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.

So in this problem I have two i.i.d random variables $X$ and $Y$, which are uniformly distributed over interval $[0,1]$.

I know that $S = X + Y$. To find the density of $S$, I find the convolution of $X$ and $Y$. However I am struggling conceptually to understand what $p_{X \vert S}(x \vert s)$ is, for a given real value of $S$.

The conflict is that $X$ is defined as a uniform over [0,1], so it seems that the pdf should remain just that. However clearly if we condition on $S$ being 0.5 for example, then $X$ cannot sample values greater than that.

I am inclined to say

$p_{X \vert S}(x \vert s) = \begin{cases} \frac{1}{s} & \mathrm{for}\ 0 \le x \le s, \\[8pt] 0 & \mathrm{otherwise}\ \end{cases} $

But this is clearly wrong if s > 2, or s < 0, and I am not sure it is correct. Could someone offer advice on how $p_{x \vert S}(x \vert s)$ would be derived?

share|improve this question
    
Thanks, could you clarify how to go from the random point on X,Y to the conditional density. –  zenna Feb 12 '12 at 23:53
    
Thanks, if you care to put this as an answer I can accept. –  zenna Feb 13 '12 at 20:12
    
OK, here goes. I am deleting my comments. –  Dilip Sarwate Feb 13 '12 at 20:52
add comment

1 Answer

up vote 2 down vote accepted
  • Given $S = X+Y = s$ where $0 \leq s \leq 1$, the random point $(X,Y)$ is constrained to be somewhere on the straight-line segment with endpoints $(0,s)$ and $(s,0)$.

  • Given $S = X+Y = s$ where $1 < s \leq 2$, the random point $(X,Y)$ is constrained to be somewhere on the straight-line segment with endpoints $(1,s-1)$ and $(s-1,1)$.

In both cases, $(X,Y)$ is uniformly distributed on the line segment. Note that conditioned on $S$, $X$ and $Y$ are both continuous random variables uniformly distributed on $[0,s]$ (or $[s-1,1]$ as the case may be), but they are not jointly continuous; they do not have a joint density function. In fact, given $S=X+Y=s$, we have that $Y = s-X$ and the two random variables are conditionally linearly related.

So your conditional density $p_{X|S}(x|s) = 1/s$ for $0 \leq x \leq s$ is correct for $0 \leq s \leq 1$, but not for $1 < s \leq 2$. The conditional density is undefined for $s < 0$ or $ > 2$ because $S$ cannot take on values outside the interval $[0,2]$.

share|improve this answer
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.