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I am trying to show that the sample variance is an unbiased estimator of $\lambda$ for a Poisson distribution.

Let $(X_1, \dots, X_n)$ be a random sample from a Poisson distribution with mean $\lambda > 0$.

The sample variance is given as

$$S^2 = \dfrac{\sum_{i=1}^{n}(X_i - \bar{X})^2}{n-1}$$

For $S^2$ to be an unbiased estimator, I need to show that $\mathbb{E}[S^2] = \lambda$

My attempt:

I know that $\mathbb{E}[\bar{X}] = \lambda$ and $\sum_{i=1}^{n}{X_i} \sim \text{Poisson}(n\lambda)$

Using linearity of expectations as in this similar question:

$$ \begin{align*} \mathbb{E}[S^2] &= \mathbb{E}\left[ \dfrac{\sum_{i=1}^{n}(X_i - \bar{X})^2}{n-1} \right] \\ &= \dfrac{1}{n-1}\mathbb{E} \left[ \sum_{i=1}^{n} \left( {X_{i}}^{2} + {\bar{X}}^2 - 2X_i\bar{X} \right) \right] \\ &= \dfrac{1}{n-1} \left( n \lambda^2 + \lambda^2 - 2\lambda^2 \right) \\ &= \dfrac{1}{n-1}(n + 1 -2) \lambda^2 \\ &= \dfrac{1}{n-1} (n-1) \lambda^2 \\ &= \lambda^2 \end{align*} $$

I end up with $\lambda^2$ but I need to end up with $\lambda$ to show that $S^2$ is an unbiased estimator of $\lambda$. Where have I gone wrong?

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    $\begingroup$ Using your approach, you should have \begin{align} &E\left[\sum_{i=1}^n(X_i^2+\bar X^2-2X_i\bar X)\right] \\&=\sum_{i=1}^nE(X_i^2)-nE(\bar X^2) \\&=n(\lambda+\lambda^2)-n\left(\frac{\lambda}{n}+\lambda^2\right) \\&=n\lambda-\lambda \end{align} $\endgroup$ Aug 17, 2018 at 19:45
  • $\begingroup$ Thank you, that makes sense now. $\endgroup$
    – meenaparam
    Aug 17, 2018 at 20:06
  • $\begingroup$ math.stackexchange.com/q/1701626/321264 $\endgroup$ May 29, 2020 at 18:13
  • $\begingroup$ Maybe it won't hurt to remember that the variance of a Poisson distribution is the same as its expected value. $\endgroup$ Dec 10, 2021 at 4:31

1 Answer 1

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You asked where you went wrong. There are several mistakes you made. Firstly,

$$ \mathbb{E} \left[ \sum_{i=1}^{n} {X_{i}}^{2} \right] \ne n \lambda^2 $$

Handled correctly,

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

Also you incorrectly did this evaluation:

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

The correct evaluation is:

$$ \mathbb{E} \left[ \sum_{i=1}^{n} \left( {\bar{X}}^2 - 2X_i\bar{X} \right) \right] =-n\mathbb{E} \left[ {\bar{X}}^2 \right] = -\lambda - n\lambda^2 $$ since $$ \mathbb{E}[\bar{X}^2] = \left(\mathbb{V}[\bar{X}]+\mathbb{E}[\bar{X}]^2\right)=\lambda/n+\lambda^2 $$

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  • $\begingroup$ Thanks, that's helpful, I realise where I went wrong now. $\endgroup$
    – meenaparam
    Aug 17, 2018 at 20:06
  • $\begingroup$ When I tried this, summing together $n\lambda + n\lambda^{2} + (n\lambda^{2} - 2n\lambda^{2}) = n\lambda$. Then multiplying by $\frac{1}{n-1}$ still doesn't get me to $\lambda$. What am I missing? $\endgroup$
    – meenaparam
    Aug 17, 2018 at 20:14
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    $\begingroup$ Sorry, corrected a mistake above. $\endgroup$
    – Dean
    Aug 17, 2018 at 21:30
  • $\begingroup$ Thanks, makes sense now, much appreciated. $\endgroup$
    – meenaparam
    Aug 17, 2018 at 21:40

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