Expected value of a random variable Max

Question

You choose a uniformly random element, $a$, from $\{1, 2, . . . , 100\}$, and you choose a uniformly random element, $b$, from the same set $\{1, 2, . . . , 100\}$. ($a$ and $b$ are chosen independently)

Define the random variable X to be $X = max(a, b)$, where $X$ is the maximum of $a$ and $b$.

The answer is:

$\mathbb{E}(X) = \sum_{k=1}^{100} k \cdot \frac{1+2(k-1)}{100^2}$

I know that the formula for the expected value is $\mathbb{E}(X) = \sum_{}^{} k \cdot Pr(X = k)$.

I understand that the sample space is $100^2$, however I do not understand the numerator $1+2(k-1)$.

Thank you.

Answer 1

Have a look at what $Pr(X=k)$ is, $X=k$ if:

• $a=b=k$ (which happens with probability $\frac{1}{100^2}$)
• $a=k$ (probability $\frac{1}{100}$), $b<k$ ($k-1$ possibilities so probability $\frac{k-1}{100}$), in the end that has a probability of $\frac{k-1}{100^2}$
• $b=k$ (probability $\frac{1}{100}$), $a<k$ ($k-1$ possibilities so probability $\frac{k-1}{100}$), in the end that has a probability of $\frac{k-1}{100^2}$

In the end, the total probability is the sum of the probabilities of those three cases.

Answer 2

When $ X = k $, we have $ 3 $ possibilities. We can have $ a < b $ and $ b = k $, we can have $ a > b $ and $ a = k $, and we can have $ a = b = k $. There are $ k - 1 $ ways in which each of the first two scenarios can happen, since, in the first case for instance, we can have $ a = 1, ..., k - 1 $. This gives us $ 2(k-1) + 1 $ distinct possibilities overall.

Answer 3

$X=k$ for all the cases $a = k$ and $b$ is chosen from $\{1,\dots, k-1\}$ so $k-1$ cases.

Same for $b=k$ so $k-1$ more cases.

Plus the case $a = k$ and $b=k$.

$2(k-1) +1$ favorable cases