# Finding expressions for probability and $E[Y]$ as a function of $\lambda$

Let $X$ represent the number of people arriving at an open house to view it during a designated period of time. Assume that $X$ has a Poisson distribution with parameter $\lambda > 0$. You believe that $\lambda$ is a small number since most people in there know how bad the house and neighborhood are. You convince your friend, who recently moved and has no knowledge of the house, to play the following game with you: If more than $2$ people arrive at the residence in a designated period of time, you will pay your friend $X$ dollars. Your friend will pay you a fee of $50(2-X)$ if two or fewer people arrive at the residence in that period of time. $Y$ denotes your profit from playing the game once.

1) Find the expression (as a function of $\lambda$) for the probability that you do not lose money on the game.

2) Find the expression for $E[Y]$ as a function of $\lambda$. Note that the $E[Y]$ is not just $E[50(2-X)]$.

I am pretty sure I got number one: $$P(X\leq2) = P(X=0) + P(X=1) + P(X=2) = \frac{e^{-\lambda}\lambda^{0}}{0!} + \frac{e^{-\lambda}\lambda^{1}}{1!} + \frac{e^{-\lambda}\lambda^{2}}{2!}\\ = e^{-\lambda} + e^{-\lambda}\lambda + \frac{e^{-\lambda}\lambda^{2}}{2!}.$$ I am stuck on trying to find an expression for $E[Y]$ but I am assuming I need to split it into cases where $X \leq 2$ and $2 < X$. Any help is greatly appreciated.

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Note that $E[X]=\lambda$ and that the series sum that makes up $E[Y]$ is largely the same as $-E[X]$ except:

• if $2$ people come you get $0$ rather than $-2$;
• if $1$ person comes you get $50$ rather than $-1$; and
• if $0$ people come you get $100$ rather than $0$.

So

$$E[Y]= -E[X] + (100-0) \Pr(X=0) + (50+1) \Pr(X=1) + (0+2) \Pr(X=2)$$ which you can turn into a function of $\lambda$.

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Hint: $Y=50(2-X)\mathbf 1_{X\leqslant2}+X\mathbf 1_{X\geqslant3}$.