Distributome Colorblindness Activity

Overview

Colorblindness – Can you see the number in this image? 

This Distributome Activity illustrates an application of probability theory to study Colorblindness, typically a genetic disorder which results from an abnormality on the X chromosome. The condition is thus rarer in women since a woman would need to have the abnormality on both of her X chromosomes in order to be colorblind (whether a woman has the abnormality on one X chromosome is essentially independent of having it on the other).

Goals

The goal of this activity is to demonstrate an efficient protocol of estimating the probability that a randomly chosen individual may be colorblind.

Hands-on Activity

Suppose that \(p\) is the probability that a randomly selected ”man” is colorblind.

• 100 men are selected at random. What is the distribution of \(X_m\) = the number of these men that are colorblind?

Answer
\(X_m\) ~ \(Binomial(100,p)\)

• 100 women are selected at random. What is the distribution of \(X_f\) = the number of these women that are colorblind?

See Hint
The chance that an individual woman is colorblind is \(p^2\), why?

Answer
\(X_f\)~\(Binomial(100,p^2)\)

• To estimate the probability that a randomly selected woman is colorblind, you might use the proportion of colorblind women in a sample of n women. What is the variance of this estimator?

Answer
\(X_f\)~\(Binomial(n,p^2)\).
Thus \(Var(\frac{X_f}{n}) \equiv \frac{p^2(1-p^2)}{n}\).

• Alternatively, to estimate the probability that a randomly selected woman is colorblind, you might use the square of the proportion of colorblind men in a sample of n men. Explain why this estimate makes sense. What is the variance of this estimator?

  See Hint
  The moment generating function can be used to find the fourth moment about the origin.

  Answer
  We want to estimate \(p^2\) and \(\frac{X_m}{n}\) estimates \(p\) so it makes sense to use \((\frac{X_m}{n})^2\) as the estimator (in fact it will be the maximum likelihood estimate. We have \(Var[( \frac{X_m}{n} )^2 ] \equiv n^{-4}[E(X_m^4 ) – (E(X_m^2 ))^2 ]\). Take \(q \equiv 1-p\). Then the fourth moment about the origin of a binomial is \(E(X^4) \equiv np(q-6pq^2+7npq-11np^2q+6n^2p^2q+n^3p^3)\) and the second moment is \(E(X^2) \equiv np(q+np)\). Thus \(Var[( \frac{X_m}{n} )^2 ] \equiv n^{-3}(pq + 6(n-1)p^2q^2 + 4n(n-1)p^3q)\).
  

• For large samples, is it better to use a sample of men or a sample of women to estimate the probability that a randomly selected women is colorblind? Explain.

See Hint
Show that a normal approximation is valid for both and then compare the variances.

Answer
For large n the ratio of the variances for the estimate in part c to the estimate in part d is \(\frac{Var(\frac{X_f}{n})}{Var((\frac{X_m}{n})^2 )} \sim \frac{p^2(1-p^2)}{4p^3q} \equiv \frac{1+ p}{4p}\). When this ratio is greater than 1, the estimator based on the sample of men will be better. Since this happens for any \(p < \frac{1}{3}\), which is clearly the case for colorblindness, it is better to use a sample of men to estimate the probability that a random woman is colorblind.

Alternate approach

You can also use the delta method to find the approximate variance for the estimator above.

Conclusions
In practice, it may difficult to obtain reliable parameter estimates when the event at hand is very rare (such as with colorblindness in women). The use of a valid probability model such as the relationship between the chance of colorblindness in men and the chance in women may improve these estimates.

Distributome Homicide Trends Activity

Overview

A Columbus Dispatch newspaper story on Friday January 1, 2010 discussed a drop in the number of homicides in the city the previous year. Here are the first few paragraphs from the article:

• Homicides take big drop in city: Trend also being seen nationally, but why is a mystery.
• The number of homicides in Columbus dropped 25 percent last year after spiking in 2008. As of last night, the city was expected to close out 2009 with 83 homicides, 27 fewer than in 2008, according to records kept by police and The Dispatch. In 2007, 79 people were slain in Columbus. “I don’t know that there’s one reason for homicides going up or down,” said Lt. David Watkins, supervisor of the Police Division’s homicide unit.
• Why one year do we have 130, and then the next year we have 80?
• “You just can’t explain it,” Sgt. Dana Norman said. He supervises the third-shift squad that investigated 44 of last year’s homicides, which occurred at a rate of 11.1 for every 100,000 people in Columbus, based on recent population estimates …

A table appearing with the article showed that there were 568 homicides in the previous 6 years.

Hands-on Activity

Sargent Norman’s statement that “”You just can’t explain it”” presents an intriguing probability question – Is it possible that natural random fluctuation might be a good explanation? Let’s consider probability models for the number of observed crimes and how they might fluctuate to see if the data mentioned in the article is unusual.

• If homicides are rare events that might be independently perpetrated by individuals in a large population – what distribution would approximately describe the number of murders in a year?

Answer
A reasonable model would be the Poisson distribution (since the mean is quite large, a normal model with equal mean and variance would be an alternative approximation).

• Suppose the expected annual number of homicides in the city is denoted by \(\lambda\) and that the number of homicides is independent from year to year. The article notes that 2008 saw a “spike” in the number of homicides and in fact that was the highest number in the last six years. If nothing is going on except random fluctuations – we want to know if observing 27 fewer homicides in 2009 after the peak year is unusual (peak here meaning the highest in the last 6 years).

Use the Distributome Poisson simulator for the model you specified above to examine the distribution of the change in the number of homicides you would see following a peak of a six year stretch. Does the 27 murder drop seem unusual? Explain.

See a Hint
To get started, you will need
i) to find an estimate of \(\lambda\) to use in your simulations, and
ii) to examine groups of 7 years of simulated homicide data and isolate those cases that satisfy the conditions of the problem.

See the First part of the Answer
There were 568 homicides in the preceding six years so a reasonable estimate of \(\lambda\) would be \(\lambda \equiv \frac{568}{6} \equiv \frac{284}{3} \equiv 82.67\). From a simulation of 100,000 sets of six independent Poisson variables, we find the maximum would have a distribution with a histogram that looks like this image.

See the Second part of the Answer
The difference between this maximum and another independent \(Poisson(\lambda \equiv 82.67)\) variable would have a histogram that looks like the following:

The shaded region corresponds to values of at least 27, which happens about 12% of the time so the drop of homicides in Columbus would not be particularly unusual when nothing is happening but regular random fluctuations.

Alternative Approach

This problem might also be viewed as an example of the regression effect where you should expect a regression to the mean following a very high observed value.

Conclusions
When viewing a random process over time it is the extremes that make the headlines – so the probability models we should use to answer the question “What is unusual?” should be probability models about extremes.

Distributome Activity on Sample Sizes and the Accuracy of Polls

Introduction

Surveys about public opinions on controversial social issues are becoming increasingly frequent as topics such as the legalization of marijuana, abortion policy, marriage rights for homosexuals, and immigration policy are hotly debated in the media.

For example, both the Opinion Research Corporation (polling for CNN) and the Pew Research Center for The People and The Press conducted surveys of American adults in spring of 2011 to estimate the percentage of the public that favors the legalization of marijuana. The sample sizes in the two polls were 824 for the Opinion Research Corporation poll and 1504 for the Pew poll.

Goals

This activity illustrates the inter-distribution relationships between Cauchy, Student’s T and Standard Normal (Gaussian) distributions.  These relationships are used to provide a better understanding of how strongly sample sizes are related to the accuracy of polls.

Hands-on Activity

In this activity you may assume that both of these pollsters use similar techniques that involve telephone interviews and weighting the answers given by individuals to align the respondents demographics with population values and finally averaging to produce unbiased and essentially normally distributed estimates. Below are 4 related, but complementary, problems regarding this study.

Note: The problems below may be appropriate for an undergraduate course in probability. The last part (Problem 4) would be more appropriate for masters level course (and should have a General Cauchy distribution tag and a tag for its relationship to the bivariate normal).

Specific Problems and their Solutions

Problem 1: Difference in Poll Accuracies?

The Pew poll had almost twice the sample size of the Opinion Research Corporation poll. What is the chance that it was more accurate than that poll for estimating p = the percentage of American adults that favored the legalization of marijuana in spring, 2011? Be sure to clearly define how you are interpreting “more accurate.” Also state and justify any assumptions you make in solving for this probability.

See a Hint
Decide on the measure of the accuracy of a poll and then evaluate \(P[\frac{Accuracy_{poll1} }{Accuracy_{poll2} } < 1]\) to gauge the chance that one poll s more accurate than the other.

See a Solution: Step 1
We want the probability that the Pew estimate based on a sample size of \(n_1=1504\), which comes closer to the true value of \(p\) than the Opinion Research poll based on \(n_2=824\) respondents. If \(\hat{p_1}\) = the estimate of \(p\) from Pew, and \(\hat{p_2}\) = the estimate of \(p\) from Opinion Research. With this notation, the problem asks for \(P(\vert\hat{p_1}-p\vert < \vert\hat{p_2}-p\vert ) \)

See a Solution: Step 2
Taking \(X \equiv \hat{p_1}-p\) and \(Y\equiv \hat{p_2}-p\), the problem statement indicates that \(X\sim N(0, \sigma_1^2)\) and \(Y\sim N(0, \sigma_2^2)\).

See a Solution: Step 3
Since the two polls use a similar methodology we may assume that the technique produces estimates with variance \(\frac{\sigma^2}{n}\) when a sample size of \(n\) is used (for \(n\) large enough for the normal approximation to be valid as in these cases).

See a Solution: Step 4
Thus, \(\frac{\sigma_1^2}{\sigma_1^2}\equiv \frac{824}{1504}\equiv 0.548\). Finally, we should assume that the estimates from the two polls are stochastically independent, which is reasonable here since they were conducted by two separate companies and the population of U.S. adults is much larger than either sample size.

See a Solution: Step 5
Now \(P(\vert X\vert < \vert Y\vert)\equiv P(-1 < \frac{X}{Y} < 1)\equiv F(1)-F(-1)\), where F is the CDF of the [http://socr.ucla.edu/htmls/dist/GeneralCauchy_Distribution.html Cauchy distribution] with scale parameter 0.548 so \(F(u)\equiv 0.5+\frac{1}{\pi}\arctan(\frac{u}{\sqrt{0.548} })\), and the answer is \(F(1)-F(-1)\equiv \frac{1}{\pi}[\arctan(\frac{1}{\sqrt{0.548} }) – \arctan(\frac{-1}{\sqrt{0.548} })]\approx 0.594.\) Many people are surprised by how low this answer is – despite having nearly double the sample size, the chance that the Pew poll is more accurate is only 0.594.

Alternative approaches

Alternative 1: Ratio of bivariate Normal variables.

See a Solution to First Problem: Alternative 1
You may solve \(P(-1 < \frac{X}{Y} < 1)\equiv P(-\frac{\sigma_2}{\sigma_1} < \frac{X/\sigma_1}{Y/\sigma_2} < \frac{\sigma_2}{\sigma_1})\)

\(\equiv P(-\sqrt{0.548}< \frac{Z_1}{Z_2} < \sqrt{0.548})\), where \(Z_1\) and \(Z_2\) are standard normal variables, and their ratio is the standard Cauchy that can then be used to get the correct numerical answer.

Alternative 2: Direct calculation of the marginal distribution

See a Solution to First Problem: Alternative 2
A second, more cumbersome, alternative is to work directly to find the marginal distribution of \(\frac{X}{Y}\) from the bivariate normal – see the solution to the last part below with \(\rho\equiv 0\).

Problem 2: Pooling Data across Polls?

Describe how you would combine the data from these two polls to form a single estimate of \(p\).

The obvious choice is to propose a linear combination of the two estimates weighting inversely proportional to the variances to get the smallest overall variance amongst such linear combinations.

See a Solution to Problem 2
\(\hat{p}\equiv \frac{1504}{2328}\hat{p_1}+\frac{824}{2328}\hat{p_2}\). Of course, this is just the estimate that comes from combining the samples into one large sample of \(n\equiv 2328\) and has variance \(\frac{\sigma^2}{2328}\).

Problem 3: Are these probability estimates correlated?

What is the correlation between your estimate above and the individual estimate produced by the Pew poll?

Note that \(Cov(\hat{p},\hat{p_1})=Cov(\frac{1504}{2328}\hat{p_1}+\frac{824}{2328}\hat{p_2}, \hat{p_1})=\frac{1504}{2328}\sigma_1^2=\frac{\sigma^2}{2328}\).

See a Solution to Problem 3
Thus, \( Corr(\hat{p},\hat{p_1})\equiv \frac{\sigma^2/2328} { \sqrt{\sigma^2(\frac{1504}{2328^2}+\frac{824}{2328^2})} \sqrt{\frac{\sigma^2}{1504} } } \equiv \sqrt{\frac{1504}{2328} } \equiv 0.8038.\)

Problem 4: Accuracy of probability estimates?

What is the probability that your combined estimate (from the second problem) is more accurate than the estimate based only on the Pew poll?

We want \(P[|\hat{p}-p| < |\hat{p_1}-p| ]\).

See a Solution to Problem 4
Taking \(X\equiv \hat{p}-p\) and \(Y\equiv \hat{p_1}-p\), then \(X\sim N(0, \frac{\sigma^2}{2328})\), \(Y\sim N(0, \frac{\sigma^2}{1504})\) and \(Corr(X,Y)\equiv 0.8038\). <br/> Hence, \(P(\vert X\vert < \vert Y\vert) \equiv P(-1< \frac{X}{Y} < 1) \equiv G(1) – G(-1)\), where G is the CDF of the generalized Cauchy distribution (see below):
\(G(u) \equiv 0.5 +\frac{1}{\pi}\arctan(\frac{u-\rho\frac{\sigma_X}{\sigma_Y} }{\frac{\sigma_X}{\sigma_Y}\sqrt{1-\rho^2} })\). <br/> So, the answer is \(\frac{1}{\pi}(\arctan(\sqrt{\frac{824}{1504} }) -\arctan(\frac{-3832}{\sqrt{1504 \times 824} }))\approx 0.613\).

Generalized Cauchy distribution CDF derivation

To derive the generalized Cauchy distribution CDF directly, we start with the bivariate normal distribution of ”X” and ”Y”:

See the Derivation of the Cauchy CDF
We start with \(f(x,y) \equiv \frac{1}{\pi\sigma_x\sigma_y\sqrt{1-\rho^2} } exp(-\frac{(x/\sigma_x)^2-2\rho xy/(\sigma_x\sigma_y)+(y/\sigma_y)^2}{2(1-\rho^2)})\).
Then the pdf of \(U\equiv \frac{X}{Y}\) is
\(g(u) \equiv \int_{-\infty}^{\infty}{\vert y\vert f(uy,y)dy}\equiv \) \(\frac{1}{\pi\sigma_x\sigma_y\sqrt{1-\rho^2} } \int_0^{\infty} {yexp(-y^2\frac{(u/\sigma_x)^2-2\rho uy/(\sigma_x\sigma_y)+(1/\sigma_y)^2}{2(1-\rho^2)})dy}\).
Since \(\int_0^{\infty}{y \exp(-ay^2)dy} \equiv \frac{1}{2a}\), we find that
\(g(u)\equiv \frac{1}{\pi\sigma_x\sigma_y\sqrt{1-\rho^2} } (\frac{1-\rho^2}{(u/ \sigma_x)^2 – 2\rho u / (\sigma_x\sigma_y) +(1/\sigma_y)^2})\equiv \) \(\equiv \frac{1}{\pi}\frac{\frac{\sigma_x}{\sigma_y}\sqrt{1-\rho^2} }{u^2-2\rho u \frac{\sigma_x}{\sigma_y} (\frac{\sigma_x}{\sigma_y})^2}.\)
Thus, \(g(u)\equiv \frac{1}{\pi}\frac{\frac{\sigma_x}{\sigma_y}\sqrt{1-\rho^2} }{(u-\rho \sigma_x / \sigma_y)^2 +(\sigma_x / \sigma_y)^2(1-\rho^2)}.\)
Finally, we know that \(\int_0^y {\frac{b}{(u-m)^2+b^2}du} \equiv \arctan(\frac{y-m}{b})\).
Therefore, the generalized Cauchy distribution CDF is
\(G(u) \equiv 0.5 +\frac{1}{\pi}\arctan(\frac{u-\rho\frac{\sigma_X}{\sigma_Y} }{\frac{\sigma_X}{\sigma_Y}\sqrt{1-\rho^2} })\).

Cauchy, Student’s T and Gaussian distribution interrelations

The Student’s T-distribution represents a one-parameter homotopy path connecting Cauchy and Gaussian Distribution:

• \(Cauchy=T_{(df=1)} \longrightarrow T_{(df)}\longrightarrow N(0,1)=T_{(df=\infty)}\).
• See the Distributome Navigator, Cauchy, Student’s T and Gaussian distribution calculators.
• Explore the relations between these and many other probability distributions using the interactive graphical Distributome Navigator.

Conclusions

Increasing the sample size may help significantly in certain situations – but not as much as intuition often suggests.