# How to identify the limits of triple integral where variables are functionally related?

I have to calculate $$P(X > Y+Z, X > Y)$$, where $$X, Y$$ and $$Z$$ are i.i.d. exponential (1).

So, I know that the joint pdf is just the product of marginal pdfs: $$e^{-(x + y + z)}$$

I have been given that the probability evaluates to $$1/4$$, but when I work out the triple integral, I get answers that are either functions of $$y$$ or $$z$$.

I guess I am taking the wrong limits for the 3 variables in the triple integral. Currently, I am using

$$x$$: from 0 to infinity,
$$y$$: from 0 to $$x-z$$ (I also tried $$y$$: from 0 to $$x$$),
$$z$$: from 0 to $$x-y$$

The event for which I need the probability is: $$X > Y+Z$$ and $$X > Y$$

I'd be grateful for any advice on where I am going wrong with the limits for the triple integral...

• What is your integrand? (The thing inside the integrals) By the way, one of the two limits is redundant-- can you see why? Sep 16, 2023 at 5:52
• Thanks! Yes, when X > Y + Z, then the additional condition that X > Y is redundant, since X > Y+Z automatically means X > Y as well, when X, Y and Z are all positive. The integrand is e^-(x+y+z) which is just the product of e^-x, e^-y, and e^-z. I guess I am stuck at correctly identifying the limits... Sep 16, 2023 at 5:57

In case you aren't compelled to compute the integrals, here is a fun way of solving the problem:

We know that $$X, Y, Z \sim \text{Expo}(1)$$ independently, and we are interested in $$P(X>Y+Z)$$.

The intuition is that if $$X>Y$$ (with half chance), then $$X$$ has a half chance (again) to be $$>Z$$.

Formally, \begin{align} P(X>Y+Z) &= P(X>Y)\cdot P(X>Y+Z|X>Y) \quad \text{by conditioning} \\ &= \frac12 \cdot E(P(X>Y+Z | X>Y, Y, Z)) \quad \text{by symmetry and the tower rule} \\ &= \frac12 \cdot E(P(X > Z | Y, Z)) \quad \text{by memoryless property of Expo} \\ &= \frac12 \cdot P(X>Z) \quad \text{by tower rule (in reverse)}\\ &= \frac12 \cdot \frac12 = \frac14 \quad \text{by symmetry} \end{align}

This technique easily generalises to arbitrarily many random variables on the right hand side, so if all random variables are i.i.d $$\text{Expo}(1)$$ then

$$P\left(X > \sum_{i=1}^n Y_i\right) = \frac{1}{2^n}.$$

• Beautiful solution! This would be the the proof that goes in “The Book.” Sep 16, 2023 at 16:33
• That's very interesting! Thanks for showing this other way of answering this... The following explanation is very helpful in understanding this: "The intuition is that if X>Y (with half chance), then X has a half chance (again) to be > Z". Sep 16, 2023 at 22:50

You have to be careful when there are multiple constraints. Observe that $$X$$ dominates $$Y+Z$$ and $$Y$$, so $$X$$ is free. After considering $$X>Y$$, we see that $$X-Y>Z>0$$, which means $$Z$$ is dominated by $$X-Y$$. \begin{align} &P(X > Y+Z, X > Y)\\ &=\int_0^\infty\int_0^x\int_0^{x-y}e^{-x-y-z}dzdydx\\ &=\int_0^\infty e^{-x}dx\int_0^x e^{-y}dy\int_0^{x-y}e^{-z}dz\\ &=\int_0^\infty e^{-x}dx\int_0^x(1-e^{y-x})e^{-y}dy\\ &=\int_0^\infty e^{-x}dx\int_0^x(e^{-y}-e^{-x})dy\\ &=\int_0^\infty e^{-x}dx\cdot(1-e^{-x}-xe^{-x})\\ &=\int_0^\infty e^{-x}dx-\int_0^\infty e^{-2x}dx-\int_0^\infty xe^{-2x}dx\\ &=1-\frac12+\frac12\int_0^\infty xd(e^{-2x})\\ &=\frac12+\frac12\left[xe^{-2x}|_0^\infty-\int_0^\infty e^{-2x}dx\right]\\ &=\frac14 \end{align}

• I tried working this out and I can solve it now! Thanks so much for clarifying the limits of the integral. That's very helpful. Sep 16, 2023 at 6:35