A Probability Problem in Heredity – Part 2

In my last post I solved a problem from chapter 2 of M.G. Bulmer’s Principles of Statistics. In this post I work through a problem in chapter 11 that is basically a continuation of the chapter 2 problem. If you take a look at the previous post, you’ll notice we were asked to find probability in terms of theta. I did it and that’s nice and all, but we can go further. We actually have data, so we can estimate theta. And that’s what the problem in chapter 11 asks us to do. If you’re wondering why it took 9 chapters to get from finding theta to estimating theta, that’s because the first problem involved basic probability and this one requires maximum likelihood. It’s a bit of a jump where statistical background is concerned.

The results of the last post were as follows:

purple-flowered red-flowered
long pollen \( \frac{1}{4}(\theta + 2)\) \( \frac{1}{4}(1 – \theta) \)
round pollen \( \frac{1}{4}(1 – \theta) \) \( \frac{1}{4}\theta \)

The table provides probabilities of four possible phenotypes when hybrid sweet peas are allowed to self-fertilize. For example, the probability of a self-fertilizing sweet pea producing a purple-flower with long pollen is \( \frac{1}{4}(1 – \theta)\). In this post we’ll estimate theta from our data. Recall that \( \theta = (1 – \pi)^{2} \), where \( \pi \) is the probability of the dominant and recessive genes of a characteristic switching chromosomes.

Here’s the data:

Purple-flowered Red-flowered
Long pollen 1528 117
Round pollen 106 381

We see from the table there are 4 exclusive possibilities when the sweet pea self-fertilizes. If we think of each possibility having its own probability of occurrence, then we can think of this data as a sample from a multinomial distribution. Since chapter 11 covers maximum likelihood estimation, the problem therefore asks us to use the multinomial likelihood function to estimate theta.

Now the maximum likelihood estimator for each probability is \( \hat{p_{i}} = \frac{x_{i}}{n} \). But we can’t use that. That’s estimating four parameters. We need to estimate one parameter, theta. So we need to go back to the multinomial maximum likelihood function and define \( p_{i}\) in terms of theta. And of course we’ll work with the log likelihood since it’s easier to work with sums than products.

\( \log L(\theta) = y_{1} \log p_{1} + y_{2} \log p_{2} + y_{3} \log p_{3} + y_{4} \log p_{4} \)

If you’re not sure how I got that, google “log likelihood multinomial distribution” for more PDF lecture notes than you can ever hope to read.

Now let’s define the probabilities in terms of one parameter, theta:

\( \log L(\theta) = y_{1} \log f_{1}(\theta) + y_{2} \log f_{2}(\theta) + y_{3} \log f_{3}(\theta) + y_{4} \log f_{4}(\theta) \)

Now take the derivative. Once we have that we can set equal to 0 and find a solution for theta. The solution will be the point at which theta obtains its maximum value:

\( \frac{d \log L(\theta)}{d\theta} = \frac{y_{1}}{f_{1}(\theta)} f’_{1}(\theta) + \frac{y_{1}}{f_{2}(\theta)} f’_{2}(\theta) + \frac{y_{1}}{f_{3}(\theta)} f’_{3}(\theta) + \frac{y_{1}}{f_{4}(\theta)} f’_{4}(\theta)\)

Time to go from the abstract to the applied with our values. The y’s are the data from our table and the functions of theta are the results from the previous problem.

\( \frac{d \log L(\theta)}{d\theta} = \frac{1528}{1/4(2 + \theta)} \frac{1}{4} – \frac{117}{1/4(1 – \theta)}\frac{1}{4} – \frac{106}{1/4(1 – \theta)}\frac{1}{4} + \frac{381}{1/4(\theta)} \frac{1}{4} \)
\( \frac{d \log L(\theta)}{d\theta} = \frac{1528}{2 + \theta} – \frac{117}{1 – \theta} – \frac{106}{1 – \theta} + \frac{381}{\theta} \)
\( \frac{d \log L(\theta)}{d\theta} = \frac{1528}{2 + \theta} – \frac{223}{1 – \theta} + \frac{381}{\theta} \)

Set equal to 0 and solve for theta. Beware, it gets messy.

\( \frac{[1528(1 – \theta)\theta] – [223(2 + \theta)\theta] + [381(2 + \theta)(1 – \theta)]}{(2 + \theta)(1- \theta)\theta} = 0\)

Yeesh. Fortunately the denominator cancels out when we start multiplying terms and solving for theta. So we’re left with this:

\( 1528\theta – 1528\theta^{2} – 446\theta – 223\theta^{2} + 762 – 381\theta – 381\theta^{2} = 0\)

And that reduces to the following quadratic equation:

\( -2132\theta^{2} + 701\theta + 762 = 0\)

I propose we use an online calculator to solve this equation. Here’s a good one. Just plug in the coefficients and hit solve to find the roots. Our coefficients are a = -2132, b = 701, and c = 762. Since it’s a quadratic equation we get two answers:

\( x_{1} = -0.4556 \)
\( x_{2} = 0.7844 \)

The answer is in terms of a probability which is between 0 and 1, so we toss the negative answer and behold our maximum likelihood estimator for theta: 0.7844.

Remember that \( \theta = (1 – \pi)^{2}\). If we solve for pi, we get \( \pi = 1 – \theta^{1/2}\), which works out to be 0.1143. That is, we estimate the probability of characteristic genes switching chromosomes to be about 11%. Therefore we can think of theta as the probability of having two parents that experienced no gene switching.

Now point estimates are just estimates. We would like to know how good the estimate is. That’s where confidence intervals come in to play. Let’s calculate one for theta.

It turns out that we can estimate the variability of our maximum likelihood estimator as follows:

\( V(\theta) = \frac{1}{I(\theta)}\), where \( I(\theta) = -E(\frac{d^{2}\log L}{d \theta^{2}}) \)

We need to determine the second derivative of our log likelihood equation, take the expected value by plugging in our maximum likelihood estimator, multiply that by -1, and then take the reciprocal. The second derivative works out to be:

\( \frac{d^{2}\log L}{d \theta^{2}} = -\frac{1528}{(2 + \theta)^{2}} -\frac{223}{(1 – \theta)^{2}} -\frac{381}{\theta^{2}} \)

The negative expected value of the second derivative is obtained by plugging in our estimate of 0.7844 and multiplying by -1. Let’s head to the R console to calculate this:

th <- 0.7844 # our ML estimate
(I <- -1 * (-1528/((2+th)^2) - 223/((1-th)^2) - 381/(th^2))) # information
[1] 5613.731

Now take the reciprocal and we have our variance:

(v.th <- 1/I)
[1] 0.0001781347

We can take the square root of the variance to get the standard error and multiply by 1.96 to get the margin of error for our estimate. Then add/subtract the margin of error to our estimate for a confidence interval:

# confidence limits on theta
0.784+(1.96*sqrt(v.th)) # upper bound
[1] 0.8101596
0.784-(1.96*sqrt(v.th)) # lower bound
[1] 0.7578404

Finally we convert the confidence interval for theta to a confidence interval for pi:

# probability of switching over
th.ub <- 0.784+(1.96*sqrt(v.th))
th.lb <- 0.784-(1.96*sqrt(v.th))
1-sqrt(th.ub) # lower bound
[1] 0.09991136
1-sqrt(th.lb) # upper bound
[1] 0.1294597

The probability of genes switching chromosomes is most probably in the range of 10% to 13%.

One thought on “A Probability Problem in Heredity – Part 2

  1. Pingback: A Probability Problem in Heredity – Part 3 | statistics you can probably trust

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