Question

The number of automobile accidents a driver will be involved in during a one-year period is a random variable Y having a Poisson distribution with parameter X, where X depends on the driver. Suppose a driver is chosen at random from some population and hence X itself is a

continuous random variable with p.d.f f(x) , 0 < x < ∞

(a) Show that P(Y=k)= int o ^ infty P(Y=k|X=x)f(x)dx

(b) Suppose that x has an exponential distribution with mean 1/c where c is a

positive constant. Obtain the distribution of Y, hence find the expectation of Y.

Answer 1

$a:\\P\left( Y=k \right) =EI\left\{ Y=k \right\} =E\left( E\left( I\left\{ Y=k \right\} |X \right) \right) \\E\left( I\left\{ Y=k \right\} |X \right) =P\left( Y=k|X \right) =h\left( X \right) \\P\left( Y=k \right) =Eh\left( X \right) =\int_0^{+\infty}{h\left( x \right) f\left( x \right) dx}=\int_0^{+\infty}{P\left( Y=k|X=x \right) f\left( x \right) dx}\\b:\\P\left( Y=k \right) =\int_0^{+\infty}{\frac{x^ke^{-x}}{k!}\cdot ce^{-cx}dx}=\frac{c}{k!}\int_0^{+\infty}{x^ke^{-\left( c+1 \right) x}dx}=\\=\left[ \left( c+1 \right) x=y \right] =\frac{c}{\left( c+1 \right) ^{k+1}k!}\int_0^{+\infty}{y^ke^{-y}dy}=\frac{c}{\left( c+1 \right) ^{k+1}k!}\varGamma \left( k+1 \right) =\frac{c}{\left( c+1 \right) ^{k+1}}=\\=\left( 1-\frac{c}{c+1} \right) ^k\left( \frac{c}{c+1} \right) \sim Geom\left( \frac{c}{c+1} \right) \Rightarrow \\\Rightarrow EY=\frac{1-\frac{c}{c+1}}{\frac{c}{c+1}}=\frac{1}{c}$