Q7.59 - A First Course in Probability

Question

$A$There are $n + 1$ participants in a game. Each person independently is a winner with probability $p$. The winners share a total prize of 1 unit. (For instance, if $4$ people win, then each of them receives $1 4$ , whereas if there are no winners, then none of the participants receives anything.) Let A denote a specified one of the players, and let $X$ denote the amount that is received by $A$.

(a) Compute the expected total prize shared by the players. 

(b) Argue that $E\left[X\right] =\frac{ 1 - {(1 - p)}^{n+1}}{ n + 1}$.

(c) Compute E[X] by conditioning on whether  is a winner, and conclude that $E\left[\right(1 + B\left){ }^{-1}\right] =\frac{ 1 - {(1 - p)}^{n+1}}{(n + 1)p}$ when $B$ is a binomial random variable with parameters $n$ and $p$ 

Step-by-Step Solution

Answer

Use the law of total expectation to obtain the required.

Given Information

(a) Observe that the total amount of prize that is shared is equal to $0$ or $1$ . It is equal to 0 if and only if there is no person that has been declared as the winner. Because of the independence, we have that the expected amount of prize that is shared is equal to

        $0.{\left(1-p\right)}^{n+1}+1.\left(1-{\left(1-p\right)}^{n+1}\right)=1-{\left(1-p\right)}^{n+1}$

(b)Observe that the player $A$ is able to win $1 /(k+1)$ of prize, where $k$ marks the number of remaining people that have been declared as winners. But notice, in order to win a prize, (player$A$ also has to be declared as winner). Say that $B~${Binom}$(n, p)$ represents the number of remaining people that have been declared as winners. In that case, we have that

$$\begin{gathered} E\left(X\right)=\sum _{k=0}^n\frac{1}{k+1}p.\left(\begin{array}{c}n\\ k\end{array}\right)p^k{\left(1-p\right)}^{n-k} \\ =\sum _{k=0}^n\frac{1}{k+1}p.\frac{n!}{k!\left(n-k\right)!}{p}^k{\left(1-p\right)}^{n-k} \\ =\frac{1}{n+1}\sum _{k=0}^n\frac{\left(n+1\right)!}{\left(k+1\right)!\left(\left(n+1\right)-\left(k+1\right)!\right)}{p}^{k+1}1-p^{\left(n+1\right)-\left(k+1\right)} \\ =\frac{1}{n+1}\sum _{k=0}^n\left(\begin{array}{c}n+1\\ k+1\end{array}\right)p^{k+1}{\left(1-p\right)}^{\left(n+1\right)-\left(k+1\right)} \\ =\frac{1}{n+1}\left(1-{\left(1-p\right)}^{n+1}\right) \end{gathered}$$

The last sum has to be equal to one since we add up all probabilities of Binomial with parameters $n+1$ and $p$

Final Answer

(c) If player $A$ is a winner, the expected number of money that player $A$ wins is equal to ${\left(1+B\right)}^{-1}$. If player $A$is not a winner, he wins $0$ . Hence, $E\left(X\right)$ can be written as

                          $E\left(X\right)=E\left({\left(1+B\right)}^{-1}\right).p$

which implies (using part (b))

                             $E\left({\left(1+B\right)}^{-1}\right)=\frac{E\left(X\right)}{p}=\frac{1-{\left(1-p\right)}^{n+1}}{\left(n+1\right)p}$