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Let $Y_1,\cdots,Y_n$ an independent random sample such that $Y_i\sim\text{ Binomial}(m_i,p)$, where $m_i$ is constant for each $i$. Let $T=\frac{\sum_{i=1}^nY_i}{\sum_{i=1}^nm_i}$ and $V=\frac{1}{n}\sum_{i=1}^n\frac{Y_i}{m_i}$ estimators of $p$, Find Mean squared error of each estimator.

My attempt

I'm trying to use the equation $$MSE(T)=Var(T)+(E[T]-p)^2, \ \ \ \ \ (1)$$ and $$Var(T)=E[T^2]-E^2[T], \text{ where}$$

$$E[T]=E\left[\frac{\sum_{i=1}^nY_i}{\sum_{i=1}^nm_i}\right]=\frac{\sum_{i=1}^nm_ip}{\sum_{i=1}^nm_i}=p$$ so, the 2nd term of the sum (1) is null.

But I can't find the value of $$E[T^2]=E\left[\frac{(\sum_{i=1}^nY_i)^2}{(\sum_{i=1}^nm_i)^2}\right]=\frac{E[(\sum_{i=1}^nY_i)^2]}{(\sum_{i=1}^nm_i)^2}$$

and, by the other hand:

$$Var[V]=Var\left(\frac{1}{n}\sum_{i=1}^n\frac{Y_i}{m_i}\right)=\frac{1}{n^2}\sum Var(\frac{Y_i}{m_i})=\frac{1}{n^2}\sum\frac{m_ip(1-p)}{m_i^2}.$$

$$Var[V]=\frac{p(1-p)}{n^2}\sum \frac{1}{m_i} \ \ \ \ \ (2)$$ is correct?

I would appreciate suggestions.

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    $\begingroup$ Use $Var(c\sum Y_i)=c^2Var(\sum Y_i)=c^2\sum Var(Y_i)$ where $c$ is a constant. $\endgroup$ Commented Feb 28, 2020 at 16:32
  • $\begingroup$ Independence in the $Y_i$´s implies independence in the $Y_i^2$'s? $\endgroup$
    – User 2014
    Commented Feb 28, 2020 at 16:43
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    $\begingroup$ Yes, but the above follows from the independence of $Y_i$'s. $\endgroup$ Commented Feb 28, 2020 at 16:46
  • $\begingroup$ Thanks, I added the expression of $Var(T)$ above in (2), it's correct? $\endgroup$
    – User 2014
    Commented Feb 28, 2020 at 17:39
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    $\begingroup$ Yes, variance of $V$ is correct. Similarly do this for variance of $T$. $\endgroup$ Commented Mar 2, 2020 at 13:29

1 Answer 1

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Both $T$ and $V$ are unbiased, so indeed the second term in (1) is 0 in both cases. Now, $$\mathbb{E}[T^2] = \frac{1}{(\sum_{i=1}^Nm_i)^2}(\sum_{i=1}^N\mathbb{E}[Y_i^2] + \sum_{j \neq i}\mathbb{E}[Y_i]\mathbb{E}[Y_j])$$ where linearity of expectation, squared sum property, and independence were used. Continuing, we can write: $$\frac{1}{(\sum_{i=1}^Nm_i)^2}(\sum_{i=1}^N(m_ip(1-p) + m_i^2p^2) + \sum_{j \neq i}m_im_jp^2) = $$ $$\frac{p(1-p)}{\sum_{i=1}^Nm_i} + \frac{p^2(\sum_{i=1}^Nm_i^2 + \sum_{j \neq i}m_im_j)}{(\sum_{i=1}^Nm_i)^2} = \frac{p(1-p)}{\sum_{i=1}^Nm_i} + p^2$$ Substracting $\mathbb{E}[T]^2 = p^2$ from this gives $$\mathbb{V}[T] = \frac{p(1-p)}{\sum_{i=1}^Nm_i}$$ The second one seems correct to me.

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