Convolution and Probability Distributions

[O_o]

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]

Homework Equations

The Attempt at a Solution

So what I'm trying to do is create a new random variableZ = X_1 + X_2 When I do this I get the following convolution formula for its densityg(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.

 

[mfb]

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]

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).

 

[O_o]

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

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

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

 

[mfb]

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]

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

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

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 &lt; X_2)\\<br /> = P(X_1+X_2 \leq t \,|\, X_1 &gt; X_2) \\<br /> = 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!

 

[O_o]

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 &lt; X_2)\\<br /> = P(X_1+X_2 \leq t \,|\, X_1 &gt; X_2) \\<br /> = 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.