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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 ?

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  • $\begingroup$ 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
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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, $$P(Y\leq y) = P(aX\leq y) = P(X\leq y/a).$$

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  • $\begingroup$ 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

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