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?


2 Answers 2


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. $E[X^2]$)?

  • 1
    $\begingroup$ Alright, thanks for the hint. Another follow up question is what is the relationship between $E[X^2]$ and $E[\hat\lambda^2]$ $\endgroup$ Mar 14, 2019 at 2:39
  • 1
    $\begingroup$ 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, 2019 at 2:51

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.