I am learning basic statistics and I am trying to solve the (example) problems but I can't figure out how to solve the following problem.
I understand how to use the MGF to find expected value and variance etc, but I have no clue how to start with this problem. Any help would be great.
Given an exponential distribution $X_1$ with $f(x) = \frac{1}{6}\cdot e^{(-x/6)}$ for $x > 0$ and a chi-squared distribution $X_2$ with $f(x) = \frac{1}{4\,\Gamma(2)}\cdot e^{(-x/2)}$ for $x > 0$.
Find the distribution of $X_1 + 3X_2$ and its parameters.
My approach:
From the density function I know that for an exponential distribution, the MGF is $\frac{\alpha}{\alpha - t} , t < \alpha$.
For the first function I have $\alpha = \frac{1}{6}$
and
For the second function, the MGF is: $(1-\beta t)^{-\gamma}$ where $t < \frac{1}{\beta}$. Values of $\beta = \gamma = 2$. (I think I can read it directly from the functions)
To solve $X_1 + 3X_2$ , I have:
$\frac{0.1666}{0.16666 - t} + 3\cdot(1-2t)^{-2}$ or maybe I can write :
$\frac{\frac{1}{6}}{\frac{1}{6}-t} \rightarrow \frac{1}{1-6t} + 3\cdot(1-2t)^{-2}$.
So that I get:
$(1-6t)^{-1} + 3\cdot (1-2t)^{-2}$ .
Here I can't simplify any more :( and don't know whether I am right or wrong!
