So, dealing with this problem and in need of experts opinion.
As far as my computations follow, it results that my result is different from the book I'm looking at. So please, help me out with this.
I want to compute the expectation of $\bar{X}= \frac{1}{n} \sum_{i=1}^n X_i$, where each $X_i \sim \Gamma(\alpha,\beta)$.
Here my reasoning:
$ E(\bar{X}) = E[ \frac{1}{n} \sum_{i=1}^n X_i ] $
This is:
$ E(\bar{X}) = \frac{1}{n} E[ \sum_{i=1}^n X_i ] $
Now, since all $X_i$ are distributed as $\Gamma(\alpha,\beta)$, we have that $\sum_{i=1}^n X_i \sim \Gamma(n\alpha,\beta) $.
Therefore, going back to the expectation operation we get:
$ E(\bar{X}) = \frac{1}{n} n\alpha\beta = \alpha\beta $.
Then, somehow, the book I'm reading says this result is $ E(\bar{X}) = \beta $.
What am I missing??
Thanks for your help.
Assuming you and the book are both using a parametrization where the PDF is$$ f_X(x) = \frac{1}{\beta^\alpha\Gamma(\alpha)}x^{\alpha-1}e^{-x/\beta}$$ (which is one of the two most common), I would say you aren't missing anything and the book is simply wrong. To simplify what you wrote somewhat, we always have $E(\bar X) = E(X)$ (i.e. the sample mean is an unbiased estimator of the true mean), so $E(\bar X) = E(X) = \alpha \beta.$ My guess is that since $\beta$ is a scale parameter (so the mean is proportional to it given that it's nonzero) your book simply slipped up and assumed it was equal to the mean.