# Collective Risk Model

As an acturial student, I have to do :

Show that for any two random variables V, W one has

Var(W) = E(Var(W|V )) + Var(E(W|V )).

Use the formula to solve the following problem: In a tourist office trips are organized at days with sunny weather only. Assume that on each day the weather is sunny with probability 0.7, independent of the other days. If a new guide starts to work on a sunny day, and if this guide faces N consecutive sunny days, the total number of customers until the first non-sunny day is Poisson distributed with parameter 5N, what is the expectation and variance of total customers that this guide faces until the first non-sunny day?

Is it right to use a geometric distribution for the sunny or non-sunny day ? If not which one do I need to use ?

For the first part of the total variance equation, take $$Var(W) = E[W^2] - E^2[W],$$ by using the law of total expectation $$Var(W) = E[E[W^2|X]] - (E[E[W|X]])^2= E[Var(W|X) + E^2[W|X]] - (E[E[W|X]])^2,$$ slightly rearranging the equation you'll get the formula \begin{align} Var(W) = & E[Var(W|X)] + E( E^2[W|X]) - (E[E[W|X]])^2\\ = & E[Var(W|X)] + Var[E(W|X)] \, . \end{align}
For the second part, denote $N$ the number of consecutive sunny day, that is $N \sim Geo(0.3)$ and the number of costumers $X$ each day $i$ is $Poisson(5)$, thus $$E(\sum_{i=1}^NX_i) =EX_i EN=\frac{5}{0.3}.$$ For the variance, use the law of total variance that you proved in the first part.
• 2) Sorry, it is $0.3$ indeed. I've changed the answer. 1) It $5N$ for $N$ day, hence $5$ for each day and $E(5N) = 5EN=EXEN=5/0.3$. – V. Vancak Mar 21 '17 at 12:15