girls will be exactly 48.54%, the same as if everyone had one child, as long as infanticide and selective abortion are not occurring.
But the main point here is that a small number (15%) of couples having second children and not reporting the first girl leads to the warped percentages of baby girls, if there is high under-reporting of these first children. You do not need to assume that infanticide or selective abortion plays a role at all.
Tuesday, April 15, 2008
What's the chance of rain?
Everyday probability barely shows up in weather forecasts these days. For example, Yahoo's weather will say something like "a few showers" to mean that there is a chance of showers. However, if you are a purist, go to the National Weather Service (NWS) site, where they still make predictions using probabilities. See the Chicago forecast for this week in Yahoo and at the NWS site (the online NY times can barely be bothered to give any information at all, unless you can interpret their icons).
When it comes to data, I've always felt more is more, and so if I am really interested in the weather, I go to the NWS site where I'll get more than just "chance of rain" or "a few showers."
But what does it mean when the weather forecast says there is a 30% chance of rain Wednesday and a 50% chance of rain Thursday? If we focus on a single time period, say Thursday, the conclusion if pretty clear: there's a 50-50 chance of rain. Put another way, when encountering conditions like this in the past, the NWS model data shows rain half the time and no rain half the time.
The inference becomes more difficult when we want to ask a more complex question. For example, suppose I'm going to Chicago Wednesday and returning Thursday night. I want to know whether to bring an umbrella. Since I hate lugging an umbrella along, I only want to bring one if there is at least a 75% chance of rain at some point while I'm there.
It turns out that the answer to this question cannot be determined with the information given (don't you just love that choice on multiple choice tests?).
Before we explain why, though, we need some definitions and notation.
To do the math for this, we generally define each possible outcome as an event. In this case, we have the following events:
Event A: Rains Wednesday
Event B: Rains Thursday
We are interested in the chance that either Event A or Event B occurs. We have a shorthand for expressing the probability of Event A: "P(A)".
There is a simple probability formula that is very useful here:
P(A or B) = P(A) + P(B) - P(A and B)
This formula says that the probability of Event A or Event B happening is the probability of A plus the probability of B minus the probability that A and B both happened (the event that A and B occurred is called the intersection of Events A and B). This makes sense because if we just added them (as you might intuitively do) we are double counting the times both events occur, and thus we need to subtract out the intersection once at the end.
In some cases P(A and B)=0. In other words, Events A and B never occur together. You may have noticed this comes up when you toss a coin: it is never both heads and tails at the same time (except for that time I got the coin to stand on its side). Events like these are called mutually exclusive.
In other cases P(A and B)=P(A)*P(B). This means the probability of A and B is the product of the probabilities A and B. In this case, the two events can both occur, but they have nothing to do with each other. Events like these are called independent events.
In still other cases, P(A and B) is neither P(A) + P(B) or P(A)*P(B).
If we assume the events A and B are mutually exclusive, then there's an 80% chance (50+30) of rain either Wednesday or Thursday. This seems unlikely though, because most storms could last more than an evening.
If we assume the events A and B are independent, then there's an 65% chance of rain either Wednesday or Thursday. This is a little more complicated to calculate, because we need to figure out the chances of it raining both Wed. and thursday, which we assume is independent and thus is P(A)*P(B)=30%*50%=15%. Thus:
P(A or B)=P(A) + P(B) - P(A and B)=50% + 30% -15%=65%.
[we could also figure out the chances of not raining either night. Since the chance of rain is independent, the chance of no rain is also independent. Also, the chance of rain plus the chance of no rain must be one. Thus P(no rain Wednesday)=1-P(A)=100%-30%=70%. Similary, P(not B)=100%-50%=50%. Then, the chance of no rain either time period = P(not A and not B)=70%*50%=35%. Thus, there is a 35% chance it will not rain either night, and we can conclude there would be a 65% chance of rain one of the nights, of course all hinging on the independence assumption].
okay. So finally, we have two probabilities 80% and 65%, based on two different and rather extreme assumptions. On the high side, 80% is the most extreme. We can see this by seeing that in order to get a larger number, we'd have to plug in a negative probability for the value of P(A and B) in the general formula (which does not assume independence or anything else):
P(A or B) = P(A) + P(B) - P(A and B)
Since probabilities must always be at least 0 and at most 100%, we cannot have a negative number for P(A and B). So at most, the chance of rain Wednesday or Thursday is 80%.
But what about the least the chance might be? Independence seems a pretty extreme assumption in the other direction, but in fact it is not. What would lead to the smallest probability is if the two events A and B were highly related--in fact so related that P(B)=1 if A occurs. This would mean the P(A and B)=30% (the smaller of P(A) and P(B)). This would lead to a probability that it rains either Wednesday or Thursday of just 50%:
P(A or B) = P(A) + P(B) - P(A and B) = 30% + 50% - 30% = 50%
So now that we've got rain down, let me go back to the original impetus for this blog: it is easy to make the wrong inference when given information about the chances of a series of events.
The recent Freakonomics blog about chances of birth defects addresses this issue. In it, Steven Levitt describes a couple who was told that there was a 1 in 10 chance that a test, which showed an embryo was not vailable, was wrong. The test was done twice on each of two embryos, and all four times the outcome was that the embryos were not viable. Thus, the lab told them that the chances of having healthy twins from two such embryos was 1 in 10,000. Of course, after reading about rain above, you recognize this as the application of the independence assumption (1/10 times 1/10 times 1/10 times 1/10 equals 1 in 10,000). The couple didn't listen to the lab though, and, nine months later, 2 very vaiable and very healthy babies were born.
Post hoc, it seems the lab should have (at least) said the chances were somewhere between 1 in 10 and 1 in 10,000. In addition, the 1 in 10 seems like an awfully round number--could it have been rounded down from some larger probability (1 in 8, 1 in 5, 1 in 3, who knows?). Levitt wonders whether the whole test is just nonsense in the first place.
So what do you do when confronted with a critical medical problem and a slew of probabilities? There's no easy answer, of course, but I believe gathering as much hard data as possible is important. Then make sure you distinguish between the inferences made by your doctor, nurse, or lab technician (which are more subject to error) and the underlying probabilities associated with the drug, the test, or the procedure (which are less subject to error).
When it comes to data, I've always felt more is more, and so if I am really interested in the weather, I go to the NWS site where I'll get more than just "chance of rain" or "a few showers."
But what does it mean when the weather forecast says there is a 30% chance of rain Wednesday and a 50% chance of rain Thursday? If we focus on a single time period, say Thursday, the conclusion if pretty clear: there's a 50-50 chance of rain. Put another way, when encountering conditions like this in the past, the NWS model data shows rain half the time and no rain half the time.
The inference becomes more difficult when we want to ask a more complex question. For example, suppose I'm going to Chicago Wednesday and returning Thursday night. I want to know whether to bring an umbrella. Since I hate lugging an umbrella along, I only want to bring one if there is at least a 75% chance of rain at some point while I'm there.
It turns out that the answer to this question cannot be determined with the information given (don't you just love that choice on multiple choice tests?).
Before we explain why, though, we need some definitions and notation.
To do the math for this, we generally define each possible outcome as an event. In this case, we have the following events:
Event A: Rains Wednesday
Event B: Rains Thursday
We are interested in the chance that either Event A or Event B occurs. We have a shorthand for expressing the probability of Event A: "P(A)".
There is a simple probability formula that is very useful here:
P(A or B) = P(A) + P(B) - P(A and B)
This formula says that the probability of Event A or Event B happening is the probability of A plus the probability of B minus the probability that A and B both happened (the event that A and B occurred is called the intersection of Events A and B). This makes sense because if we just added them (as you might intuitively do) we are double counting the times both events occur, and thus we need to subtract out the intersection once at the end.
In some cases P(A and B)=0. In other words, Events A and B never occur together. You may have noticed this comes up when you toss a coin: it is never both heads and tails at the same time (except for that time I got the coin to stand on its side). Events like these are called mutually exclusive.
In other cases P(A and B)=P(A)*P(B). This means the probability of A and B is the product of the probabilities A and B. In this case, the two events can both occur, but they have nothing to do with each other. Events like these are called independent events.
In still other cases, P(A and B) is neither P(A) + P(B) or P(A)*P(B).
If we assume the events A and B are mutually exclusive, then there's an 80% chance (50+30) of rain either Wednesday or Thursday. This seems unlikely though, because most storms could last more than an evening.
If we assume the events A and B are independent, then there's an 65% chance of rain either Wednesday or Thursday. This is a little more complicated to calculate, because we need to figure out the chances of it raining both Wed. and thursday, which we assume is independent and thus is P(A)*P(B)=30%*50%=15%. Thus:
P(A or B)=P(A) + P(B) - P(A and B)=50% + 30% -15%=65%.
[we could also figure out the chances of not raining either night. Since the chance of rain is independent, the chance of no rain is also independent. Also, the chance of rain plus the chance of no rain must be one. Thus P(no rain Wednesday)=1-P(A)=100%-30%=70%. Similary, P(not B)=100%-50%=50%. Then, the chance of no rain either time period = P(not A and not B)=70%*50%=35%. Thus, there is a 35% chance it will not rain either night, and we can conclude there would be a 65% chance of rain one of the nights, of course all hinging on the independence assumption].
okay. So finally, we have two probabilities 80% and 65%, based on two different and rather extreme assumptions. On the high side, 80% is the most extreme. We can see this by seeing that in order to get a larger number, we'd have to plug in a negative probability for the value of P(A and B) in the general formula (which does not assume independence or anything else):
P(A or B) = P(A) + P(B) - P(A and B)
Since probabilities must always be at least 0 and at most 100%, we cannot have a negative number for P(A and B). So at most, the chance of rain Wednesday or Thursday is 80%.
But what about the least the chance might be? Independence seems a pretty extreme assumption in the other direction, but in fact it is not. What would lead to the smallest probability is if the two events A and B were highly related--in fact so related that P(B)=1 if A occurs. This would mean the P(A and B)=30% (the smaller of P(A) and P(B)). This would lead to a probability that it rains either Wednesday or Thursday of just 50%:
P(A or B) = P(A) + P(B) - P(A and B) = 30% + 50% - 30% = 50%
So now that we've got rain down, let me go back to the original impetus for this blog: it is easy to make the wrong inference when given information about the chances of a series of events.
The recent Freakonomics blog about chances of birth defects addresses this issue. In it, Steven Levitt describes a couple who was told that there was a 1 in 10 chance that a test, which showed an embryo was not vailable, was wrong. The test was done twice on each of two embryos, and all four times the outcome was that the embryos were not viable. Thus, the lab told them that the chances of having healthy twins from two such embryos was 1 in 10,000. Of course, after reading about rain above, you recognize this as the application of the independence assumption (1/10 times 1/10 times 1/10 times 1/10 equals 1 in 10,000). The couple didn't listen to the lab though, and, nine months later, 2 very vaiable and very healthy babies were born.
Post hoc, it seems the lab should have (at least) said the chances were somewhere between 1 in 10 and 1 in 10,000. In addition, the 1 in 10 seems like an awfully round number--could it have been rounded down from some larger probability (1 in 8, 1 in 5, 1 in 3, who knows?). Levitt wonders whether the whole test is just nonsense in the first place.
So what do you do when confronted with a critical medical problem and a slew of probabilities? There's no easy answer, of course, but I believe gathering as much hard data as possible is important. Then make sure you distinguish between the inferences made by your doctor, nurse, or lab technician (which are more subject to error) and the underlying probabilities associated with the drug, the test, or the procedure (which are less subject to error).
Wednesday, March 19, 2008
Stocks or Bonds?
A couple of posts ago, I talked about the question of whether to rent or buy (the answer: in the long run, buy; in the short run, rent). With all the turmoil in the financial market, it seems a reasonable time to visit the question of whether stocks or bonds are a better investment.
If you look at data from 1929-2007 for the Dow Jones Industrial Average (DJIA), a 20 year investment in stocks yielded an inflation-adjusted return of about 2.6% annually. This is before taking dividends into account, which add at least a couple percent (recently, the dividend yield has been closer to 2% while in the past 4-5% was the norm, see this article for relevant charts). The net return for stocks, after inflation, is around 6% annually.
For bonds, the returns also generally beat inflation, but are not as good. Their average is around 2% annually, after subtracting out inflation.
The following chart shows the DJIA inflation adjusted for 10 and 20 year investments (in blue and purple, downloaded from Yahoo Finance), versus Treasury bonds (in yellow). The US Treasury Bond return is based on 20-year bonds when available from the Fed and 10 year bonds or estimating using the article sited above when 20-year returns were not available.
The year shown is the final year of the investment. Thus, if you made a 20 year investment beginning in 1985, you can look at the points corresponding to 2005 to find out that you would have earned approximately 10% annually, whether it was in bonds (yellow) or stocks (purple for 20 year) by the start of 2005, and that is after inflation.
[Click on the graph see it in higher resolution]
If you look at data from 1929-2007 for the Dow Jones Industrial Average (DJIA), a 20 year investment in stocks yielded an inflation-adjusted return of about 2.6% annually. This is before taking dividends into account, which add at least a couple percent (recently, the dividend yield has been closer to 2% while in the past 4-5% was the norm, see this article for relevant charts). The net return for stocks, after inflation, is around 6% annually.
For bonds, the returns also generally beat inflation, but are not as good. Their average is around 2% annually, after subtracting out inflation.
The following chart shows the DJIA inflation adjusted for 10 and 20 year investments (in blue and purple, downloaded from Yahoo Finance), versus Treasury bonds (in yellow). The US Treasury Bond return is based on 20-year bonds when available from the Fed and 10 year bonds or estimating using the article sited above when 20-year returns were not available.
The year shown is the final year of the investment. Thus, if you made a 20 year investment beginning in 1985, you can look at the points corresponding to 2005 to find out that you would have earned approximately 10% annually, whether it was in bonds (yellow) or stocks (purple for 20 year) by the start of 2005, and that is after inflation.
[Click on the graph see it in higher resolution]
The yellow line, denoting Treasury bond returns, is mostly below the blue and purple lines, indicating that, for most years, a 10 or 20 year investment in the Dow Jones Index is better than a similar investment in Treasury bonds. In fact, for 57 of the 68 ten-year investments, stocks do better. For 20 year investments, the numbers are even more promising for stocks, which perform better for 56 out of 57 20-year investments.
Treasury Bonds, on the other hand, are considered risk-free. The idea is that there is no risk that the U.S. Treasury will default on its loan (or, at least, if it does default, there are far more serious problems to worry about). On the other hand, with stocks, there are no guarantees that they will not go down.
This basic idea, that stocks have broadly out-performed bonds and beaten inflation in the long run, seems to be well-understood. This fact, however, does not imply that individual stock investments will outperform bonds, or that a shorter term investment will outperform.
The "truth" about stocks that is implied by this graph, however, is a little deceiving, for three reasons.
1. The graph tells you little about what might happen to a 20-year investment beginning in a particular year
The graph shows that for most time periods, stocks do well, but there is an enormous amount of variation, even for the long-term investments considered here. If you happened to make some long-term investments in 1962, at the end of a recession and the beginning of a long boom, you'd still be out of luck if you needed that money in 1982, when its real value would have been far less than when you invested it (the pink point corresponding to 1982 shows a 1% annual loss for the 20 prior years, amounting to about an 18% total loss in real dollars).
[The "safe" Treasury bond portfolio, however, would have done far worse, losing about twice as much over the same period. This is not because the US defaulted on its bonds. Instead, it's because of inflation. The 20-year bond you bought in 1962, yielding 4%, did not keep up with inflation. This is the long-term, somewhat hidden risk for bonds.]
2. The graph averages over a portfolio of stocks
The other issue is that despite the graphs very clear implication that stocks are better, even in bad times, this does not necessarily imply that individual stocks are better. Returns on an individual stock, or even on a small portfolio of stocks, vary much more wildy than the Dow Jones average shown in the graph. Also, in recent years, the Treasury has issued inflation-indexed bonds, which guarantee a real return above 0, thus insuring the yellow line in a future graph will be more than 0 (see information on inflation-indexed bonds).
3. The future is not now
While it seems convincing that 80 years of history show stocks in a very positive light, we only need to look at Japan, whose Nikkei average, since 1985, has lost about 15% adjusting for inflation (it's far more than 50% for an investment made near Japan's stock market peak). There are some indications that the current U.S. problems (real estate boom and bust, credit problems) are worse than Japan's. Much of U.S. investment in the last few decades has been fueled by the safety of the dollar. The rise of the Euro and of globalization has already begun to change that, and a continuing fall in the dollar will almost certainly cause inflation, which was devastating to the stock market in the 1970's.
Bottom Line
There's a lot of evidence that, in the past, a long-term stock investment paid off, relative to both risk-free bonds and inflation. However, there is no guarantee this party will continue, especially if there is a sea-change in dollar investments.
Where's my retirement money? Almost all in stocks...but almost all foreign stocks.
Sunday, February 24, 2008
7 letter words and 8 card suits
A recent commenter asked whether the following is true: "ETAERIO is the most likely seven-letter word to get in scrabble."
As all scrabble players know, if you use all 7 of your letters, you get a bonus. However, except for the first turn, you'd need an 8-letter word to achieve this--ETAERIO would not do. Thus, I am going to try to answer the question: "What are the chances of getting the letters ETAERIO on your initial turn in scrabble (you also have to hope you are first)?" I have no hope of finding out whether this is the most likely seven letter word, because I can't automatically check all letter combinations, but I will try to give some guidance there as well.
To find the chances of gettting ETAERIO, we need the number of combinations that produce these letters divided by the total number of combinations. In other words, we have to go back to 12-th grade math, where we all learned (or sort-of learned) permutations and combinations.
There are 100 tiles in scrabble and we are choosing 7. Thus, there are 100 ways to to choose the first tile, 99 ways to choose the second, and so forth down to 94. If we chose them in order, we'd have 100*99*98*97*96*95*94 permutations. However, we don't care about the order, so we have to take the above product and divide by the number of ways we can permute the 7 tiles, which is 7*6*5*4*3*2*1. The shorthand way to express this number of combinations is "100 choose 7" or
By dividing the two ratios above ((100*99*98*97*96*95*94) / (7*6*5*4*3*2*1)), we come up with 16,007,560,800. Since most letters appear multiple times, the number of possible letter combinations is far less, and to know the chances of getting ETAERIO, I need to know how many times each letter appears.
Thus, I found the letter distributions on Wikipedia (counting our own scrabble pieces would probably not do with a three-year old distributing them around the house). The most common ones as follows:
E - 12 tiles
A, I - 9 tiles
O - 8 tiles
N, R, T - 6 tiles
D, L, S, and U - 4 tiles
other letters - 3 or less, but not relevant here
To figure out the chances of getting ETAERIO, we need to know the number of combinations that produce it. We need 2 E's, 1 T, 1 A, 1 R, 1 I, and 1 O. It turns out that the number of ways is the product of each of these implied combinations. Thus, it is "12 choose 2" (E's) times "6 choose 1" (T) times "9 choose 1" (A) and so forth. This comes out to 1,539,648 ways to get the letters in ETAERIO. If we divide this by the total number of combinations (16,007,560,800), we find that there is about a 1 in 10,000 chance of getting ETAERIO as your first 7 letters. Of course, from there, you have to know it is a word and figure out that you can make that word from those letters, since they are not likely to appear in that order.
I could not find another word with a higher probability, but I did find TREASON and TRAINED (both about 1 in 20,000). However, It's clear from the distribution of letter tiles that in order to find a word that beats ETAERIO, you can only use letters appearing in 6 or more tiles.
BRIDGE HANDS
Now that we all remember the mechanics of combinations (or at least, we are on the subject of them), let's investigate another oft-asked question around here: what's the chance of being dealt a 7 card suit in bridge? This would be 4 (number of suits) times "13 choose 7" (ways to choose 7 from a suit) times "39 choose 6" (ways to choose the other 6 from the other 3 suits) divided by "52 choose 13" (ways to choose 13 cards from 52). This comes out to about 3.5%, or 3 or 4 times in every 100 hands.
For an 8-card suit, it is 1 in about 200. For a 9-card suit, it is about 1 in 2,700. Of course, my kids are always asking about the chances of being dealt a 10 card suit or even a 13-card suit:
10-card suit: 1 in 60,738
11-card suit: 1 in 2,746,693 (less than 1 in million)
12-card suit: 1 in 313,123,057 (less than 1 in 300 million)
13-card suit: 1 in 158,753,389,900 (less than 1 in 150 billion)
The chances aren't too great, but with some really poor shuffling, they've managed the 13-card suit once or twice.
As all scrabble players know, if you use all 7 of your letters, you get a bonus. However, except for the first turn, you'd need an 8-letter word to achieve this--ETAERIO would not do. Thus, I am going to try to answer the question: "What are the chances of getting the letters ETAERIO on your initial turn in scrabble (you also have to hope you are first)?" I have no hope of finding out whether this is the most likely seven letter word, because I can't automatically check all letter combinations, but I will try to give some guidance there as well.
To find the chances of gettting ETAERIO, we need the number of combinations that produce these letters divided by the total number of combinations. In other words, we have to go back to 12-th grade math, where we all learned (or sort-of learned) permutations and combinations.
There are 100 tiles in scrabble and we are choosing 7. Thus, there are 100 ways to to choose the first tile, 99 ways to choose the second, and so forth down to 94. If we chose them in order, we'd have 100*99*98*97*96*95*94 permutations. However, we don't care about the order, so we have to take the above product and divide by the number of ways we can permute the 7 tiles, which is 7*6*5*4*3*2*1. The shorthand way to express this number of combinations is "100 choose 7" or
By dividing the two ratios above ((100*99*98*97*96*95*94) / (7*6*5*4*3*2*1)), we come up with 16,007,560,800. Since most letters appear multiple times, the number of possible letter combinations is far less, and to know the chances of getting ETAERIO, I need to know how many times each letter appears.
Thus, I found the letter distributions on Wikipedia (counting our own scrabble pieces would probably not do with a three-year old distributing them around the house). The most common ones as follows:
E - 12 tiles
A, I - 9 tiles
O - 8 tiles
N, R, T - 6 tiles
D, L, S, and U - 4 tiles
other letters - 3 or less, but not relevant here
To figure out the chances of getting ETAERIO, we need to know the number of combinations that produce it. We need 2 E's, 1 T, 1 A, 1 R, 1 I, and 1 O. It turns out that the number of ways is the product of each of these implied combinations. Thus, it is "12 choose 2" (E's) times "6 choose 1" (T) times "9 choose 1" (A) and so forth. This comes out to 1,539,648 ways to get the letters in ETAERIO. If we divide this by the total number of combinations (16,007,560,800), we find that there is about a 1 in 10,000 chance of getting ETAERIO as your first 7 letters. Of course, from there, you have to know it is a word and figure out that you can make that word from those letters, since they are not likely to appear in that order.
I could not find another word with a higher probability, but I did find TREASON and TRAINED (both about 1 in 20,000). However, It's clear from the distribution of letter tiles that in order to find a word that beats ETAERIO, you can only use letters appearing in 6 or more tiles.
BRIDGE HANDS
Now that we all remember the mechanics of combinations (or at least, we are on the subject of them), let's investigate another oft-asked question around here: what's the chance of being dealt a 7 card suit in bridge? This would be 4 (number of suits) times "13 choose 7" (ways to choose 7 from a suit) times "39 choose 6" (ways to choose the other 6 from the other 3 suits) divided by "52 choose 13" (ways to choose 13 cards from 52). This comes out to about 3.5%, or 3 or 4 times in every 100 hands.
For an 8-card suit, it is 1 in about 200. For a 9-card suit, it is about 1 in 2,700. Of course, my kids are always asking about the chances of being dealt a 10 card suit or even a 13-card suit:
10-card suit: 1 in 60,738
11-card suit: 1 in 2,746,693 (less than 1 in million)
12-card suit: 1 in 313,123,057 (less than 1 in 300 million)
13-card suit: 1 in 158,753,389,900 (less than 1 in 150 billion)
The chances aren't too great, but with some really poor shuffling, they've managed the 13-card suit once or twice.
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