By Prof.@Did thought https://math.stackexchange.com/a/490550/91036 as following below:

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

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 $U$ conditionally on $V=v$.

If i have a case:

From the pdf:

$$f(x,y)= \frac{\beta^{\alpha+\gamma}x^{\alpha-1} (y-x)^{\gamma-1}\exp^{-\beta y}}{\Gamma(\alpha) \Gamma(\gamma)}; 0<x<y<\infty$$

and marginal pdf of $X$ with

$$f_X(x)= \frac{\beta^{\alpha}x^{\alpha-1}\exp^{-\beta x}}{\Gamma(\alpha) }; 0<x<\infty$$

and marginal pdf of $Y$ with

$$f_Y(y)= \frac{\beta^{\alpha+\gamma}y^{\alpha+\gamma-1}\exp^{-\beta y}}{\Gamma(\alpha+\gamma) }; 0<y<\infty$$

conditional pdf of $X$ given $Y=y$

$$f_{X|Y}(x|y)= \frac{x^{\alpha-1}(y-x)^{\alpha-1}\Gamma(\alpha+\gamma)}{y^{\alpha+\gamma-1}\Gamma(\alpha)\Gamma(\gamma) }; 0<x<\infty$$

conditional pdf of $Y$ given $X=x$

$$f_{Y|X}(y|x)= \frac{\beta^{\gamma}(y-x)^{\gamma-1}\exp^{-\beta y}}{\Gamma(\gamma)\exp^{-\beta x} }; 0<y<\infty$$

My Question:

  1. How to applied (@Did thought) formula on Conditional pdf of $Y-X$ given $X=x$ by using cases above?.
  2. Then, how about (@Did thought) formula's on Conditional distribution of $X/Y$ given $Y=y$ by using cases above?.

Thanks in advanced!

  • $\begingroup$ Well... apply the formulas in my post. (What is your question, actually?) $\endgroup$ – Did Sep 11 '13 at 16:32

@Mary, thanks for your participate in answer, anyway may be you have made a mistake in your answer, let me show:

Conditional P.D.F. for $Z=Y-X$, which is given by $X=x$

$$g(z)= \frac{\beta^{\gamma}(z)^{\gamma-1}\exp^{-\beta z}}{\Gamma(\gamma)}; 0<(z+x)<\infty$$

Conditional Distribution for $T=X/Y$, which is given by $Y=y$

$$h(t)=|y|\frac {(ty)^{\alpha-1}(y-ty)^{\gamma-1}\Gamma^(\alpha+\gamma)}{y^{\alpha+\gamma-1} \Gamma(\alpha) \Gamma(\gamma)};0<ty<\infty$$

  • $\begingroup$ @Did how do you think about my correction to answer? is it OK? $\endgroup$ – user91036 Sep 12 '13 at 13:02

For Question 1:

$$g(z)= \frac{\beta^{\gamma}((z+x)-x)^{\gamma-1}\exp^{-\beta (z+x)}}{\Gamma(\gamma)\exp^{-\beta x} }; 0<(z+x)<\infty$$

For Question 2:

$$h(t)=|y|\,\frac {ty^{\alpha-1}(y-ty)^{\gamma-1}\Gamma^(\alpha+\gamma)}{\gamma^{\alpha+\gamma-1} \Gamma(\alpha) \Gamma(\gamma)};0<x<\infty$$

  • 1
    $\begingroup$ First formula correct but can be hugely simplified. Second formula wrong. Two reasons to accept answer after 9 minutes, I guess. // @lpchristine: What are you doing exactly? $\endgroup$ – Did Sep 11 '13 at 17:12

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