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I'm reading through the textbook "All of Statistics" and one of the problems gives the following estimator for the lambda parameter of the Poisson distribution:

$\hat{\lambda} = \frac{\sum_{i=1}^n x_i}{n}$

I have already shown that this is an unbiased estimator, but I would like to find the standard error, which involves finding the variance. I was trying to use the following variance definition to do this:

$Var(\hat\lambda) = E[\hat\lambda^2] - E[\hat\lambda]^2$

$Var(\hat\lambda) = E[\hat\lambda^2] - \lambda^2$ since it is unbiased

However I am a bit unsure about the left-hand term. My initial thought was the it $\lambda^2 = (\frac{\sum_{i=1}^n x_i}{n})^2$ but wouldn't this lead to variance that is equal to zero? $\lambda^2 = (\frac{\sum_{i=1}^n x_i^2}{n})$ seems more reasonable, but I'm not sure how you could get this. Could anyone provide some guidance?

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It can't be true that $\lambda^2 = (\frac{\sum_{i=1}^n x_i}{n})^2$ because the left hand side is a parameter, and the right hand side is a random variable.

Also $(\sum_i X_i)^2 \neq \sum_i x_i^2$.

Hint: can you find the second moment of one Poisson random variable (i.e. $E[X^2]$)?

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  • $\begingroup$ Alright, thanks for the hint. Another follow up question is what is the relationship between $E[X^2]$ and $E[\hat\lambda^2]$ $\endgroup$ – user9933193 Mar 14 at 2:39
  • $\begingroup$ using linearity $E[\hat\lambda^2] = n^{-2}E[Y^2]$ where $Y=\sum_i X_i$ is a Poisson random variable (assuming independence) $\endgroup$ – Taylor Mar 14 at 2:51
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You can use the fact that when observations i.i.d.

$$ \mathbb{E}\left[\left( \sum_{i=1}^{n}X_i \right)^2 \right]=n \mathbb{E}[X^2]$$

Don't forget how factor $1/n$ acts on the result.

You will also have to use that fact that when $X \sim \text{Poisson}(\lambda)$ $$\mathbb{E}[X] = \lambda, Var(X) = \lambda$$

Now think of how variance is defined... Substitute. And you've got your answer .

But if you try hard and still have difficulties, I could provide you with an answer.

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