Solution: 2022 Winter Final - 20

Author: Michiel Smid

Question

Let $X_1,X_2,X_3$ be three numbers chosen independently and uniformly from the set $\{1,\ldots,50\}$. Let $Z=\max\{X_1,X_3,X_4\}$. What is $ E(Z) $?

(a)

$50^{-3}\cdot\sum_{i=1}^{50} i\cdot i^2$

(b)

$50^{-3}\cdot\sum_{i=1}^{50} i\cdot (3(i-1)^2 + 3(i-1) + 1)$

(c)

$99/2-49.5$

(d)

$33$

(e)

$50^{-3}\cdot\sum_{i=1}^{50} i\cdot i^3$

Solution

Although all the answers in the question look really yucky, I feel like by explaining conceptually what the answer represents (and therefore how to get there), it should make more sense.

To find the expected value of the max of 3 numbers, we first need to know what is the range of possible values for Z. Since the smallest number in the set is 1 and the largest is 50, that means the range of values is: $1 - 50$.

Let $Z_i$ represent the random variable where:

$ Z_i = \begin{cases} 1 & max{X_1, X_2, X_3} = i
0 & \text{otherwise.} \enspace \end{cases} $

Using the linearity of expectation, we just need to find out the expected value of $Z_i$, or $Pr(Z_i = i)$.

We know that for a number to be a “maximum”, there are 3 possible cases:

Either that number is chosen 3 times
That number is chosen 2 times and the other number is a number less than the maximum
The maximum number is chosen once but the other 2 numbers are less than the maximum

We can convert these cases into a formula, which corresponds to: $3(i - 1)^2 + 3(i - 1) + 1$, where $i$ is the “maximum” number. We also know that the total sample space (aka, all combinations of 3 numbers chosen) are: $(\frac{1}{50})^3 = 50^{-3}$ Now we can formulate the entire formula together:

$E(Z) = E(\sum_{i=1}^{50} Z_i)$
$E(Z) = \sum_{i=1}^{50} E(Z_i)$
$E(Z) = \sum_{i=1}^{50} i \cdot (3(i - 1)^2 + 3(i - 1) + 1) \cdot 50^{-3}$
$E(Z) = 50^{-3} \cdot \sum_{i=1}^{50} i \cdot (3(i - 1)^2 + 3(i - 1) + 1)$