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problems

I think it is easy to find $p_{S}(s)$ in the first question, which is the convolution of $p_{X}(x)$ and $p_{Y}(y)$.

In the second question, I try to use Jacobian to find out $p_{S,X}(s,x)$ by adding auxiliary variable. But how to deal with the intervals when evaluating $p_{S,X}(s,x) = \frac{p_{X,Y}(w,s-w)}{\lvert J(x,y)\rvert}$, where $w=x$. Since variable $w$ needs to be in $[0,1]$ and $s-w$ to be in $[0,1]$ but $s$ to be in $[0,2]$, I am in trouble with the expression of $p_{X,Y}(w,s-w)$.

There's obviously something I don't understand well, please tell me where my mistake is. Any help would be great! Thanks!

1 Answers 1

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Finding the conditional density directly:

For $0 \le x \le \min(1,s)$, $$p_{X \mid S=s}(x \mid S=s) = \frac{p_X(x)\cdot p_Y(s-x)}{\int_0^{\min(1,s)} p_X(x') \cdot p_Y(s-x') \mathop{dx'}} = \frac{1}{\min(1,s)}.$$

Finding the joint density:

Let $g(x,y) = (x,x+y)$. Then $g^{-1}(x,s) = (x,s-x)$ and its Jacobian is $$\begin{bmatrix}1 & 0 \\ -1 & 1\end{bmatrix}$$ which has determinant $1$. So, $$p_{X,S}(x,s) = p_X(x) \cdot p_Y(s-x) = \begin{cases} 1 & x \in [0,1], s-x \in [0,1] \\ 0 & \text{otherwise}.\end{cases}= \begin{cases} 1 & s \in [0,2], x \in [0,\min(1,s)] \\ 0 & \text{otherwise}.\end{cases}$$

Computing $$p_{X \mid S=s} (x \mid S=s) = \frac{p_{X,S}(x,s)}{p_S(s)}=\frac{p_{X,S}(x,s)}{\int p_{X,S}(x',s)\mathop{dx'}}$$ is the first line of my answer.

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    Thanks for your help! But I'm not sure how you transform x∈[0,1],s−x∈[0,1] to s∈[0,1],x∈[0,min(1,s)]. Could you please explain in detail? – 2017-01-12
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    @AracWu Sorry, I made a typo, it should be $s \in [0,2]$. Anyway, it's just another way of describing the set of valid $(x,s)$ pairs, which forms a parallelogram if you plot it on a plane. Drawing a picture may help. – 2017-01-12