I have a gamma distributed random variable $x$ with pdf $$p_x(x)=\frac{\lambda^r}{\Gamma(r)}x^{r-1}\exp(-\lambda x),$$ where $r$ and $\lambda$ are shape and rate parameters respectively.

If for example I multiply $x$ by a constant value $a$ so $y=ax$, what will be the distribution of $y$? How can we introduce the mean value of $y$ in the pdf expression ?

  • $\begingroup$ Do you know what a scale parameter is? This would help out immensely. Otherwise, use $$p_{Y}(y) = p_{X}\left(\dfrac{y}{a}\right)\left|\dfrac{\text{d}}{\text{d}y}\left[ \dfrac{y}{a} \right]\right|$$ for $a \neq 0$. $\endgroup$ – Clarinetist Mar 15 '16 at 10:43
  • $\begingroup$ @Clarinetist actually the scale parameter $m$ is nothing but the inverse of the rate $\lambda$ : $m$=1/$\lambda$, in some references you may find either (shape,scale) or (shape,rate) representations. Thanks anyway. $\endgroup$ – Elmehdi Mar 15 '16 at 10:47
  • $\begingroup$ Yes, what you are saying is correct. Actually, one of the things about a scale parameter is that if $X$ has scale parameter, say $\tau$, then $aX$ (for $a > 0$) has scale parameter $a\tau$, with all other parameters remaining unchanged. $\endgroup$ – Clarinetist Mar 15 '16 at 11:44

If $X\sim\text{Gamma}(r,\lambda)$, by which I mean it has the density that you provided, then for $a>0$, $$Y =aX\sim \text{Gamma}(r,\lambda/a).$$ This implies that the $E[Y] = r/(\lambda/a)$.

To prove this, notice that $Y = aX$ is one to one over the support of $X$, and hence we can use the one to one transformation; $X= Y/a$, and so $$f_Y(y) = \frac{f_X(y/a)}{\left|\frac{dy}{dx}\right|_{x = y/a}}.$$

Otherwise, you can use the cdf method, $$P(Y\leq y) = P(aX\leq y) = P(X\leq y/a).$$

  • $\begingroup$ thank you so much, your solution was so helpful. $\endgroup$ – Elmehdi Mar 15 '16 at 10:55
  • $\begingroup$ I'm glad. Consider giving a check mark. Good luck. $\endgroup$ – Em. Mar 15 '16 at 10:58

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.