Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider two fair $k$-sided dice with the numbers 1 through $k$ on their faces, obtaining values $X_1$ and $X_2$. What is $\mathbb{E}[\max(X_1, X_2)]$ and what is $\mathbb{E}[\min(X_1, X_2)]$.

share|cite|improve this question
What do you know? What did you try? Why did it fail? – Did Nov 17 '11 at 6:20
It's getting started that is causing the problem for me. In the previous problem I calculated E[x] and E[sum/difference of different number of die]. I found E[x] to be (k+1)/2. The expected value of the sums/differences I was able to calculate using the linearity property; but I don't know how to go about dealing with max/min functions inside the E[]. – jamesio Nov 17 '11 at 6:32
up vote 4 down vote accepted

HINT: $\min(x_1,x_2) + \max(x_1,x_2) = x_1 + x_2$, so it is enough to evaluate either expected value. Also $\max(x_1,x_2) - \min(x_1,x_2) = \vert x_2-x_1\vert$. Therefore, finding $\mathbb{E}(\vert X_2-X1 \vert)$ allows to determine expectations needed:

$$ \begin{eqnarray} \mathbb{E}\left( \vert X_2- X_1\vert \right) &=& \sum_{n_1=1}^k \sum_{n_2=1}^k \mathbb{P}(X_1=n_1) \mathbb{P}(X_2=n_2) \vert n_2 - n_1 \vert \\ &=& 2 \sum_{n_2=1}^k \sum_{n_1=1}^{n_2-1} \mathbb{P}(X_1=n_1) \mathbb{P}(X_2=n_2) (n_2 - n_1 ) \\ &=& 2 \sum_{n_2=1}^k \sum_{n_1=1}^{n_2-1} \mathbb{P}(X_1=n_1) \mathbb{P}(X_2=n_2-n_1) n_1 \end{eqnarray} $$

Can you finish this ?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.