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Suppose I have a joint distribution with PDF $f(x,y)$, then the marginal PDF of $X$ is $f_X(x)$ and the marginal PDF of $Y$ is $f_Y(y)$.

How are the formula to determine Conditional PDF of $(Y-X)$ given $X=x$ and Conditional distribution of $(X/Y)$ given $Y=y$?

Thanks in Advanced...

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The conditional PDF $g$ of $Z=Y-X$ conditionally on $X=x$ is such that, for every $z$, $g(z)=f(x,z+x)/f_X(x)$.

The conditional PDF $h$ of $T=X/Y$ conditionally on $Y=y$ is such that, for every $t$, $h(t)=|y|\,f(ty,y)/f_Y(y)$.

Which part of this causes you trouble?

Note: Formally, the conditional distribution of some $U$ conditionally on $V=v$ is some distribution, that is, a probability measure $\mu$. If $\mu$ has a density $m$, that is, if, for every $B$, $P(U\in B\mid V=v)=\mu(B)=\int\limits_Bm(u)\mathrm du$, then one says that $m$ is the conditional density of $U$ conditionally on $V=v$.

Edit: To compute the conditional PDF $h$, note that, by definition, for every measurable function $u$, $$ E[u(X,Y)\mid Y=y]=\int u(x,y)f(x,y)\mathrm dx/f_Y(y). $$ Using this for $u(x,y)=v(x/y)$ yields $$ E[v(T)\mid Y=y]=\int v(x/y)f(x,y)\mathrm dx/f_Y(y). $$ The change of variable $t=x/y$ yields $\mathrm dx=y\mathrm dt$ hence $$ E[v(T)\mid Y=y]=\int v(t)f(ty,y)(|y|\mathrm dt)/f_Y(y), $$ which yields the conditional density $h$.

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    $\begingroup$ Hi @Did thanks for your advanced, I got the ideas..many thanks $\endgroup$ – user91036 Sep 11 '13 at 13:20
  • $\begingroup$ why in $h(t)$ there is $|y|$ ? $\endgroup$ – user91036 Sep 12 '13 at 10:53
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    $\begingroup$ See Edit. $ $ $ $ $\endgroup$ – Did Sep 12 '13 at 11:17

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