I'm trying to compute the conditional distribution of $X|Y = y$.

$X\sim Gamma(3,2)$

$Y|X = x \sim Weibull(2,x)$

I was doing this:

$f_{X|Y = y}(x) \propto f_x(x)\cdot f_{Y|X = x}(y)$

$f_x(x) = \frac{1}{\Gamma(\alpha)} \beta^{\alpha} x^{\alpha -1} e^{-\beta x}$

$f_x(x) \propto x^{\alpha-1}e^{-\beta x}$

$f_{Y|X = x}(y) = \alpha x y^{\alpha -1} e^{-x y^{\alpha}}$

$f_{Y|X = x}(y) \propto y^{\alpha -1}e^{-xy^\alpha}$

Using proportionality

$ f_{X|Y = y}(x) \propto x^{\alpha-1}e^{-\beta x} \cdot y^{\alpha -1}e^{-xy^\alpha}$

$ f_{X|Y = y}(x) \propto y^{\alpha -1} x^{\alpha-1} e^{-(\beta+y^\alpha)x}$

This looks like a Gamma distribution with the new $\beta = \beta+y^\alpha$ but I don't know how to compute the alpha.

Sorry If there's something obvious there that I'm not watching.

Could it be $Gamma(\alpha,\beta + y^\alpha)$ since $y^{\alpha-1}$ doesn't depend on $x$ and I'm using proportionality?


1 Answer 1


$X \sim Gamma(\alpha_g, \beta_g)$

$Y|X = x \sim W(\alpha_w, x)$

$f_x(x)\cdot f_{Y|X = x} (y) = \frac{1}{\Gamma(\alpha_g)}\beta_g^{\alpha_g}x^{\alpha_g-1}e^{-\beta_{g}x} \cdot \alpha_w xy^{\alpha_w -1}e^{-xy^{\alpha_w}}$

using proportionality

$f_{X|Y = y}(x) \propto x^{\alpha_g -1}\cdot x \cdot e^{-\beta_g x - xy^{\alpha_w}}$

$f_{X|Y = y}(x) \propto x^{(\alpha_g + 1)-1}\cdot e^{-(\beta_g + y^{\alpha_w})x}$

so $X|Y = y \sim Gamma(\alpha_g +1,\beta_g + y^{\alpha_w})$

Now we can replace for the initial values


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