According to the Wikipedia article on the Gamma distribution:

If $X\sim\mathrm{Gamma}(a,\theta)$ and $Y\sim\mathrm{Gamma}(b,\theta)$, where $X$ and $Y$ are independent random variables, then $X+Y\sim \mathrm{Gamma}(a+b, \theta)$.

But I don't see any proof. Can anyone point me to its proof please?

Edit: Thanks to Zen a lot, and also I found the answer as an example on the Wikipedia page about characteristic functions.

  • 3
    $\begingroup$ Intuition: Gamma$(n)$ distributions arise as the sums of $n$ independent Exponential distributions, whence it is immediate in this context that $X+Y$ will have a Gamma$(a+b,\theta)$ distribution provided $a$ and $b$ are both positive integers. $\endgroup$
    – whuber
    Mar 7 '13 at 16:54

The proof is as follows: (1) Remember that the characteristic function of the sum of independent random variables is the product of their individual characteristic functions; (2) Get the characteristic function of a gamma random variable here; (3) Do the simple algebra.

To get some intuition beyond this algebraic argument, check whuber's comment.

Note: The OP asked how to compute the characteristic function of a gamma random variable. If $X\sim\mathrm{Exp}(\lambda)$, then (you can treat $i$ as an ordinary constant, in this case)

$$\psi_X(t)=\mathrm{E}\left[e^{itX}\right]=\int_0^\infty e^{itx} \lambda\,e^{-\lambda x}\,dx = \frac{1}{1-it/\lambda}\, .$$

Now use Huber's tip: If $Y\sim\mathrm{Gamma}(k,\theta)$, then $Y=X_1+\dots+X_k$, where the $X_i$'s are independent $\mathrm{Exp}(\lambda = 1/\theta)$. Therefore, using property (1), we have $$ \psi_Y(t) = \left( \frac{1}{1-it\theta}\right)^k \, . $$

Tip: you won't learn these things staring at the results and proofs: stay hungry, compute everything, try to find your own proofs. Even if you fail, your appreciation of somebody else's answer will be at a much higher level. And, yes, failing is OK: nobody is looking! The only way to learn mathematics is by fist fighting for each concept and result.

  • $\begingroup$ The referenced statement explicitly states "provided all Xi are independent." $\endgroup$
    – whuber
    Mar 7 '13 at 16:52
  • 1
    $\begingroup$ One thing I don't get though, is how did we arrive at the characteristic functions? $\endgroup$
    – Dexter12
    Mar 7 '13 at 17:58
  • $\begingroup$ I'll add it to the answer. Take a look. $\endgroup$
    – Zen
    Mar 7 '13 at 18:02
  • $\begingroup$ Perhaps you can include a reference for the characteristic function of a $\Gamma(a,\theta)$ for non-integer values of $a$? $\endgroup$ May 20 '17 at 14:26

Here is an answer that does not need to use characteristic functions, but instead reinforces some ideas that have other uses in statistics. The density of the sum of independent random variables is the convolutions of the densities. So, taking $\theta = 1$ for ease of exposition, we have for $z > 0$, $$\begin{align} f_{X+Y}(z) &= \int_0^z f_X(x)f_Y(z-x)\,\mathrm dx\\ &=\int_0^z \frac{x^{a-1}e^{-x}}{\Gamma(a)}\frac{(z-x)^{b-1}e^{-(z-x)}}{\Gamma(b)}\,\mathrm dx\\ &= e^{-z}\int_0^z \frac{x^{a-1}(z-x)^{b-1}}{\Gamma(a)\Gamma(b)}\,\mathrm dx &\scriptstyle{\text{now substitute}}~ x = zt~ \text{and think}\\ &= e^{-z}z^{a+b-1}\int_0^1 \frac{t^{a-1}(1-t)^{b-1}}{\Gamma(a)\Gamma(b)}\,\mathrm dt & \scriptstyle{\text{of Beta}}(a,b)~\text{random variables}\\ &= \frac{e^{-z}z^{a+b-1}}{\Gamma(a+b)} \end{align}$$

  • 3
    $\begingroup$ (+1) It is ideal to have more than one way to prove everything. Maybe someone will post an answer considering the transformation $(X,Y)\mapsto(U,V)=(X+Y,X)$. $\endgroup$
    – Zen
    Mar 7 '13 at 18:37
  • $\begingroup$ Can we similarly find the density of $X-Y$ in a closed form expression? I'm unable to simplify the integrals in that case. $\endgroup$ Feb 26 '16 at 20:33
  • $\begingroup$ @pikachuchameleon See this answer of mine. $\endgroup$ Feb 27 '16 at 2:42

On a more heuristic level: If $a$ and $b$ are integers, the Gamma distribution is an Erlang distribution, and so $X$ and $Y$ describe the waiting times for respectively $a$ and $b$ occurrences in a Poisson process with rate $\theta$. The two waiting times $X$ and $Y$ are

  1. independent
  2. sum up to a waiting time for $a+b$ occurrences

and the waiting time for $a+b$ occurrences is distributed Gamma($a+b,\theta$).

None of this is a mathematical proof, but it puts some flesh on the bones of the connection, and can be used if you want to flesh it out in a mathematical proof.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.