Q.4.13 - A First Course in Probability

Question

Let X be a binomial random variable with parameters (n, p). What value of p maximizes P{X = k}, k = 0, 1, ... , n? This is an example of a statistical method used to estimate p when a binomial (n, p) random variable is observed to equal k. If we assume that n is known, then we estimate p by choosing that value of p that maximizes P{X = k}. This is known as the method of maximum likelihood estimation. 

Step-by-Step Solution

Answer

In the given information the answer is $p=\frac{k}{n}$

Step 1:Given Information

We are require to find$p\in \left(0,1\right)$ which maximize the probability

$P(X=k)=\left(\begin{array}{l}n\\ k\end{array}\right)p^k·(1-p{)}^{n-k}$

where $k$ is some fixed number. consider the logarithm of expression we have

$\log P(X=k)=\log \left(\begin{array}{l}n\\ k\end{array}\right)+k·\log p+(n-k)\log (1-p)$

Step 2:Calculation

We need to maximize the function $p\to k$.

$\log p+(n-k)\log (1-p)$.call the function$g$.

we have that 

$\frac{dg}{dp}=\frac{k}{p}-\frac{n-k}{1-p}=0$

since, equation is equal to $k(1-p)=(n-k) p$.which is satisfied for$p=\frac{k}{n}$.

Step 3:Final Answer

The answer is$p=\frac{k}{n}$.