Determine the expected value of an estimator of a Bernoulli distribution $\hat p_{\alpha,\beta} = \frac{n\bar{X}+\alpha}{n+\beta}$

Question

Determine the expected value of the following family of estimators, where $X_1, ..., X_n$ are Bernoulli distributed:

$\hat p_{\alpha,\beta} = \frac{n\bar{X}+\alpha}{n+\beta}$

I tried:

$E(\hat p_{\alpha,\beta}) = E(\frac{n\bar{X}+\alpha}{n+\beta}) = E(\frac{n\bar{X}}{n+\beta}) + \frac{\alpha}{n+\beta} = E(\frac{n \cdot \frac{1}{n}\sum_{i=1}^nX_i}{n+\beta}) + \frac{\alpha}{n+\beta}= E(\frac{X_1 + ... X_n}{n+\beta}) + \frac{\alpha}{n+\beta}= \frac{1}{n+\beta}E(X_1 + ... X_n) + \frac{\alpha}{n+\beta}= \frac{np}{n+\beta} + \frac{\alpha}{n+\beta}= \frac{np+\alpha}{n+\beta}$.

But apparently the result is: $E(\hat p_{\alpha,\beta}) = p + \frac{\alpha - \beta p}{n+\beta}$.

Can anyone explain how this result comes up?

Answer 1

$p + \frac{\alpha -\beta p}{n + \beta} = \frac{np+ \beta p + \alpha -\beta p}{n + \beta} = \frac{np + \alpha}{n + \beta} $

Your proof is correct.