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I'm working through a practice problem for an exam and I would like to verify that I've done it correctly. Additionally I'd like some insight on the intuition behind the numbers I'm getting.

Problem

For $X$ and $Y$ independent $Exp(1)$ r.v.'s compute $P(X<3|X+Y)$.

Attempt

Since the sum of exponentials is a scaled gamma, $S=X+Y$~$Ga(1,2)$, where the rate is 1 and the shape is 2.

When $X+Y\leq 3$, the probability is 1 that $\{X<3\}$. I checked this also by solving:

$$P(X<3|X+Y\leq 3)=\frac{P(X<3,Y<3-X)}{P(S\leq 3)}=\frac{\int_0^3\int_0^{3-x} e^{-x}e^{-y}dy dx}{\int_0^3 se^{-s}ds}=1$$

Now when $X+Y>3$, I compute a similar quantity, namely, $$P(X<3|X+Y> 3)=\frac{P(X<3,Y>3-X)}{P(S> 3)}=\frac{3}{4}$$

Interpretation

Assuming that the above quantities are correct, could I interpret the quantity $P(X<3|X+Y>3)$ as follows:

"Given that the sum is greater than 3, the chances that the one of the variables is less than 3 is distributed uniformly (due to memoryless property of exponential r.v.'s). That is, if the sum is >3, and if we fix e.g. Y=3, then on average the sum is $Y + E(X)=4$, and we get that X<3, 3/4ths of the time."

The above interpretation is loosely based on my finding that for $k\in\mathbb{N}$ $$P(X<k|X+Y>k)=\frac{k}{k+1}=\frac{k}{k+E(X)}$$

Thanks for any help.

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Denote $S=X+Y$. Then

$$f_S(s)=se^{-s} \text{; } f_{X,S}(x,s)=1\{s\ge x\}e^{-s}$$

and

$$P\{X<c|S=s\}=\int_0^{c}\frac{f_{X,S}(x,s)}{f_{S}(s)}dx=\int_0^{c}\frac{1\{s\ge x\}e^{-s}}{se^{-s}}dx$$ $$=\frac{1}{s}\int_0^c 1\{s\ge x\}dx=\frac{c}{s}\wedge 1$$

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  • $\begingroup$ could you explain how you got $f_{X|S}$? $\endgroup$ – user251257 Jul 4 '15 at 2:37
  • $\begingroup$ It was a typo... $\endgroup$ – d.k.o. Jul 4 '15 at 2:40
  • $\begingroup$ @user251257 $f_{X,S}(x,s)=f_{S|X}(s|x)f_X(x)$ $\endgroup$ – d.k.o. Jul 4 '15 at 2:43
  • $\begingroup$ @d.k.o. Thank you! It actually makes sense that the probability depend on the sum. I'm unsure how you got $f_{S|X}$. Could you explain? $\endgroup$ – probtheory Jul 4 '15 at 3:53
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    $\begingroup$ @probtheory $F_{S|X}(s|x)=P\{Y\le s-x\}=1\{s\ge x\}(1-e^{x-s})$ $\endgroup$ – d.k.o. Jul 4 '15 at 4:01

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