An Introduction to Probability and Statistical Inference, 2nd Edition

2.34 Let the r.v. Y ~ P(λ), and suppose that the conditional distribution of the r.v. X, given Y = y, is B(y, p); i.e., XǀY = y ~ B(y, p). Then show that the (unconditional) p.d.f. of X is P(λp); i.e., X ~ P(λp).Hint. For x = 0, 1, …, write $P(X=x)={\displaystyle {\sum }_{y=0}^{\infty }P(X=x,\text{?}Y=y)}={\displaystyle {\sum }_{y=0}^{\infty }P(X=x|Y=y)P(Y=y)}$, use what is given, recall that $\left(\begin{array}{c}y\\ x\end{array}\right)=0$ for y < x, and that ${\displaystyle {\sum }_{k=0}^{\infty }\frac{u^k}{k!}}={\text{e}}^u$.

2.35 Consider the r.v.'s X and N, where X ~ B(n, p) and N ~ P(λ). Next, let X(k) be the r.v. ...