# Confusion over Gamma distribution CDF

Here's a probability question (probably really simple) I'm not sure how to solve:

Gamma distribution $X\sim \mathcal{G}(\alpha,\beta)$ with $\mu = 20$ and $\sigma^2 = 80$
$P(X \le 24)$ = ?

The previous question was finding the values of $\alpha$ and $\beta$, which I did using $\mu$ = $\alpha$$\beta and \sigma^2 = \alpha$$\beta^2$.

For the gamma distribution cdf, my textbook says $P(X \le x) = F(x; \alpha, \beta) = F(x / \beta; \alpha,1)$ where $F(x / \beta; \alpha,1)$ is the standard gamma distribution cdf $$F(x;\alpha,1) = \frac{1}{\Gamma(\alpha)}\int_0^x {y^{\alpha-1}e^{-y}} \text{d}y$$

To integrate that, it looks like I need to use the chain rule, but our professor never did an example. Is there a shortcut method? We've never used integration in a real example, only to define the pdf and get the cdf for different distributions.

Edit

The examples in my textbook involving standard gamma distribution problems say to look up the values for $F(x;\alpha)$ in Table A.4 of the appendix. When I looked, Table A.4 was missing, which really disappoints me. Are there any standard gamma distribution tables online that I can print out and hand in with the assignment? I checked Wolfram Alpha but they didn't have one. Casio has something, but I'm not sure what the shape and scale parameters are.

Edit 2

Found that table. In the front of the book, it Table A.5 came right after A.3, which is why I thought A.4 was missing. I went to the library to see if they had the same textbook; they did, and someone had the common sense (which I didn't have) to look in the back of the book, and there it was. No more help is needed.

• You need to integrate by parts repeatedly beginning with $u = y^{\alpha-1}$ and $v=-e^{-y}$, $dv = e^{-y}dy$, and $$\int u dv = uv - \int v du.$$ Each time you do so, you will get an integral with a smaller exponent for $y$. If $\alpha$ is an integer, you will be able to finish up the process. If $\alpha$ is not an integer, things are more complicated. – Dilip Sarwate Feb 4 '12 at 23:06
• @dilip you should post your comment as an answer. – probabilityislogic Feb 5 '12 at 6:32
• @DilipSarwate, there is no closed form solution for $\alpha$ non-integer, this cdf is then the incomplete gamma function. – Xi'an Feb 5 '12 at 7:48
• And I strongly doubt integration by part was the aim of the exercise. – Elvis Feb 5 '12 at 7:53
• wolframalpha.com/input/?i=CDF[GammaDistribution[5%2C+4]%2C+24] – whuber Feb 6 '12 at 16:00

You need to integrate by parts repeatedly beginning with $u=y^{\alpha-1}$, $v=−e^{−y}$, $\mathrm dv=e^{−y}\mathrm dy$, and using $$\int_0^x u\ \mathrm dv= uv\biggr|_0^x − \int_0^x v\ \mathrm du.$$ Since $\mathrm du = (\alpha-1)y^{\alpha-2}\mathrm dy$, each time you do an integration by parts, you will get an integral with a smaller exponent for $y$ on the right hand side. If $\alpha$ is an integer (as it is in this particular case), you will be able to finish up the process with a $\int_0^x e^{-y}\mathrm dy$. If $\alpha$ is not an integer, things are more complicated because there is no general closed-form expression for $\int_0^x y^{\gamma}e^{-y}\mathrm dy$ where $0 < \gamma < 1$. As noted by Xi'an, the cdf is the incomplete gamma function, and its numerical values have been tabulated.