Question

4. A lifetime X of an animal (counted in hours) has a Poisson distribution with parameter 4. The animal hunts during some part of its life, which is described by the following Bernoulli distribution: if X = k, then the ())'(£)*=0,1,2,. . . ,k. number Y of hours spent on hunting has the distribution (3 p.) Find the distribution of Y (7 p.) Suppose that the animal hunted for exactly 12 hours during its life; what is the most probable value of its lifetime? More precisely, find k> 0 for which P(X k|Y = 12 ) is the biggest possible. Hint: compare P(X= kIY = 12 ) and P (X = k+ 1|Y = 12)

Answer 1

$The~pmf~of~X~is\\ P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!};~x=0,1,2,3,.....\\ =0~otherwise\\ The~conditional~pmf~of~Y~given~X=k~is\\ P(Y=j|X=k)=\binom{k}{j}\left(\frac{1}{5} \right )^j\left(\frac{4}{5} \right )^{k-j};~j=0,1,2,....,k\\ =0~otherwise\\ P(Y=j)=\sum_{k=j}^\infty P(Y=j|X=k)P(X=k)\\ =\sum_{k=j}^\infty \binom{k}{j}\left(\frac{1}{5} \right )^j\left(\frac{4}{5} \right )^{k-j}\times \frac{e^{-\lambda}\lambda^k}{k!}=\frac{\lambda^je^{-\lambda}}{j!}\times \frac{1}{5^j}\sum_{k=j}^\infty \frac{\left(\frac{4\lambda}{5} \right )^{k-j}}{(k-j)!}\\ =\frac{\lambda^je^{-\lambda}}{j!}\times \frac{1}{5^j}\times e^{\frac{4\lambda}{5}}=\frac{\left(\frac{\lambda}{5} \right )^je^{-\frac{\lambda}{5}}}{j!};~j=0,1,2,.....\\ Hence~Y\sim Poisson\left(\frac{\lambda}{5} \right ).\\ Since~X=k|Y=12\sim Binomial(12,~1/5).\\ The~mode~of~binomial(12,~1/5)=[12/5]=2~([x]=greatest~integer~contained~in ~x)\\$