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Let $X \sim \mathsf{Gamma}(2,3)$ and $Y = 2X.$ Find the pdf of $Y$ at $y=13.5.$

Attempt: $f_X(x)= 2*[1/9*\Gamma(2)]*x*e^{-x/3}.$ Do I have to integrate now?

BruceET
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user417179
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  • I don't understand. You want the probability that $Y=13.5$? This is surely $0$ since this is a continuous distribution. – Dave Mar 06 '17 at 01:42
  • That's what confuses me. Is the question worded incorrectly? Or is it a trick question? The instructions say that my answer needs to have 4 decimal places. – user417179 Mar 06 '17 at 01:44
  • Well, I would start by finding the pdf of $Y $. Do you know the method of CDF, MGF or transformation? – Dave Mar 06 '17 at 01:47
  • I'm not sure how to find the pdf of Y. – user417179 Mar 06 '17 at 01:51
  • No need to hypothetize a "trick question": in full generality, $$f_{2X}(y)=\frac12,f_X\left(\frac{y}2\right)$$ – Did Mar 06 '17 at 08:22

2 Answers2

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$X\sim \text{Gamma}(2,3)\implies f_X(x)=\frac{1}{9\Gamma(2)}xe^{-x/3}$ for $x>0$. The method of CDF yields (for $y>0$): $$\begin{align}F_Y(y)&=P(Y\leq y)\\&=P\left(X\leq\frac{y}{2}\right)\\&=\int_0^{y/2}\frac{1}{9}xe^{-x/3}dx\\&=\frac{-1}{3}\left[(x+3)e^{-x/3}\right]^{y/2}_0\\&=1-\frac{y+6}{6}e^{-y/6}\end{align}$$ Using the fact that $f_Y(y)=\frac{d}{dy}F_Y(y)$, we see that: $$f_Y(y)=\begin{cases}\frac{1}{36}ye^{-y/6}\qquad x>0\\0\qquad\qquad\text{otherwise}\end{cases}$$ Thus, $Y\sim\text{Gamma}(2,6)$. Hope this helps.

Dave
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  • Why go through PDF$\to$CDF$\to$CDF$\to$PDF when PDF$\to$PDF is elementary? – Did Mar 06 '17 at 08:24
  • @Did I figured the CDF method provides more information than the transformation method, both of which are unknown to the OP apparently. – Dave Mar 06 '17 at 16:15
  • "more information" How? – Did Mar 06 '17 at 16:33
  • @Did it provides the CDF function of $Y$. – Dave Mar 06 '17 at 16:38
  • Which is not needed for the PDF... Providing the direct way (possibly completed, if really one insists, by the détour in your answer) would seem mandatory in such situations, no? – Did Mar 06 '17 at 16:40
  • @Did I wouldn't say it is mandatory to use the more direct method, although it may be easier. I simply just offer a solution using the CDF method which yields the distribution of $Y $, as well as provides the CDF function which may be needed to answer other questions. – Dave Mar 06 '17 at 17:42
  • "Providing" (not "using") the direct method seems mandatory, yes. – Did Mar 06 '17 at 18:53
  • @Did I would not say it is "mandatory" to provide such a solution. – Dave Mar 06 '17 at 19:31
  • And this opinion indicates a peculiar approach to pedagogy... – Did Mar 06 '17 at 19:32
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Another option:

Suppose $X $~ Gamma(2,3). Then the moment generating function for $X$ is:

$$f(t) = E(e^{tX}) = (1-\theta t)^{-k} = (1-3t)^{-2}$$

Also, the moment generating function for $2X$ is: $$E(e^{t2X}) = E(e^{(2t)X}) = f(2t) = (1-3(2t))^{-2}$$

So we see that $2X$ has moment generating function $(1 - 6t)^{-2}$, so that $2X$ ~ Gamma(2,6).

OldGodzilla
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