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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?

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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
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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 $[0,2]$.

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