# Suppose $Y_i \overset{\text{iid}}{\sim}\text{Poisson}(X_i \lambda)$, $X_i$ are known. Show $\hat{\lambda}_{\text{MLE}}$ is consistent for $\lambda$

Suppose $Y_i \overset{\text{iid}}{\sim}\text{Poisson}(X_i \lambda)$, where $X_i$ are known covariates. Give a condition on $\{X_i\}$ such that $\hat{\lambda}_{\text{MLE}}$ is a consistent estimator of $\lambda$, and give a counterexample of $\{X_i\}$ for which $\hat{\lambda}_{\text{MLE}}$ is not a consistent estimator of $\lambda$ when that condition is violated.

I have already showed that $$\hat{\lambda}_{\text{MLE}} = \dfrac{\sum_{i=1}^{n}Y_i}{\sum_{i=1}^{n}X_i}$$ and obviously $\sum_{i=1}^{n}Y_i \sim \text{Poisson}(\lambda\sum_{i=1}^{n}X_i)$.

In general, I know that the MLE is a consistent estimator, but I'm not sure how to go about showing that for this particular situation. I was initially considering dividing by $n$ for both the numerator and denominator to get $$\hat{\lambda}_{\text{MLE}} = \dfrac{\bar{Y}_n}{\bar{X}_n}$$ and we know $\bar{Y}_n \overset{p}{\to} \lambda\sum_{i=1}^{\infty}X_i$ (is this right?) by the Weak Law of Large Numbers (WLLN), so I would guess that if $0 < \sum_{i=1}^{\infty}X_i < \infty$, then $$\hat{\lambda}_{\text{MLE}}\overset{p}{\to}\dfrac{\lambda\sum_{i=1}^{\infty}X_i}{\sum_{i=1}^{\infty}X_i} = \lambda\text{.}$$ If $\sum_{i=1}^{\infty}X_i = \infty$, then I would guess the condition would be violated, though I wouldn't know how to show this.

Is my work correct? If not, how can I fix it?

Edit: I've just realized my work is definitely wrong, as $\bar{X}_n \overset{p}{\to} \sum_{i=1}^{\infty}X_i$ is obviously not true.

• This is surely overkill, but one way would be to calculate $E[\hat{\lambda}]$ and $\text{Var}(\hat{\lambda})$ and give an argument that a standardized $\hat{\lambda}$ converges to a standard normal in the limit, from which the required result follows easily. – Glen_b -Reinstate Monica May 8 '17 at 2:14

You need to use the fact that $$\sum_{i=1}^{n}Y_i \sim \text{Poisson}(\lambda\sum_{i=1}^{n}X_i).$$
• Perhaps I'm being nitpicky, but could you please clarify something? I started reviewing this: essentially, what you have here is a sufficient condition for consistency of $\hat{\lambda}_{\text{MLE}}$. A sufficient condition for $\hat{\lambda}_{\text{MLE}}$ to be consistent for $\lambda$ is that $\sum_{i=1}^{\infty}x_i = \infty$, which I understand. But this isn't a necessary condition for consistency. It is true that if $\hat{\lambda}_{\text{MLE}}$ were NOT consistent, then $\sum_{i=1}^{\infty}x_i < \infty$, but lack of consistency is exactly what we're trying to show. – Clarinetist May 8 '17 at 12:14
• What would be a condition that would imply that $\hat{\lambda}_{\text{MLE}}$ is not consistent? – Clarinetist May 8 '17 at 12:15
• Take $x_i = i^{-2}$ then – Taylor May 8 '17 at 15:13
• If we take $x_i = i^{-2}$, then we get a geometric series so that $\sum_{i=1}^{\infty}x_i < \infty$, and thus the asymptotic variance is nonzero. What I don't understand is that how this implies that $\hat{\lambda}_{\text{MLE}}$ isn't consistent. We have an unbiased estimator with non-zero asymptotic variance, but we see from this answer that there are examples of unbiased, consistent estimators with nonzero asymptotic variance. – Clarinetist May 8 '17 at 15:18