# Convolution and Probability Distributions

## Homework Statement

Have 2 iid random variables following the distribution $$f(x) = \frac{\lambda}{2}e^{-\lambda |x|}, x \in\mathbb{R}$$

I'm asked to solve for $$E[X_1 + X_2 | X_1 < X_2]$$

## The Attempt at a Solution

So what I'm trying to do is create a new random variable$$Z = X_1 + X_2$$ When I do this I get the following convolution formula for its density$$g(z) = \int_{-\infty}^{\infty} \frac{\lambda^2}{4} e^{-\lambda |z- x_1|} e^{-\lambda |x_1|} dx_1$$

I'd really only like some advice on how to go about attacking this integral. It looks to me like I need to break it down into cases depending on z<x1 or z>x1 but that doesn't seem like it will produce a clean solution to me.

Or if you can see that I'm attacking this problem completely the wrong way and I shouldn't even be trying to do this please let me know. No alternative method of attack needed. I can try to figure out other ways if this is completely off base.

Thanks

edit:
I've had a thought. If X1 > 0 then $$Z = X_1 + X_2 \gt 2X_1 \gt X_1$$ So now if I can do somthing similar for X1 < 0 I can evaluate the integral.

Last edited:

mfb
Mentor
It is possible to use an integral, but you can use the symmetry of the problem. Consider $$E[X_1 + X_2 | X_1 > X_2]$$

haruspex
Homework Helper
Gold Member
It is possible to use an integral, but you can use the symmetry of the problem. Consider $$E[X_1 + X_2 | X_1 > X_2]$$
I don't see how that gets one out of doing the integral. But certainly it is worth considering symmetries. Forgetting the (irrelevant) X1<X2, there are six orderings of 0, z, x1. Symmetries get it down to only two integrals. E.g. conside g(-z).

Thanks guys. I hadn't considered the symmetry of the problem. Does this look alright:

$$E[X_1 + X_2]\\ = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} (x_1 + x_2) f_{X_1,X_2}(x_1, x_2) dx_1 dx_2 \\ = \int_{-\infty}^{\infty}\int_{-\infty}^{x_2} (x_1 + x_2)f_{X_1,X_2}(x_1, x_2) dx_1 dx_2 + \int_{-\infty}^{\infty}\int_{x_2}^{\infty} (x_1 + x_2) f_{X_1,X_2}(x_1, x_2) dx_1 dx_2 \\ = P(X_1 < X_2) \int_{-\infty}^{\infty}\int_{-\infty}^{x_2} (x_1 + x_2) \frac{f_{X_1,X_2}(x_1, x_2)}{P(X_1 < X_2)} dx_1 dx_2 + P(X_2 < X_1)\int_{-\infty}^{\infty}\int_{x_2}^{\infty} (x_1 + x_2) \frac{f_{X_1,X_2}(x_1, x_2)}{P(X_2 < X_1)} dx_1 dx_2 \\ = P(X_1 < X_2) E[X_1 + X_2 | X_1 < X_2] + P(X2 < X1) E[X_1 + X_2 | X_2 < X_1] \\ = \frac{1}{2} \left(E[X_1 + X_2 | X_1 < X_2] + E[X_1 + X_2| X_2 < X_1] \right) \\ = E[X_1 + X_2 | X_1 < X_2]$$

So now I can solve for $$E[X_1 + X_2] = 2E[X_1]$$ instead to get my answer, which looks easier.

Last edited:
mfb
Mentor
I think you can skip the integration steps because ##P(X_1 < X_2) = P(X_1 > X_2) = \frac{1}{2}## follows from symmetry and the formula where it appears is simply the weighted average, but it looks possible and the result is right.

Ray Vickson
Homework Helper
Dearly Missed
Thanks guys. I hadn't considered the symmetry of the problem. Does this look alright:

$$E[X_1 + X_2]\\ = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} (x_1 + x_2) f_{X_1,X_2}(x_1, x_2) dx_1 dx_2 \\ = \int_{-\infty}^{\infty}\int_{-\infty}^{x_2} (x_1 + x_2)f_{X_1,X_2}(x_1, x_2) dx_1 dx_2 + \int_{-\infty}^{\infty}\int_{x_2}^{\infty} (x_1 + x_2) f_{X_1,X_2}(x_1, x_2) dx_1 dx_2 \\ = P(X_1 < X_2) \int_{-\infty}^{\infty}\int_{-\infty}^{x_2} (x_1 + x_2) \frac{f_{X_1,X_2}(x_1, x_2)}{P(X_1 < X_2)} dx_1 dx_2 + P(X_2 < X_1)\int_{-\infty}^{\infty}\int_{x_2}^{\infty} (x_1 + x_2) \frac{f_{X_1,X_2}(x_1, x_2)}{P(X_2 < X_1)} dx_1 dx_2 \\ = P(X_1 < X_2) E[X_1 + X_2 | X_1 < X_2] + P(X2 < X1) E[X_1 + X_2 | X_2 < X_1] \\ = \frac{1}{2} \left(E[X_1 + X_2 | X_1 < X_2] + E[X_1 + X_2| X_2 < X_1] \right) \\ = E[X_1 + X_2 | X_1 < X_2]$$

So now I can solve for $$E[X_1 + X_2] = 2E[X_1]$$ instead to get my answer, which looks easier.

Even easier: you can use the surprising result that for iid continuous ##X_1, X_2## and for all ##t \in \mathbb{R}## we have
$$P(X_1+X_2 \leq t \,|\, X_1 < X_2)\\ = P(X_1+X_2 \leq t \,|\, X_1 > X_2) \\ = P(X_1 + X_2 \leq t)$$
In other words, the random variables ##X_1 + X_2##, ##[X_1 + X_2 | X_1 < X_2]## and ##[X_1+X_2|X_1 > X_2]## all have the same distribution, hence the same expectation!

Even easier: you can use the surprising result that for iid continuous ##X_1, X_2## and for all ##t \in \mathbb{R}## we have
$$P(X_1+X_2 \leq t \,|\, X_1 < X_2)\\ = P(X_1+X_2 \leq t \,|\, X_1 > X_2) \\ = P(X_1 + X_2 \leq t)$$
In other words, the random variables ##X_1 + X_2##, ##[X_1 + X_2 | X_1 < X_2]## and ##[X_1+X_2|X_1 > X_2]## all have the same distribution, hence the same expectation!
That's really neat, thanks for sharing.