Variance of Poisson distribution of two variables

Question

I was doing practice questions of probability. I was able to do the first 2 parts of this question but I don't understand how to go about the last part.

An instructor gets her students A and B to type out his research papers. Both students make errors while typing according to a Poisson distribution. A makes errors at the rate of j per page, while B makes errors at the rate of k per page. Let the random variable F be the number of errors on a randomly chosen page.

a) If the students each do half of the total typing, what is the PMF of F?

b) What is the PMF of F if B types 70% of the pages?

c) What is var(F) for the PMF in part (a)?

My answers for the first two parts:

F = number of erros on a randomly chosen page

a) 0.5*(exp(-j) . (j) ^ f)/ f! + 0.5*(exp(-k) . (k) ^ f)/ f!

b) 0.3*(exp(-j) . (j) ^ f)/ f! + 0.7*(exp(-k) . (k) ^ f)/ f!

Answer 1

Your answers for the first two parts are correct. As for the third part, the random variable for the error on a page would be:

$Z = \frac12 X\space + \frac12 Y$

so,

$Var(Z) = (\frac12)^2 *(k+j)$

Answer 2

Let rv $X$ have Poisson distribution with parameter $\lambda=j$ and let rv $Y$ have Poisson distribution with parameter $\lambda=k$.

Let $B_p$ have Bernoulli distribution with parameter $p$.

Let $X,Y,B_p$ be independent.

Let $F_p=B_pX+(1-B_p)Y$.

In a) and b) you calculated the distribution of $F_{0.5}$ and $F_{0.3}$

Observe that: $$F_p^2=B_p^2X^2+2B_p(1-B_p)XY+(1-B_p)^2Y^2=B_pX^2+(1-B_p)Y^2$$

so that:$$\mathbb EF_p^2=\mathbb E(B_pX^2+(1-B_p)Y^2)=p\mathbb EX^2+(1-p)\mathbb EY^2$$

This enables you to find $\mathsf{Var}F_p=\mathbb EF_p^2-(\mathbb EF_p)^2$.

I leave the rest to you.