Problem 9 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

An examination of 1000 people showed that 41 were carriers (heterozygotic) of the gene for cystic fibrosis. In a second, independent examination of 2000 people, 79 were found to be carriers of cystic fibrosis. Let \(p\) be the proportion of all people who are carriers of cystic fibrosis. For any number \(p\) in \([0,1],\) let \(L(p)\) be the likelihood of finding that 41 of 1000 people in one study and 79 out of 2000 people in a second independent study are carriers of cystic fibrosis given that the probability of being a carrier is \(p\). Then $$ L(p)=\left(\begin{array}{c} 1000 \\ 41 \end{array}\right) p^{41} \times(1-p)^{959} \times\left(\begin{array}{c} 2000 \\ 79 \end{array}\right) p^{79} \times(1-p)^{1921} $$ where \(\left(\begin{array}{c}1000 \\ 41\end{array}\right) \doteq 1.3 \times 10^{73}\) and \(\left(\begin{array}{c}2000 \\ 79\end{array}\right) \doteq 1.4 \times 10^{143}\) are constants. a. Simplify \(L(p)\). b. Compute \(L^{\prime}(p)\). c. Find the value \(\hat{p}\) of \(p\) for which \(L^{\prime}(p)=0\). The value \(L(\hat{p})\) is the maximum value of \(L(p)\) and \(\hat{p}\) is called the maximum likelihood estimator of \(p\).

Step-by-Step Solution

Answer

The maximum likelihood estimator of \(p\) is \(\hat{p} = \frac{1}{25}\).

Step 1: Simplify the Given Likelihood Function

The likelihood function is given as:\[L(p) = \left(\begin{array}{c} 1000 \ 41 \end{array}\right) p^{41} \times(1-p)^{959} \times\left(\begin{array}{c} 2000 \ 79 \end{array}\right) p^{79} \times(1-p)^{1921}\]Substituting the given constants:\[L(p) = (1.3 \times 10^{73}) p^{41} (1-p)^{959} \times (1.4 \times 10^{143}) p^{79} (1-p)^{1921}\]Combine the powers of \(p\) and \(1-p\):\[L(p) = (1.3 \times 10^{73} \times 1.4 \times 10^{143}) p^{120} (1-p)^{2880}\]

Step 2: Differentiate the Likelihood Function

To find the maximum likelihood estimator, compute the derivative of the log-likelihood function. Start by differentiating the natural log of the likelihood function:\[ln(L(p)) = ln(1.3 \times 10^{73}) + ln(1.4 \times 10^{143}) + 120 ln(p) + 2880 ln(1-p)\]Differentiate each term with respect to \(p\):\[\frac{d}{dp} ln(L(p)) = \frac{120}{p} - \frac{2880}{1-p}\]

Step 3: Set the Derivative Equal to Zero and Solve

Setting \(L'(p) = 0\) gives:\[\frac{120}{p} - \frac{2880}{1-p} = 0\]Simplify and solve for \(p\):\[120(1-p) = 2880p\]\[120 - 120p = 2880p\]\[120 = 3000p\]\[p = \frac{120}{3000} = \frac{1}{25}\]

Step 4: Verify and Interpret the Result

The value of \(p\) that maximizes \(L(p)\) is \(\hat{p} = \frac{1}{25}\). This is the maximum likelihood estimator of the proportion of carriers.

Key Concepts

Likelihood Function

When we talk about the likelihood function, we're referring to a way of measuring how well a particular set of parameters explains the observed data. In our exercise, the likelihood function is denoted by \(L(p)\). It represents the likelihood of observing 41 carriers of the cystic fibrosis gene in a sample of 1000 individuals and 79 carriers in a second sample of 2000 individuals, given a certain probability \(p\) of being a carrier.

To compute \(L(p)\), we utilize combinations and probabilities. The formula uses binomial coefficients for each study: \( \left(\begin{array}{c} 1000 \ 41 \end{array}\right) \) and \( \left(\begin{array}{c} 2000 \ 79 \end{array}\right) \), which represent the number of ways to choose 41 carriers out of 1000 and 79 out of 2000, respectively. This is then multiplied by probabilities \(p\) for carriers and \(1-p\) for non-carriers, raised to the powers corresponding to the number of carriers and non-carriers in each sample.

• The exponents: \(p^{41}\), \((1-p)^{959}\), \(p^{79}\), and \((1-p)^{1921}\).
• It shows how combinatorics and probability theory come together to form a complex, yet meaningful expression.

This function helps statisticians and researchers estimate the true probability of being a carrier, based on the data provided.

Derivative

In mathematics and statistics, derivatives play a crucial role in optimization problems, like finding a maximum likelihood estimator. The derivative of a function gives us information about its rate of change. For maximum likelihood estimation, we focus on the derivative of the log of the likelihood function, also known as the log-likelihood. This simplifies the differentiation because logarithms turn products into sums, making the math easier to handle.

For our likelihood function \(L(p)\), the log-likelihood is \( \ln(L(p)) = \ln(1.3 \times 10^{73}) + \ln(1.4 \times 10^{143}) + 120 \ln(p) + 2880 \ln(1-p) \). By differentiating this expression with respect to \(p\), we obtain:

• \(\frac{d}{dp} \ln(L(p)) = \frac{120}{p} - \frac{2880}{1-p}\)
• This derivative helps us determine the conditions when our likelihood function is maximized.

Setting this derivative to zero and solving the resulting equation is a common procedure to find extreme points, like maximums or minimums. Here, it provides critical information about the probability \(p\) that maximizes our likelihood function.

Probability

Probability is the foundation of the likelihood function and helps quantify uncertainty. It denotes the likelihood of an event occurring. In the context of our exercise, the probability \(p\) represents the proportion of all people who are carriers of the cystic fibrosis gene.

To understand this, it's essential to grasp that probability ranges from 0 to 1, where 0 means the event never occurs, and 1 means it always occurs. For carriers, \(p\) would be between these limits, showing the portion of the population carrying the gene for cystic fibrosis.

• If \(p\) is close to 1, most individuals are carriers.
• If \(p\) is closer to 0, very few are carriers.

When calculating the likelihood function, the probabilities \(p\) and \(1-p\) are used to measure how well a certain value of \(p\) predicts the observed outcomes (number of carriers) in our samples. In summary, understanding probability is key to interpreting and calculating likelihoods, which then feed into larger concepts like maximum likelihood estimation.

Maximum Likelihood Estimator

The maximum likelihood estimator (MLE) is a fundamental concept in statistics used for estimating the parameters of a statistical model. In our example, the parameter is \(p\), which is the probability of being a carrier of the cystic fibrosis gene. The goal of MLE is to find the value of \(p\) that maximizes the likelihood function \(L(p)\).

To find the MLE, we follow these steps:

• Take the derivative of the log-likelihood function.
• Set the derivative equal to zero to find the point where the likelihood function does not increase or decrease.

In this exercise, solving \( \frac{120}{p} - \frac{2880}{1-p} = 0 \) results in \( \hat{p} = \frac{1}{25} \). This \( \hat{p} \) is what we call the maximum likelihood estimator of \( p \).

The MLE is crucial because it provides a single, best estimate that makes the observed data most probable under the model. It is widely used because of its desirable statistical properties, especially in large samples, where it tends to be unbiased and efficient.