Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a question in understanding conditional distributions.

Let $X, Y, Z$ be independent and identically distributed (i.i.d.) random variables and let $U=X+Y$. Can I claim that the conditional distributions satisfy $f_{(U\mid X+Y<b,Z<a)}(u)=f_{(U\mid X+Y<b)}(u)$, where $a, b$ are constants and $f(\cdot)$ is the p.d.f.

If so please demonstrate how to arrive at that claim.

I tried the probability $P(U<u\mid X+Y<b,Z<a)=\frac{P(X+Y<u,X+Y<b,Z<a)}{P(X+Y<b,Z<a)}$, and then

$P(X+Y<b,Z<a)=P(X+Y<b)P(Z<a)$, (I'm not sure about this step)

$P(X+Y<u,X+Y<b,Z<a)=P(X+Y<u,X+Y<b\mid Z<a)\cdot P((Z<a)$. And from substitution and differentiation with respect to $u$ I prove the claim. But I'm not sure if it is correct.

Thank you.

share|improve this question
1  
What have you tried? –  Bunder Jun 10 '13 at 18:05
    
Could some one please comment if my answer is correct and also explain the step I have marked "not sure". Thanks –  triomphe Jun 10 '13 at 18:53
add comment

1 Answer 1

You can see this claim directly since $(U|X+Y,Z)=(U|X+Y)$ since you condition it with an independent random variable $Z$.

The following step is true $$P(X+Y<b,Z<a)=P(X+Y<b)P(Z<a).$$

In fact, you use the independence of the random variables $X+Y$ and $Z$.

This comes from the fact that if $X,Y,Z$ are independent, then the random variables $f(X,Y)$ and $Z$ are independent for any measurable function $f:\mathbb{R^2}\to\mathbb{R}$. Here you simply take $f(x,y)=x+y$.

More generally, if $(X_i^j)_{i=1\ldots n,j=1\ldots m}$ is an array of $nm$ independent random variables then for any sequence $(f_i)_{i=1\ldots n}$ of measurable functions $f_i:\mathbb{R^m}\to\mathbb{R}$, we obtain the independence of the sequence of random variables $(Y_i)_{i=1\ldots n}$ given by $$Y_i=f_i(X_i^1,\ldots,X_i^m)\qquad i=1\ldots n.$$

share|improve this answer
    
Hello. Does this mean my initial claim is correct? Thank you –  triomphe Jun 10 '13 at 21:35
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.