Convolution and Probability Distributions

In summary: So in summary, we can use the symmetry of the problem and the surprising result that for iid continuous random variables X1 and X2, P(X1+X2≤t|X1<X2)=P(X1+X2≤t|X1>X2)=P(X1+X2≤t) to simplify the calculation of E[X1+X2|X1<X2]. This leads to the conclusion that E[X1+X2]=2E[X1].
  • #1
O_o
32
3

Homework Statement


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

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

Homework Equations

The Attempt at a Solution


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

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 [tex] Z = X_1 + X_2 \gt 2X_1 \gt X_1[/tex] So now if I can do somthing similar for X1 < 0 I can evaluate the integral.
 
Last edited:
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  • #2
It is possible to use an integral, but you can use the symmetry of the problem. Consider $$E[X_1 + X_2 | X_1 > X_2]$$
 
  • #3
mfb said:
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).
 
  • #4
Thanks guys. I hadn't considered the symmetry of the problem. Does this look alright:

[tex] 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]
[/tex]

So now I can solve for [tex] E[X_1 + X_2] = 2E[X_1][/tex] instead to get my answer, which looks easier.
 
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  • #5
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.
 
  • #6
o_O said:
Thanks guys. I hadn't considered the symmetry of the problem. Does this look alright:

[tex] 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]
[/tex]

So now I can solve for [tex] E[X_1 + X_2] = 2E[X_1][/tex] 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
[tex]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) [/tex]
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!
 
  • #7
Ray Vickson said:
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
[tex]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) [/tex]
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.
 

1. What is convolution in probability distributions?

Convolution is a mathematical operation that combines two probability distributions to produce a third distribution. It is used to find the probability of the sum of two random variables, where the random variables are independent and identically distributed.

2. Why is convolution important in probability and statistics?

Convolution is important because it allows us to model the behavior of complex systems by breaking them down into simpler components. It is also used to calculate the probability of events that depend on multiple random variables, such as the sum or product of those variables.

3. How is convolution related to the central limit theorem?

The central limit theorem states that the sum of a large number of independent and identically distributed random variables will be approximately normally distributed. Convolution is used to calculate the probability distribution of the sum of these random variables, making it a key tool in understanding and applying the central limit theorem.

4. Can convolution be applied to non-probabilistic functions?

Yes, convolution can be applied to any two functions, not just probability distributions. It is commonly used in signal processing and image processing to combine two functions to produce a third function.

5. How does convolution differ from correlation?

While convolution and correlation are both mathematical operations used to combine two functions, they differ in the way they weight the values of each function. Convolution weights the values of the two functions being combined, while correlation weights the values of one function against the other. Additionally, convolution is commutative (the order of the functions being combined does not matter), while correlation is not.

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