I can buy one lottery ticket out of two available. In the first lottery I can win \$100 with probability 0.1, and the price of ticket is 10. In the second lottery, I can win \$50 with probability 0.1 and $500 with probability 0.01. The price of ticket is \$20. To decide which ticket to buy I toss a fair coin once. I chose first ticket in case of head and second otherwise. Let X be random variable that denotes my net payout (taking into account price of a ticket). Find probability mass function of X (hint: use law of total probability). Show that expected value of X is an average of expected values of net payouts for each of two lotteries. Explain, why. Will it still hold if lotteries has different payouts or probabilities? Prove it.
Could you please check if my calculations and assumptions are correct?
Let L be a lottery. After tossing a coin once we have P(L1)=0.5 and P(L2)=0.5
| L1 | L2 | L2 | |
|---|---|---|---|
| X | 100-10 | 50-20 | 500-20 |
| P | 0.1 | 0.1 | 0.0.1 |
According to the law of total probability:
$pmf(X)=P(L1) \cdot P(x)+P(L2) \cdot P(x)+P(L2) \cdot P(x)$
$pmf(X)=0.5 \cdot 0.1+0.5 \cdot 0.1+0.5 \cdot 0.01=0.105$
Expected value:
$E(X) = \sum_{i=1}^n x_ip_i$
$E(X) = (0.1 \cdot 90) \cdot 0.5 + (0.1 \cdot 30) \cdot 0.5 + (0.01 \cdot 480) \cdot 0.5 = 8.4$
I don't quite understand what should I do next concerning the last questions.
Show that expected value of X is an average of expected values of net payouts for each of two lotteries. Explain, why. Will it still hold if lotteries has different payouts or probabilities? Prove it.
Best Answer
Recall: Expectation is a probability-weighted average.
So you are being asked to show that: $\mathsf E(X)=\mathsf E(X\mid L_1)\,\mathsf P(L_1)+\mathsf E(X\mid L_2)\,\mathsf P(L_2)$
Where $\mathsf E(X\mid L_1)$ and $\mathsf E(X\mid L_2)$ are the expected net payout in the event of each lottery.
As to the probability mass function of $X$. You have to calculate for each value of the support (thus it is a piecewise function, not a single value).
Also do not forget that sometimes the net payout is a loss (no return but still down the cost of the ticket ).
$$\begin{align}p_{\small X\mid L_1}(x) &=\begin{cases}0.9 &:& x=-10\\0.1&:&x=90\\0&:&\text{otherwise}\end{cases}\\[1ex]p_{\small X\mid L_2}(x) &=\begin{cases}0.89 &:& x=-20\\0.1&:&x=30\\0.01&:&x=480\\0&:&\text{otherwise}\end{cases}\\[1ex]p_{\small X}(x) & = p_{\small X\mid L_1}(x)\,\mathsf P(L_1)+p_{\small X\mid L_2}(x)\,\mathsf P(L_2) \end{align}$$