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.

Short answer

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

Step-by-step solution

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$p=\frac{k}{n}$.