So I was trying to prove the mean result of gamma distribution which is $\frac{\alpha}{\lambda}$.

My attempt,

$E(X)=\int_{0}^{\infty }x f(x)dx$

$=\int_{0}^{\infty } \frac{\lambda^{\alpha}}{\Gamma (\alpha)}x^{\alpha}e^{-\lambda x}dx$

After integrating it, I got the result $$\frac{\lambda^{\alpha}}{\Gamma (\alpha)} \cdot\frac{\alpha}{\lambda}(\int_{0}^{\infty } x^{\alpha-1}e^{-\lambda x}dx)$$. I'm stuck here. Could anyone continue it for me and explain? Thanks a lot.

  • $\begingroup$ Use the substitution $\lambda x=t $ Then the definition of the Gamma function. $\endgroup$ – Zaid Alyafeai Mar 16 '17 at 12:23
  • $\begingroup$ Note that $$\Gamma (\alpha)=\int ^\infty_0 e^{-t} t^{\alpha-1} dt$$ $\endgroup$ – Zaid Alyafeai Mar 16 '17 at 12:26
  • $\begingroup$ So I got $\frac{\lambda^{\alpha}}{\Gamma (\alpha)}*\frac{\Gamma(\alpha +1)}{\alpha+1}$ How should I continue? $\endgroup$ – Mathxx Mar 16 '17 at 12:31
  • $\begingroup$ I think Harry's answer should clear your doubts. $\endgroup$ – Zaid Alyafeai Mar 16 '17 at 12:38
  • $\begingroup$ Very similar to this post. $\endgroup$ – EditPiAf Mar 20 '17 at 22:58

Let us consider a random variable with Gamma distribution $X\sim \text{Gamma}(\alpha,\lambda)$. Its expected value is \begin{equation} E(X) = \frac{\lambda^\alpha}{\Gamma(\alpha)} \int_0^\infty x^{\alpha} \, e^{-\lambda x} \, dx \, . \end{equation} Making the change of variable $y=\lambda x$ in the integral, one has \begin{aligned} E(X) &= \frac{\lambda^\alpha}{\Gamma(\alpha)} \int_0^\infty \left(\frac{y}{\lambda}\right)^{\alpha} \, e^{-y} \, \frac{dy}{\lambda} \\ &= \frac{1}{\lambda\Gamma(\alpha)} \int_0^\infty y^{\alpha} \, e^{-y} \, dy\\ &= \frac{\Gamma(\alpha+1)}{\lambda\Gamma(\alpha)} \, . \end{aligned} Due to the relationship $\Gamma(\alpha+1) = \alpha\Gamma(\alpha)$, one obtains $E(X) = \alpha/\lambda$.

  • $\begingroup$ Am I correct in assuming that when you set $y=\lambda x$ that you implicitly set $dy=\lambda dx$ which means that $dx=dy / \lambda$? $\endgroup$ – Frank Shmrank Mar 20 '18 at 3:53
  • $\begingroup$ @FrankShmrank Yes, correct. $\endgroup$ – EditPiAf Mar 27 '18 at 6:27

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.