# Conditional distribution using Gamma and Weibull

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?

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

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

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

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