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As the first answer in a link shown , $T\sim {\rm Gamma}(2,1), S\sim {\rm Exp}(1)$, we have two properties:

  1. $T-S\sim {\rm Exp}(1)$,

  2. $T-S$ is independent of $S$.

I cant prove them. Esperically for Question 1, I find that,

$Z = T-S$,

${\rm pdf}(z)=\dfrac{1}{4}e^{z}$.

Obviously, Z does not conform to ${\rm Exp}(1)$. So I am not sure where it is wrong.

Thank you very much!

Community
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William_Third
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1 Answers1

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The joint distribution given in your link is $$f_{S,T}(s,t) = e^{-t} \mathbf{1}_{\{0<s<t\}}.$$ (The $\mathbf{1}$ is shorthand for the two cases "$0<s<t$" and "otherwise" in your link.)

Then, $$f_{S,T-S}(s,z) = f_{S,T}(s,s+z) = e^{-(s+z)} \mathbf{1}_{\{0 < s < s+z\}} = e^{-s} \mathbf{1}_{\{s>0\}}\cdot e^{-z} \mathbf{1}_{\{z>0\}},$$ which shows that $S$ and $T-S$ are independent $\operatorname{Expon}(1)$ random variables.

angryavian
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  • But I can't understand ${\rm pdf}(z) = \dfrac{1}{4}e^z$, which is not an exponential distribution. – William_Third Feb 07 '17 at 04:00
  • Furthermore, I have a question. Whether are the two properties based on $S=UV$ and $T=V$ in the link? Because the two equations show that $S$ and $T$ are dependent, while in my original question, $S$ and $T$ are independent. – William_Third Feb 07 '17 at 04:03
  • @William_Third I do not know where you got that $e^z/4$, which is not even a pdf, let alone the pdf of an exponential distribution. Regarding your second question, the two properties 1 and 2 hold in the setting of that link, where indeed $S$ and $T$ are dependent; this is what my answer is addressing. The question where $S \sim Expon(1)$ and $T \sim Gamma(2,1)$ are independent is completely different, and the two properties 1 and 2 probably do not hold. – angryavian Feb 07 '17 at 06:03