Probability question: two doors, select the correct door to win money, find expected earning

Question

The question is as follows:

• A contestant pays a dollar to enter the game.

• Suppose that there are two doors in this game, labelled $d_1$, $d_2$. A contestant tosses a coin which has probability $\Pr(\text{head}) = p$. If the coin lands on heads, then he enters door 1, otherwise, enters door 2.

• If contestant end up entering $d_1$, then he or she gets to spin the wheel of fortune and earn money according to the resulting angle from the starting angle, i.e., $x \in [0, 2\pi]$, where $x$ is the amount earned.

• If contestant end up entering $d_2$, then he or she loses the dollar!

• Let $X$ denote the amount of money the contestant has earned at the end of a game, find the expected value of $X$. (Assume that $X = -1$ if he or she enters door $2$).

I am provided with a hint, which states it would help to calculate the CDF of $X$.

---

Let $F_X(x) = \Pr[X \leq x] = \Pr[X \leq x|\text{tail}]\Pr[\text{tail}] + \Pr[X \leq x|\text{head}]\Pr[\text{head}]$

Since $x \in [0, 2\pi]$, therefore $\Pr[X \leq x|\text{head}] =\dfrac{x}{2\pi}$.

Since $x = -1 < y \in [0, 2\pi]$, therefore $\Pr[X \leq x|\text{tail}] = 1$.

Then, $F_X(x) = \Pr[X \leq x] = \Pr[X \leq x|\text{tail}]\Pr[\text{tail}] + \Pr[X \leq x|\text{head}]\Pr[\text{head}] = (1-p)+p\dfrac{x}{2\pi}.$

However, since $\lim\limits_{x \to -\infty} F_X(x) \neq 0$, this is not a valid distribution. Can anyone provide me with some assistance on calculating this CDF and the expectation?

Answer 1

The contestant has to pay $1$ dollar regardless of whether he wins or loses.

Let $X=-1+Q$ where $Q= \begin{cases} x & \text{with probability } \frac{p}{2\pi} \\ 0 & \text{with probability } {1-p}\end{cases}$ where $x \in [0, 2\pi]$

Let us first find the CDF of $Q$,

\begin{align} P(Q \leq q) &= \begin{cases} 0 & , q < 0\\1-p+\frac{pq}{2\pi} &, q \in [0,2\pi] \\ 1 & , q \ge 2\pi\end{cases} \end{align} Now, let us find the CDF of $X$, \begin{align} P(X \leq y) &= P(-1+Q \leq y) \\ &=P(Q \leq y+1) \\ &= \begin{cases} 0 & , y < -1\\1-p+\frac{py}{2\pi} &, y \in [-1,2\pi-1] \\ 1 & , y\ge 2\pi-1\end{cases} \end{align}

Guide for computing expectation:

Note that we have $\mathbb{E}[X]=-1 + \mathbb{E}[Q]$,

hence finding the expectation of $Q$ would give us the expectation of $X$.

Since support of $Q$ is nonnegative, we can use the formula.

$$\mathbb{E}[Q] = \int_0^\infty (1-F_Q(q) \, dq$$