Solution: 2018 Winter Final - 23

To prove that this statement is false, we need to provide a counterexample:

Let X, Y, Z all represent the random variables where $X, Y, Z = 1$ if a coin toss is heads and $X, Y, Z = 0$ if a coin toss is tails.

The expected value of $X, Y, Z$ in this case (as\suming that $\frac{1}{2}$ chance of hitting heads or tails) is: $E(X) = E(Y) = E(Z) = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = \frac{1}{2}$.

This means that the $min(E(X), E(Y), E(Z) = min( \frac{1}{2}, \frac{1}{2}, \frac{1}{2}) = \frac{1}{2}$ is $\frac{1}{2}$.

For the LHS, we need to calculate the expected value for $min(X, Y, Z)$. There are only two possible values in this case: $min(X, Y, Z) = 0$ and $min(X, Y, Z) = 1$

• Pr(min(X, Y, Z) = 1) = every coin has to come up with a heads = $( \frac{1}{2})^{3} = \frac{1}{8}$
• Pr(min(X, Y, Z) = 0) = 1 - Pr(min(X, Y, Z) = 1) = $1 - \frac{1}{8} = \frac{7}{8}$
• $E(min(X, Y, Z)) = 0 \cdot Pr(min(X, Y, Z) = 0) + 1 \cdot Pr(min(X, Y, Z) = 1) = 0 \cdot \frac{7}{8} + 1 \cdot ( \frac{1}{8}) = \frac{1}{8}$.

Since $E(min(X, Y, Z)) = \frac{1}{8}$ and $min(E(X), E(Y), E(Z)) = \frac{1}{2}$, the two are not the same, so the statement is false.