Given $X \sim Pois(\lambda)$, is the MLE of $P(X=3)$ consistent?

Question

Given $X \sim Pois(\lambda)$, is the MLE of $P(X=3)$ consistent?

I've defined $z = P(X=3)$, and thus $$ \hat{z}_{MLE} = \hat{P(X=3)} = \hat{\frac{\lambda^3e^\lambda}{3!}} = \frac{(\hat{\lambda}_{MLE})^3e^{(\hat{\lambda}_{MLE})}}{3!} = \frac{\bar{X}^3e^{\bar{X}}}{3!} $$

But how can I proceed from here? I know that $$ MSE(\hat{z}_{MLE}, z) = Var(\hat{z}_{MLE}) + Bias(\hat{z}_{MLE}, z)^2 $$ If I can show that the limit of that is 0 when $n$ goes to $\infty$ that'll prove what I want, but I can't seem to understand how to derive it.

Is it the only way to prove $\hat{z}_{MLE} = \hat{P(X=3)}$ is consistent?

Answer 1

Using the MSE convergence criteria is not the best approach in such problems as mostly you will not be able to find a closed analytical form for the estimator's distribution. Therefore, just use the WLLN and the continuous mapping theorem (and Slutsky, if required). In your case $$ g(x) = \frac{x^3 e^{-3x}}{3!} $$ is everywhere continuous transformation, thus you have that $$ g(\bar{X}_n)\xrightarrow{p} g(\mathbb{E}[X]=\lambda) = \frac{\lambda^3e^{-3\lambda}} {{3!}} = P(X=3). $$