Having $ \hat\theta= \frac{1}{N}\sum_{n=1}^{N} D_n $, with $ D_1 ... D_N $ independent random variables, which would be the solution of $ E[{\hat\theta}^2] $?

Would it be just $\frac{1}{N^2}\sum_{n=1}^{N}E[{D_n}^2] $ ?

And if so, knowing that $ E[\hat\theta] = \theta $, and therefore $ E[D_n] = \theta $, would the result of $ E[{\hat\theta}^2] $ be $\frac{1}{N^2}N\theta^2 = \frac{\theta^2}{N} $?

Any help is appreciated!


1 Answer 1


Notice that,

$$\hat{\theta}^2 = \left[\dfrac{1}{N} \sum_{n=1}^{N}D_n \right]^2 = \dfrac{1}{N^2}\left[ \sum_{n=1}^{N}D_n \right]^2 \ne \dfrac{1}{N^2}\left[ \sum_{n=1}^{N}D_n^2 \right]\,. $$ Notice that the last equality doesn't hold because in general $(a + b)^2 \ne a^2 + b^2$. Because of this reason,

$$E[\hat{\theta}^2] \ne \dfrac{1}{N^2}\left[ \sum_{n=1}^{N}E[D_n^2] \right] \,.$$

To find $E[\hat{\theta}^2]$, you will have to find the distribution of $\hat{\theta}$, and then calculate its second moment. Alternatively, if you know the variance of the estimator, then $$E[\hat{\theta}^2] = Var(\hat{\theta}) + E[\hat{\theta}]^2 = Var(\hat{\theta}) + \theta^2 \,.$$

  • $\begingroup$ That is really helpful! I have the p.d.f. of D so I can find its variance, and therefore the variance of $\hat\theta$. Thanks for the answer! @greenparker $\endgroup$
    – Ragnar
    Oct 26, 2017 at 10:17

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