Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $X\sim \mathrm{Gamma}(a_1,b)$ and $Y \sim \mathrm{Gamma}(a_2,b)$, I need to prove $X+Y\sim(a_1+a_2,b)$ if $X$ and $Y$ are independent.

I am trying to apply formula for independence integral and just trying to multiply the gamma function but stuck ?

share|improve this question
    
Hint: After multiplying $f_{X_1}(x)$ and $f_{X_2}(z-y)$ and making sure that the limits are correct, you will get an integral for $f_{X_1+Y_2}(z)$ that can be transformed into a Beta function whose value is $B(a_1,a_2) = \frac{\Gamma(a_1)\Gamma(a_2)}{\Gamma(a_1+a_2)}$. –  Dilip Sarwate Dec 3 '12 at 16:28

2 Answers 2

up vote 2 down vote accepted

Now that the homework deadline is presumably long past, here is a proof for the case of $b=1$, adapted from an answer of mine on stats.SE, which fleshes out the details of what I said in a comment on the question.

If $X$ and $Y$ are independent continuous random variables, then the probability density function of $Z=X+Y$ is given by the convolution of the probability density functions $f_X(x)$ and $f_Y(y)$ of $X$ and $Y$ respectively. Thus, $$f_{X+Y}(z) = \int_{-\infty}^{\infty} f_X(x)f_Y(z-x)\,\mathrm dx. $$ But when $X$ and $Y$ are nonnegative random variables, $f_X(x) = 0$ when $x < 0$, and for positive number $z$, $f_Y(z-x) = 0$ when $x > z$. Consequently, for $z > 0$, the above integral can be simplified to $$\begin{align} f_{X+Y}(z) &= \int_0^z f_X(x)f_Y(z-x)\,\mathrm dx\\ &=\int_0^z \frac{x^{a_1-1}e^{-x}}{\Gamma(a_1)}\frac{(z-x)^{a_2-1}e^{-(z-x)}}{\Gamma(a_2)}\,\mathrm dx\\ &= e^{-z}\int_0^z \frac{x^{a_1-1}(z-x)^{a_2-1}}{\Gamma(a_1)\Gamma(a_2)}\,\mathrm dx &\scriptstyle{\text{now substitute}}~ x = zt~ \text{and think}\\ &= e^{-z}z^{a_1+a_2-1}\int_0^1 \frac{t^{a_1-1}(1-t)^{a_2-1}}{\Gamma(a_1)\Gamma(a_2)}\,\mathrm dt & \scriptstyle{\text{of Beta}}(a_1,a_2)~\text{random variables}\\ &= \frac{e^{-z}z^{a_1+a_2-1}}{\Gamma(a_1+a_2)} \end{align}$$

share|improve this answer
    
Dilip, Your response assumes that parameter $B=1$ is the same for both distributions. What would the function for the sum of two independent Gamma distributions be where both sets of parameters $A$ and $B$ are different? Convolute the two functions $z^{A_i-1}\mathrm{Exp}(-B_i z)/\mathrm{Gamma}(A_i)$, where $i$ is index $1$ and $2$. –  Tomas Kollen Mar 14 at 21:26

You may use a easier method. Consider the moment generating function or probability generating function. $E(e^{(X+Y)t} )=E(e^{Xt}e^{Yt})=E(e^{Xt})E(e^{Yt})$ as they are independent then we can get a moment generating function of a gamma distribution. Then you can find the mean and variance from the Moment generating function

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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