Gamma Distribution Sum I'm trying to show that the sum of three independent gamma distributions $X_1\sim \Gamma(2,5)$, $X_2\sim \Gamma(3,5)$, and $X_3\sim\Gamma(1.5,2.5)$ are a gamma distribution. $Y=X_1+X_2+2X_3$. I've multiplied the moment generating functions of $X_1$ and $X_2$ and I've squared the moment generating function of $X_3$ and am left with gamma distributions that I cannot multiply together to give me another Gamma distribution. How would I proceed?
$\frac{1}{2.5^3\left(\frac{1}{2.5}-t\right)^3}\left(\frac{1}{5^5\left(\frac{1}{5}-t\right)^5}\right)$
 A: If $Y_1 \sim Gamma(shape = \alpha_1, rate=\lambda)$ and 
$Y_2 \sim Gamma(shape = \alpha_2, rate=\lambda)$, then
$Y_1 + Y_2 \sim Gamma(shape = \alpha_1 + \alpha_2, rate=\lambda),$
as can be seen by multiplying moment generating functions. Notice that the shape parameters add, and the rate parameters are the same.
Of course, it is possible to find the CDF and PDF of the sum of the three
distributions you mention, but I do not believe that sum has a gamma
distribution. It is easy to find the expectation $\mu_Y$ and the variance $\sigma^2_Y$ of the sum your three gammas, but I do not believe
those match the mean and variance of any gamma distribution.
Below is a histogram of a million simulated realizations of the sum $W$ of $Y_1 \sim Gamma(2, 1)$ and
$Y_2 \sim Gamma(3, 2)$ along with R's default density estimator of the sum. 
I have also found the approximate mean and SD of the sum. The mean
of a gamma distribution is $\mu = \alpha/\lambda$ and the variance is
$\sigma^2 = \alpha/\lambda^2.$ I believe you can show there is no
choice of parameters $\alpha$ and $\lambda$ that match these formulas, and also
the mean and variance of $W.$ Hence, $W$ cannot be gamma.
 m = 10^6;  y1 = rgamma(m, 2, 1);  y2 = rgamma(m, 3, 2)
 w = y1 + y2;  mean(w);  sd(w);  sqrt(2 + 3/4)
 ## 3.498231  # aprx E(W) = 3.5
 ## 1.655406  # aprx SD(W)
 ## 1.658312  # exact SD(W)
 hist(w, prob=T, col="skyblue2", 
     main="Estimated Density of Sum of 2 Gammas with Different Rates")
   lines(density(w), lwd=2, col="darkgreen")


A: I'm assuming shape/scale not shape/rate (see https://en.wikipedia.org/wiki/Gamma_distribution - I can tell by the mgf you're using).
Then $2X_3 \sim \Gamma(1.5,5)$ (see https://en.wikipedia.org/wiki/Gamma_distribution#Scaling - it might be good to prove this for yourself). Let $Z = X_1+X_2+2X_3$, then by independence the mgf of $Z$ is the product of the mgf of $X_1, X_2,$ and $2X_3$:
$$
M_Z(t) = \left[ \left( 1 - 5t  \right)^{-2} \right]
\left[ \left( 1 - 5t \right)^{-3} \right]
\left[ \left( 1 - 5t\right)^{-1.5} \right] \text{ for } t< 1/5
$$
which simplifies to $M_Z(t) = (1-5t)^{-6.5}$ for $t<1/5$. We recognize this as the mgf of a $\Gamma(6.5,5)$ random variable.
