I've been staring at this example for hours now and I can't seem to understand it. This is an example from Marcel Finan's C book.
The distribution of $X|\Lambda$ is exponential with parameter $\dfrac{1}{\Lambda}$. The distribution of $\Lambda$ is Gamma with parameters $\alpha$ and $\theta$. Find $f_X(x)$
Solution: First I try to slice things up:
Defn: $X$ is a mixture distribution if its pdf is $$f_x(x) = \int_{-\infty}^{\infty} f_{X|\Lambda}(x,\lambda)f_{\Lambda}(\lambda)d\lambda$$ where $\Lambda$ is a continous random variable.
Also
$\bullet \Lambda$~Gamma($\alpha,\theta$) if its pdf is $$f(\lambda) = \dfrac{\lambda^{\alpha-1}e^{\frac{\lambda}{\theta}}}{\Gamma(\alpha)\theta^\alpha}$$ for $\lambda\geq0$ and $0$ otherwise.
But Finan's solution kinda messes me up:
$$f_X(x) = \int_{0}^{\infty} \lambda e^{-\lambda x}\dfrac{\theta^{\alpha}}{\Gamma(\alpha)} \lambda^{\alpha-1}e^{-\lambda\theta}d\lambda = \dfrac{\theta^{\alpha}}{\Gamma(\alpha)} \int_{0}^{\infty} \lambda^{\alpha} e^{-\lambda(x+\theta)}d\lambda = \dfrac{\theta^{\alpha}}{\Gamma(\alpha)} \dfrac{\Gamma(\alpha+1)}{(x+\theta)^{\alpha+1}}$$
My questions are:
1.) What does $X|\Lambda$ having exponential distribution looks like?
2.) Also, how did $\theta^{\alpha}$ became the numerator on the first equation when in the Gamma distribution, its on the denominator?
3.) Can you please make a clarification on how did Finan arrive on the first equation? I can't seem to make a distinguish on the conditional distribution and the marginal density of $\Lambda$ on the definition of Mixture distribution.
4.) And finally, how is $$\int_{0}^{\infty} \lambda^{\alpha} e^{-\lambda(x+\theta)}d\lambda= \dfrac{\Gamma(\alpha+1)}{(x+\theta)^{\alpha+1}}$$ Is this by the Gamma function?
$f_{X|\Lambda}(x,\lambda) = \lambda e^{\lambda x}$. Here we use the rate parameter to write the p.d.f. of exponential distribution. We call $\lambda$ rate because $E[X|\Lambda] = \lambda^{-1}$ gets larger when $\lambda$ gets smaller. You may also write the p.d.f. by the scale parameter $\beta = \lambda^{-1}$. You can check this on Wikipedia.
Also, if you look at Wikipedia, you can find that there are two forms of the density function of Gamma distribution, one is given by shape-rate parameters and the other is given by shape-scale parameters. Since we use the rate parameter for the exponential distribution, we should use the shape-rate form for Gamma distribution, given by $$f_{\Lambda}(\lambda) = {\frac {\theta ^{\alpha }}{\Gamma (\alpha )}}x^{\alpha -1}e^{-\theta x}.$$
Now we are able to get the first equation. By simply plugging in the expressions into $f_X(x)$ $$f_X(x) = \int_{-\infty}^{\infty} f_{X|\Lambda}(x,\lambda)f_{\Lambda}(\lambda)d\lambda = \int_{0}^{\infty} \lambda e^{-\lambda x}\dfrac{\theta^{\alpha}}{\Gamma(\alpha)} \lambda^{\alpha-1}e^{-\lambda\theta}d\lambda$$
Recall the definition of $\Gamma(t+1) = \int_{0}^{\infty} x^{t} e^{-x}d{x}$. Now let $s = \lambda(x+\theta)$, then $$\int_{0}^{\infty} \lambda^{\alpha} e^{-\lambda(x+\theta)}d\lambda= \frac{1}{(x+\theta)^{\alpha+1}}\int_{0}^{\infty} s^{\alpha} e^{-s}ds = \dfrac{\Gamma(\alpha+1)}{(x+\theta)^{\alpha+1}}.$$