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This is from Exercise 6.15 in Canavos' Applied Probability and Statistical Methods. I cannot get my result to match answer given at the end of the book.

Given a Gamma distribution with shape and scale factors equal 2, what is the non-conditional Poisson probability distribution given one that is conditioned on the former Gamma? I tried the following:

$$ f_\gamma(\lambda, 2, 2) = \frac{1}{2}\lambda \mathrm{e}^{-\lambda/2} \\ p(x | \lambda) = \frac{\mathrm{e}^{-\lambda}\lambda^x}{x!} \\ p(x) = \int_\Lambda p(x | \lambda) \ f_\gamma(\lambda) \ d\lambda \\ p(x) = \frac{1}{2x!} \int_0^\infty \lambda^{x+1}\ \mathrm{e}^{-3\lambda/2} \ d\lambda $$ Then I do the substitution $v= 3\lambda/2, \ dv = 3d\lambda/2$ to get:

$$ p(x) = \frac{1}{2x!}\left(\frac{2}{3}\right)^{x+1} \int_0^\infty v^{x+1}\ \mathrm{e}^{-v}\ dv \\ p(x) = \frac{1}{2x!}\left(\frac{2}{3}\right)^{x+1} \Gamma(x+2) $$

The problem asks for $P(X \le 2) $, which I obtain adding $p(0) + p(1) + p(2)$, but I don't get $11/27$, which is the solution given at the end of the book.

What is wrong?

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    $\begingroup$ Note that when $x$ is an integer, $\Gamma(x+2)=(x+1)!$ hence $$p(x) = \frac{1}{2}\left(\frac{2}{3}\right)^{x+1} (x+1) $$ In particular, $$p(0)=\frac13\qquad p(1)=p(2)=\frac49$$ (Your question does not allow to know where you went wrng, though.) $\endgroup$
    – Did
    Commented May 14, 2017 at 12:11

1 Answer 1

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$f_{\gamma}(\lambda,2,2)=\frac{1}{4}\lambda e^{-\lambda/2}$. This is the distribution. You have written a wrong distribution. Also check the integration. Rest is correct.

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