It is my understanding that when you want to find the distribution of the sample mean of some rv, you can do the following.
Let $f(x)$ denote the pdf of $X$:
$$f(x)=\frac{e^{-x (\lambda +\mu )} I_1\left(2 x \sqrt{\lambda \mu }\right)}{\sqrt{\rho } x}$$
Then the mgf of the sample mean $\bar{X}$ is given by $\left[E\left(e^{\frac{t\ X}{n}}\right)]\right]^n$.
Using the above to compute the mgf, I get
$$ M_X(t) =2^{-n} \left(\frac{\lambda +\mu -\sqrt{\left(\lambda +\mu -\frac{t}{n}\right)^2-4 \lambda \mu }-\frac{t}{n}}{\sqrt{\lambda \mu \rho }}\right)^n $$
But when I take the derivative and find the moments (i.e., the mean), it turns out that no matter what the sample size of $n$ is, the sample mean, $\bar{X}$, is exactly the same as the mean of the of the pdf of $X$.
$$E[X] =\frac{2^{-n} \left(\frac{-\sqrt{(\lambda -\mu )^2}+\lambda +\mu }{\sqrt{\lambda \mu \rho }}\right)^n}{\sqrt{(\lambda -\mu )^2}}$$
So I'm confused. I've simulated this rv and I get different results.
So, the question is that is my answer correct? Or did I make an error.
Thanks.
If $X_1, \ldots, X_n$ are drawn from the same distribution with mean $\mu$, then $$E[\overline{X}] = E\left[\frac{1}{n} \sum_{i=1}^n X_i \right] = \frac{1}{n} \sum_{i=1}^n E[X_i] = \mu,$$ regardless of the value of $n$, and regardless of what the original distribution was.