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 $\alpha = \frac{1}{6}$


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!

  • $\begingroup$ The question, as written, is unanswerable, because (1) the value of $f$ given for $X_2$ does not describe a distribution (there's probably a typographical error) and (2) asking for "parameters" makes little sense. What did the original question ask? As far as your work goes, why are you adding the mgfs? $\endgroup$ – whuber Jun 25 '18 at 12:50
  • $\begingroup$ @whuber it was an old exam question , however there can be a mistake. Can you tell me in in steps please , once i find the values for the parameters of $f(x)$ (like $\alpha = 1/6$, then how should i combine them to find the distribution of the sum of the random variables! $\endgroup$ – Luai Ghunim Jun 25 '18 at 20:43
  • $\begingroup$ Please see stats.stackexchange.com/search?q=moment+generating+sum. $\endgroup$ – whuber Jun 25 '18 at 20:48

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