I know that the Gamma distribution is given by: $$\frac{1}{\beta^{\alpha}\Gamma(\alpha)}\int x^{\alpha-1}e^{-x/\beta}\,dx.$$ But when I calculate it, I'll always have to use integration by parts to solve it. Which takes so much time. I've seen a quick solution for this example like: $$\frac{1}{16}\int_{12}^{+\infty} x^2e^{-x/2}\,dx=25e^{-6}\approx 0.062.$$

But I didn't understand how they went from the left side to the right side. Is there a "quick and dirty" solution to solve such integral without having to use integration by parts? Same for Exponentional distribution.

  • $\begingroup$ this site uses MathJax for mathematical formula $\endgroup$ – JMP May 25 '17 at 5:38
  • $\begingroup$ Sorry. My first time here. $\endgroup$ – Madno May 25 '17 at 5:41

For any positive integer values of $\alpha$ and any real $\beta>0$, by using the third formula of this LIST, we obtain $$I=\frac{1}{\beta^{\alpha}\Gamma(\alpha)}\int_A^{+\infty} x^{\alpha-1}e^{-x/\beta}\,dx =e^{-A/\beta}\sum_{k=0}^{\alpha-1}\frac{(A/\beta)^k}{k!}.$$ By letting $\alpha=3$, $\beta=2$, and $A=12$, we get $I=25e^{-6}$.

  • $\begingroup$ As for Gamma distribution, b stands for lower bounds and a stands for \alpha? What about \beta? In the original question, \alpha was 3 and \beta was 2. $\endgroup$ – Madno May 25 '17 at 6:09
  • $\begingroup$ @Madno Now we have a general formula for any integer $\alpha$. $\endgroup$ – Robert Z May 25 '17 at 6:39

Well the gamma function is related to the factorial function, if you already did not know that. You can check that if you want. Also there is something called a probability distribution function and it supplies standard values for working with the normal distribution function or gamma function as you call it.


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