Question

Let $X$be a Poisson random variable with parameter $\lambda$. What value of $\lambda$maximizes $P\{X=k\},k\ge 0?$

Short answer

The solution is$\lambda =k$.

Step-by-step solution

Step 1: Given information

Let $X$ be a Poisson random variable with parameter $\lambda .$

Step 2:Explanation

We have that

$P(X=k)=\frac{{\lambda }^k}{k!}{e}^{-\lambda }$

We are required to find $\lambda >0$such that for that$\lambda$function $\lambda \mapsto \frac{{\lambda }^k}{k!}{e}^{-\lambda }$maximizes. Since $1/k!$is a constant, we can move it out of our consideration. Also, since the logarithm is strictly increasing function, it is enough to find the maximum of following function

$g\left(\lambda \right):=\log k!P(X=k)=\log {\lambda }^k{e}^{-\lambda }=k\log \lambda -\lambda$

Let's find stationary points.

We have that

$\frac{dg}{d\lambda }=\frac{k}{\lambda }-1=0\Rightarrow \lambda =k$

Since $\lambda =k$is the only stationary point, we have that for that $\lambda$density function $P(X=k)$maximizes.

Step 3:Final answer

The solution is $\lambda =k$.