If the side length isn't constrained to integers, the expected volume is 39 inches cubed.

In a bit more detail: take the integral of x^3 (the volume function) from 1 to 5 and you get 5^4/4 - 1^4/4 = 624/4 = 156. Then we divide by the length of the line segment (4) to get 39.

If the side lengths have to be integers, then it's just (1 + 8 + 27 + 64 + 125)/5 = 45 inches.

Edit: corrected division mistake, thanks to dxbydt

Edit 2: and corrected a second mistake thanks to tzs.

> Edit: corrected division mistake, thanks to dxbydt

Well, one of the mistakes:

> If the side lengths have to be integers, then it's just (1 + 8 + 27 + 125)/4 = 40.25 inches

If I'm not mistaken, 4 is an integer, and so should not have cruelly snubbed when you sent out invitations to your summation of the cubes of the integers from 1 to 5. Sure, 4 is not a cool integer like the others in 1 to 5...he's got a reputation as a square, but that doesn't mean he doesn't have feelings.

Oy. Someone needs to revoke my math license. Thanks for the correction!

My, my, 2 mistakes on a silly uniform distribution! You must atone for this sin. Explain what the volume would be if the side came from a normal distribution over [1,5] (ie. mean 3, sigma 2/3) Otherwise I will personally call the MAA director on my speed dial & haul your ass back to stat101 :)

But seriously, you write rather well on Quora & I'm sure the crowd would want to know what happens to the volume were you to choose a normal distribution instead of a uniform, so I hope you respond.

From [0]:

Let X be a random variable and g be any function.

If X is continuous, then the expectation of g(X) is defined as,

E[g(X)] = \int_−∞^∞ g(x)f(x) dx,

where f is the probability density function of X.

---

For us, with mean 3 and sigma 2/3, f = 3/2 e^(-9/8 (x-3)^2)/sqrt(2 π)

We need to integrate numerically; using Wolfram [1] we get an estimate of 30.8099.

If we ask for the expected value over the entire distribution we get 31.

[0] http://imai.princeton.edu/teaching/files/Expectation.pdf

[1] http://www.wolframalpha.com/input/?i=integrate+3%2F2+e%5E%28...

[2] http://www.wolframalpha.com/input/?i=expected+value+of+x%5E3...

156/4=39

not 59, 39.

Heh, failed at the simplest step. Thanks for the correction.

He was staring at it because he was looking for a shortcut that must make the problem obvious enough to be done in 10 secs. If you don't provide context on such tasks, that's expected.

The shortcut's E h(X) = ∫ h(x) f(x) dx. So, math.

It's stories like this why I keep coming back to Hacker News

What's the answer?

27?

Nope. To see where the subtlety lies in this problem, simplify it. Change it to a discrete problem: the side can be 1, 3, or 5, each with probability 1/3. The expected value of the side length is still 3.

1/3 of the cubes have volume 1, 1/3 have volume 27, and 1/3 have volume 125. The expected value of the volume is 51. Think about why it is 51, not 27, and you'll see why it is not 27 for the continuous case.

In general, if you have a random variable X, and some function of that random variable, f(X), then it is NOT true that expected value of f(X) = f(expected value of X).

27 would be correct if each side were independent random variables rather than the cube of a random variable. The way the problem is set up you are suppose to solve for E[a^3] (the third moment of the random variable) rather than (E[a])^3.

I'd say 45, assuming the side length is an integer.

39

    from random import random
    sum([(random()*4+1)**3 for x in xrange(1000000)])/1000000
    => ~39

Weird.

  sum(random.uniform(1, 5)**3 for i in range(10000)) / 10000

Many people don't know that the random module allows to sample from many distributions directly without having to construct your own.

I like the first version better because everyone knows what infix * and + do.

I like the second version better because it taught me something I didn't know.