Let X be an exponential random variable with parameter Λ, which is itself a random variable with distribution Exponential(β). Find a formula for the conditional density of Λ given X=x, and identify its type (i.e. its name and parameter(s)).
I made some good progress but then got stuck. I started with Bayes' Rule, and wrote
$$ f(Λ|x)=\frac {f(x|Λ)f(Λ)}{f(x)} $$ Now the numerator expressions seemed simple enough. $f(x|Λ) = Λe^{-Λx}$, and $f(Λ) = βe^{-βΛ}$. I'm having a hard time finding an expression for $f(x)$. I thought about integrating over the possible values of Λ: $$ f(x) = \int_o^\inftyΛe^{-Λx}dΛ$$ but I know this is incorrect. How to approach?
Method 1) $$ f(\lambda |x)=\frac {f(x|\lambda)f(\lambda)}{f(x)} \propto f(x|\lambda)f(\lambda)$$
$$=\lambda e^{-\lambda x} \beta e^{-\beta \lambda} \propto \lambda e^{-\lambda (x+\beta)} $$
so $\lambda \sim \Gamma(2,x+\beta)$ Gamma_distribution
$$f(\lambda|x)=\frac{(x+\beta)^2}{\Gamma(2)}\lambda^{2-1} e^{-\lambda (x+\beta)}$$
Method 2)
$$ f(\lambda |x)=\frac {\lambda e^{-\lambda x} \beta e^{-\beta \lambda}}{f(x)} $$
$$=\frac {\beta \lambda e^{-\lambda (x+\beta)} }{\beta \frac{\Gamma(2)}{(x+\beta)^2}} $$
$$=\frac{(x+\beta)^2}{\Gamma(2)}\lambda^{2-1} e^{-\lambda (x+\beta)}$$
Sine $$f(x)=\int f(x,\lambda) d\lambda=\int f(x|\lambda)f(\lambda) d\lambda$$
$$=\int \beta \lambda e^{-\lambda (x+\beta)} d\lambda$$ $$=\beta\int \lambda^{2-1} e^{-\lambda (x+\beta)} d\lambda$$
$$=\beta \frac{\Gamma(2)}{(x+\beta)^2}$$