2

If I pick two numbers in the interval (0,1), what is the expected value of the larger number?

Please help. I know that the expected value of one number is 0.5 but I'm not sure how to find the expected vlaue of the larger number here.

JMP
  • 21,771
arjun_math
  • 21
  • 1
    I am going to take a wild guess, which could easily be wrong, based on Bayes Theorem: $$\lim_{n\to\infty} \frac{\sum_{i=1}^n (i/n)^2}{\sum_{i=1}^n (i/n)}.$$ The idea is that the denominator represents the sum of the probabilities of all the intervals, and the numerator weights each interval with the actual value associated with that interval. – user2661923 Mar 18 '21 at 13:50

2 Answers2

1

Assume that the value of the larger number is $x$. The probability of this is equal to the probability that the other number is less than $x$. This probability is obviously $x $. Therefore the expected value of the larger number is: $$2\int_0^1 x\cdot x\, dx=\frac23, $$ where the factor 2 counts the ways to choose the larger number.

user
  • 26,272
0

Imagine the scenario with two die:

6 6 6 6 6 6
5 5 5 5 5 6
4 4 4 4 5 6
3 3 3 4 5 6
2 2 3 4 5 6
1 2 3 4 5 6

The expected value here is:

$$\frac{1+6+15+28+45+66}{36}=\frac{161}{36}=4.47\dot2$$

$161$ is the sixth term in the sequence of hexagonal pyramid numbers ($A002412$).

These are given by:

$$\frac{n(n+1)(4n-1)}{6}$$

On the unit grid, we can do a similar trick. However if say we use a resolution of $k$, then to calculate the expectation we use $kP(X=k)$, and over $k^2$ squares, to give the result as:

$$\lim_{k\to\infty}\frac{k(k+1)(4k-1)}{6k^3}=\frac23$$

JMP
  • 21,771