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I'm having trouble determining whether the estimator beneath is unbiased or not. First some info: $x_i \sim binom(n_i,p)$

I have a sample of $x_1,x_2,x_3,x_4$ so $n=4$

I want to find out if the following

$\tilde{p}=\frac{1}{4}\left(\frac{x_{1}}{n_{1}}+\frac{x_{2}}{n_{2}}+\frac{x_{3}}{n_{3}}+\frac{x_{4}}{n_{4}}\right)$

I know that the criteria for an unbiased estimator is that $E(\tilde{p})=p$

This is what I came up with:

$E(\tilde{p})=E\left(\frac{1}{4}\left(\frac{x_{1}}{n_{1}}+\frac{x_{2}}{n_{2}}+\frac{x_{3}}{n_{3}}+\frac{x_{4}}{n_{4}}\right)\right)=\frac{1}{4}E\left(\frac{x_{1}}{n_{1}}+\frac{x_{2}}{n_{2}}+\frac{x_{3}}{n_{3}}+\frac{x_{4}}{n_{4}}\right)$

Now, I'm not sure if my next step is valid and how to proceed if it is:

$E(\tilde{p})=\frac{1}{4}E\left(\frac{x_{1}}{n_{1}}+\frac{x_{2}}{n_{2}}+\frac{x_{3}}{n_{3}}+\frac{x_{4}}{n_{4}}\right)=\frac{1}{4}\left(\frac{E(x_{1})}{n_{1}}+\frac{E(x_{2})}{n_{2}}+\frac{E(x_{3})}{n_{3}}+\frac{E(x_{4})}{n_{4}}\right)$

I tried to look in other threads but I wasn't able to find any similar calculation.

Kristian
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1 Answers1

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You are almost there:

Now use that $E(X_i)=n_ip$, so that gives then combined with your last step that

$E(\tilde{p})=p$.

I assumed here that you actually meant that $x_i \sim Binom(n_i,p)$, otherwise the result does not follow or $n_1+n_2+n_3+n_4=4*n$ should hold then.

Alwin
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  • Awesome :)

    So the estimator is unbiased because in the end it will become:

    $E(\tilde{p})=\frac{1}{4}(\frac{n_1p}{n_1}+...+\frac{n_4p}{n_4})$

    which is similar to \frac{p+p+p+p}{4}=\frac{4p}{4}$

    Thanks :)

     – Kristian May 29 '15 at 13:32