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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?

$\frac{1}{2.5^3\left(\frac{1}{2.5}-t\right)^3}\left(\frac{1}{5^5\left(\frac{1}{5}-t\right)^5}\right)$

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  • $\begingroup$ Hint: Assuming you're using shape/scale Gamma and not shape/rate, we have $2 X_3 \sim \Gamma(1.5, 2 \cdot 2.5) = \Gamma(1.5,5)$. $\endgroup$ – user365239 Nov 24 '16 at 17:05
  • $\begingroup$ It's of the form Γ($\alpha,\beta$). As my expression above shows, the denominator includes $\beta^\alpha$. How can I multiply the beta value like you say? $\endgroup$ – bballboy8 Nov 24 '16 at 21:35
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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.

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  • $\begingroup$ Could you direct me to where in the article it's discussed that 2X3∼Γ(1.5,5)? I'm having difficulty understanding why it wouldn't be Γ(3,2.5). $\endgroup$ – bballboy8 Nov 25 '16 at 6:29
  • $\begingroup$ Also, it might be useful to note that I'm using the formula where Γ($\alpha, \beta$) and $M(t)=\frac{1}{\beta^\alpha(\frac{1}{\beta}-t)^\alpha})$. $\endgroup$ – bballboy8 Nov 25 '16 at 6:34
  • $\begingroup$ Woops i meant to link en.wikipedia.org/wiki/Gamma_distribution#Scaling. That's the same MGF as the one in wiki. . . $(1- \beta t)^{-\alpha}$. If $X \sim \Gamma(1.5, 2.5)$, then $M_{2X}(t) = E[e^{2Xt}] = M_X(2t) = (1- 2 \cdot 2.5t)^{-1.5} = (1-5 t)^{-1.5}$. $\endgroup$ – user365239 Nov 25 '16 at 7:08
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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")

enter image description here

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  • $\begingroup$ What distribution would it be then? $\endgroup$ – bballboy8 Nov 24 '16 at 21:59
  • $\begingroup$ Not all distributions have names. As far as I know, this one doesn't. (But they do all have CDFs. "By their CDFs shall ye know them.") $\endgroup$ – BruceET Nov 24 '16 at 22:05

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