I have to solve this exercise: D denotes the demand for a certain product. D is stochastic, its distribution depends on the price P. Assume that D | (P = p) ∼ Γ(1, p/10) and P ∼ Γ(2, 1). Derive the distribution of D, compute P(D ≤ 15), find the conditional distribution of the price given D = d. Compute P (P > 2) and P (P > 2 | D = 30).
So i computed the density of distribution of the vector (X, Y): which is f(p, d) = 1/10 * p^2 * e^-p*(1 + d/10)
But now i am absolutely clueless how to integrate this whole thing. Am i mistaken at any point? I am unsure. I would be thankful for every hint.
Greetings, Jim
The distribution of the price $P$ is gamma, and the conditional distribution of the demand $D$ given the price is also gamma (actually, exponential). So the marginal or unconditional distribution of the demand is found by the formula $$f_D(d) = \int_{p=0}^\infty f_{D \mid P}(d \mid p) f_P(p) \, dp,$$ where $$f_{D \mid P}(d \mid p) = \frac{p}{10} e^{-(p/10)d}, \quad d > 0,$$ and $$f_P(p) = pe^{-p}, \quad p > 0.$$ Thus your integral is $$f_D(d) = \int_{p=0}^\infty \frac{p^2}{10} e^{-(d/10 + 1)p} \, dp,$$ as you correctly noted. We can perform the integration by observing that the PDF of a gamma random variable $X$ with shape $a$ and rate $b$ is $$f_X(x) = \frac{b^a x^{a-1} e^{-bx}}{\Gamma(a)}, \quad x > 0,$$ hence the integrand is proportional to a gamma density with shape $a = 3$ and rate $b = 1 + d/10$. It follows that we need a constant of proportionality: $$f_D(d) = \frac{\Gamma(3)}{10(1 + d/10)^3} \int_{p=0}^\infty \frac{(1+d/10)^3 p^2 e^{-(1+d/10) p}}{\Gamma(3)} \, dp = \frac{1}{5(1+d/10)^3} = \frac{200}{(d+10)^3}, \quad d > 0.$$ This is a Pareto distribution.