Problem 1

a)

\[p(x) = \begin{cases} 0.08, & x=0 \\ 0.20, & x=1,3 \\ 0.26, &x=2 \\ 0.14, & x=4 \\ 0.07, & x=5 \\ 0.03, & x=6 \\ 0.02, & x=7 \\ 0, & \text{otherwise} \end{cases}\]

This PMF is valid, because all values are non-negative and the probabilities sum to \(1\).

b)

The mean is the same thing as the expected value.

\[E(X) = \sum_{x=0}^7x p(x) = 0(0.08) + 1(0.2) + 2(0.26) + 3(0.2) + 4(0.14) + 5(0.07) + 6(0.03) + 7(0.02) = 2.55\] \[E(X^2) = \sum_{x=0}^7x p(x) = 0^2(0.08) + 1^2(0.2) + 2^2(0.26) + 3^2(0.2) + 4^2(0.14) + 5^2(0.07) + 6^2(0.03) + 7^2(0.02) = 9.09\] \[\text{Var}(X) = E(X^2) - E(X)^2 = 9.09 - 2.55^2 = 2.59\]

So the mean goals per game is \(2.55\) with a variance of \(2.59\) or equivalently a standard deviation of \(1.6\) goals.

c)

We didn’t talk about CDFs for discrete RVs, so you were not required to do this problem. Here is a plot of the CDF anyways.

probs <- c(.08, .2, .26, .2, .14, .07, .03, .02)
curve(cumsum(probs)[x+1], from=0, to=8, n=1001,
      xlab='x', ylab='F(x)')

Note that this function satisfies the three properties of a CDF that we discussed in class. There is a major difference between this and the functions we discussed… it is NOT a continuous function. This makes sense because \(X\) is a discrete RV.

Problem 2.

a)

\[\begin{align*} Var(aX+b) &= E((aX+b)^2) - E(aX+b)^2 \\ &= E(a^2X^2 + 2abX + b^2) - E(aX+b)^2 \\ &= a^2E(X^2) + 2abE(X) + b^2 - (aE(X)+b)^2 && \text{(LoE)} \\ &= a^2E(X^2) + 2abE(X) + b^2 - (a^2E(X)^2 + 2abE(X) + b^2) \\ &= a^2(E(X^2) - E(X)^2) \\ &= a^2 Var(X) \\ &\square \end{align*}\]

b)

\[E(Y) = E(25X+50) = 25E(X) + 50 = 25(2.55) + 50 = 113.75\] \[Var(Y) = Var(25X+50) = 25^2Var(X) = 625(2.59) = 1618.75\]

Problem 3.

a)

We need to choose \(c\) so that the probabilties sum to 1.

\[\sum_{x=1}^4 p(x) = p(1) + p(2) + p(3) + p(4) = c/2 + c/4 + c/8 + c/16 = c(15/16) = 1\] so \(c = 16/15\).

b)

\[P(X > 2) = p(3) + p(4) = \frac{16}{15}\frac{1}{2^3} + \frac{16}{15}\frac{1}{2^4} = \frac{3}{15} = 0.2\]

c)

\[E(X) = 1\left(\frac{16}{15}\frac{1}{2^1}\right) + 2\left(\frac{16}{15}\frac{1}{2^2}\right) + 3\left(\frac{16}{15}\frac{1}{2^3}\right) + 4\left(\frac{16}{15}\frac{1}{2^4}\right) = 1.7333\] \[E(X^2) = 1\left(\frac{16}{15}\frac{1}{2^1}\right) + 4\left(\frac{16}{15}\frac{1}{2^2}\right) + 9\left(\frac{16}{15}\frac{1}{2^3}\right) + 16\left(\frac{16}{15}\frac{1}{2^4}\right) = 3.8667\] \[Var(X) = E(X^2) - E(X)^2 = 3.8667 - 1.733^2 = 0.862\]

d)

\[E(2^X) = 2^1\left(\frac{16}{15}\frac{1}{2^1}\right) + 2^2\left(\frac{16}{15}\frac{1}{2^2}\right) + 2^3\left(\frac{16}{15}\frac{1}{2^3}\right) + 2^4\left(\frac{16}{15}\frac{1}{2^4}\right) = 4\frac{16}{15} = 4.267\]

Problem 4.

a)

The probability is \(1/5\).

b)

First note that \(X \sim Bern(1/5)\) so \(E(X) = 1/5\).

\[E(Y) = E(1.25X - 0.25) = 1.25(1/5) - 0.25 = 0\]

c)

In this case, we have \(X \sim Bern(1/k)\) and \(E(X) = 1/k\). Since \(Y = X - L(1-X) = (1+L)X - L\), we have

\[E(Y) = (1+L)E(X) - L = \frac{1+L}{k} - L\]

setting equal to \(0\) and solving for \(L\) gives

\[L = \frac{1}{k-1}.\]

Problem 5.

It helps to draw out a “tree” of possibilites with this problem. Note that there are only three final outcomes for his winnings: \(\mathbb X = \{100, -100, -300\}\).

\[P(X = -300) = P(\text{lose 3 times}) = \left(\frac{20}{38}\right)^3 = 0.146\] \[P(X = -100) = P(\text{lose, win, lose OR lose, lose, win}) = \left(\frac{20}{38}\right)^2\left(\frac{18}{38}\right) + \left(\frac{20}{38}\right)^2\left(\frac{18}{38}\right) = 0.262\] \[P(X = 100) = 1 - P(X=-100) - P(X=-300) = 0.592\]

So the pmf can be written as

\[p(x) = \begin{cases} 0.592, & x = 100 \\ 0.262, & x = -100 \\ 0.146, & x=-300 \\ 0, & \text{otherwise} \end{cases}\]

b)

The probability that we win money is now \[P(X > 0) = P(X=100) = 0.592.\]

Since this is bigger than \(1/2\), this seems pretty good!

c)

The expected value is \[E(X) = 100(0.592) - 100(0.262) - 300(0.146) = -10.8.\]

So in the long run, the house always wins. On average you will lose \(\$10.80\) every time you play. The problem is that, even though you win more than half the time, your potential winnings are capped at \(\$100\). When you lose, there is a chance you “lose big”, so the expected value is still in the favor of the casino.