Suppose we have three iid draws from a uniform distribution on $[0,1]$. Call these random variables $A, B$ and $C$. Let $X=A+B$ and $Y=B+C$. I have figured out that the density of $X$ (or $Y$) is $$f_X(x) = \begin{cases} x &\mbox{ if } x \in [0,1] \\ 2-x &\mbox{ if } x \in (1,2]. \end{cases}$$ I can also see that $X=A+Y-C$. However, I am still struggling with the joint density of $X,Y$ and the conditional density of $X|Y$ and the corresponding CDFs. I am looking forward to hints!

Let me show you what I have tried and where I want to get at. Similar to $f_X$ above, I used convolution to obtain $$f_{Y|X} (y\,;\, x) = \begin{cases} 1+y-x &\mbox{ if } y \in [x-1,x] \\ 1-y+x &\mbox{ if } y \in (x,x+1]. \end{cases}$$ The joint density then should just be $f_{X,Y}(x,y) = f_X(x) f_{Y|X}(y\,;\, x)$.

My book suggests that the joint density looks like this and also that $\frac{f_{Y|X}(y\,;\, x)}{F_{Y|X}(y\,;\, x)}= \frac{2}y$. Neither coincides with what I have done. Can anyone help?

  • $\begingroup$ Since I am very certain about my marginal density and I suppose the book's joint density is correct, I assume my conditional density appears to be incorrect. I obtained it from the marginal density of $(A-C), f_{A-C}$ and I plugged in $f_{A-C}(y-x)$. If you don't like my concrete example, I would be very happy for a general approach to determine the joint density of random variables $X,Y$ with $X= g(A,B)$ and $Y=g(C,B)$ with independent draws $A,B,C$ and some function $g$. $\endgroup$ – Max Mar 27 at 12:44
  • $\begingroup$ "... and also that $\frac{f(x|x)}{F(x|x)}= $" What on earth does that $f(x|x)$ mean? TYpo? $\endgroup$ – leonbloy Mar 27 at 14:27
  • $\begingroup$ Yes, a typo. I apoligize. I edited it. It is supposed to be the conditional density over the conditional CDF evaluated at $x$. $\endgroup$ – Max Mar 27 at 14:30

Use the same method used to find the marginal.

$$\begin{align} f_{X,Y}(x,y)&=\int_0^1 f_{A,B,C}(x-b,b,y-b)~\mathrm d b \\[2ex]&=\int_{\max\{0,x-1,y-1\}}^{\min\{1,x,y\}}\mathbf 1_{0\leqslant x\leqslant 2,\max\{0,x-1\}\leqslant y\leqslant\min\{2,x+1\}}\mathrm d b \\[2ex]&\ddots \end{align}$$


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