Suppose that a random variable $X$ is distributed according to a gamma distribution with parameters $\alpha = 6$ and $\beta = 2$. Find these values. Suppose that a random variable $X$ is distributed according to a gamma distribution with parameters $\alpha = 6$ and $\beta = 2$, i.e., $X \sim \text{Gamma}(6, 2)$.
A.) Computer the mean and variance of $X$.
$E(X) = \alpha \beta = 6\times2 = 12$
$\text{Var}(X) = \alpha \beta^{2} = 6\times2^2 = 24$
B.) Find $E(X^4)$
I believe I found the correct values for the mean and variance but I am having trouble calculating $E(X^4)$. I know $E(X^2) = \text{Var}(X) + E^2(X)$ but I have no idea what to do for $E(X^4)$. Any help would be much appreciated.
 A: From wikipedia, the moments are given by the formula
$$E[X^n] = \theta^n \cdot \frac{\Gamma(n+k)}{\Gamma(k)}$$
where $\theta$ is the scale parameter and $k$ is the shape parameter.
Since you stated that you found the correct answer in the first part, you are not using the standard notation. Your $\alpha$ is the shape parameter and $\beta$ is the scale paarameter.
A: When you propose a Gamma density you must give information in order to understand which parametrization you are using. Given that you assume $E(X)=\alpha\cdot\beta$ your density is the following
$$f_X(x)=\frac{1}{\Gamma(6)2^6}x^5 e^{-x/2}$$
to calculate the 4° simple moment, simply use the definition
$$\begin{align}
\mathbb{E}[X^4] & =\int_0^{\infty}\frac{1}{\Gamma(6)2^6}x^9e^{-x/2}dx\\
& =\frac{\Gamma(10)2^{10}}{\Gamma(6)2^6} \underbrace{\int_0^{\infty}\frac{1}{\Gamma(10)2^{10}}x^9e^{-x/2}dx}_{=1}\\
& = 2^4\times9\times8\times7\times6=48{,}384
\end{align}$$

It is understood that if you are not sure about your point a) you first must check which parametrization you usually use for the gamma density. In fact, if $\beta$ is the rate parameter the solution will be different
$$\mathbb{E}[X^4]=\frac{9\times8\times7\times6}{2^4}=126$$
