Exponential random variable with parameter that is another exponential random variable.

Question

$X \sim Exp(\Lambda)$, and $\Lambda \sim Exp(\beta)$, find the conditional density for $\Lambda$ given $X = x$. This distribution should be a named distribution.

I am confused as to how to find this density. I was thinking simply:

$$ f_{\Lambda | X}\left( \lambda | x\right) = \frac{f_{\Lambda , X}\left( \lambda , x\right)}{f_{X}\left(x\right)}$$

But, I am unsure as to how to find the joint density. What is the best way to find the joint density for continuous random variables in general? If these are independent, would someone be able to explain why, because I cannot convince myself that they are since one is the parameter for the other? If so the joint density would simply be the product.

Answer 1

$X\sim\mathcal{Exp}(\Lambda)$ means $f_{X\mid\Lambda}(x\mid \lambda)=\lambda\mathsf e^{-\lambda x}\mathbf 1_{0\leqslant \lambda, 0\leqslant x}$

$\Lambda\sim\mathcal{Exp}(\beta)$ means $f_{\Lambda}(\lambda)=\beta\mathsf e^{-\beta\lambda}\mathbf 1_{0\leqslant \lambda}$

Also

• $f_{X,\Lambda}(x,\lambda)=f_{X\mid\Lambda}(x\mid\lambda)\,f_{\Lambda}(\lambda)$
• $f_{X}(x)=\int_\Bbb R f_{X,\Lambda}(x,\ell)~\mathrm d\ell$

Therefore:

$\qquad\begin{align}f_{\Lambda\mid X}(\lambda\mid x)&=\dfrac{f_{X,\Lambda}(x,\lambda)}{f_X(x)}\\[1ex]&=\dfrac{f_{X\mid\Lambda}(x\mid\lambda)~f_{\Lambda}(\lambda)}{\int_\Bbb R f_{X\mid\Lambda}(x\mid\ell)~f_{\Lambda}(\ell)\,\mathrm d\ell}\\[1ex]&=\dfrac{\lambda\beta\mathsf e^{-\lambda (x+\beta)}}{\int_0^\infty \ell\beta\mathsf e^{-\ell (x+\beta)}~\mathrm d\ell}\mathbf 1_{0\leqslant \lambda, 0\leqslant x}\\[1ex]&=\phantom{\dfrac{(x+\beta)^2\lambda\beta\mathsf e^{-\lambda(x+\beta)}}{\beta}\mathbf 1_{0\leqslant \lambda, 0\leqslant x}}\end{align}$