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$X \sim Exp(\Lambda)$, 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.

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    $\begingroup$ Is it possible to use Bayes rule and say $f_{\Lambda | X}\left(\lambda|x\right) = \frac{f_{ X|\Lambda }\left(x|\lambda\right) f_{\Lambda}(\lambda)}{f_X(x)}$ ? $\endgroup$
    – Eoin S
    Apr 2, 2020 at 23:44

1 Answer 1

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$X\sim\mathcal{Exp}(\Lambda)$ 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:

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

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  • $\begingroup$ I understand what is being done here but I am unable to find the named distribution that this becomes. The resulting density I obtain is $(x+\beta)^{2}\beta \lambda e^{-(x+\beta)\lambda}$. I thought this sort of looks like the form of the Chi Square distribution, but I could not think of the parameters that make it so. Do you think I did something wrong in my calculation? $\endgroup$
    – Eoin S
    Apr 3, 2020 at 20:09

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