Average of the smaller of three random numbers from $0$ to $1$

Question

A friend of mine claims he saw the following question on a math puzzle site:- What is the average of the smaller of three random numbers from $0$ to $1$? And they've given these options:- $A) \frac{2}{5}B)\frac{1}{10} C)\frac{1}{3} D)\frac{1}{4}$

What does the question even mean? There are infinitely many reals between 0 and 1 and there's no smallest ("smaller" seems incorrect) such triplet of reals. But if the three random numbers chosen are integers, then I believe the question is asking the average of the numbers of the smallest such triplet which is $(0,0,0)$. The average would be $0$. But that seems like a very trivial question. Besides, $0$'s not listed as one of the options. What am I missing?

The answer, my friend claims, somehow makes use of the centre of gravity of a regular tetrahedron or so. But I don't see how that even remotely fits! How would a tetrahedron even come into the picture?

Answer 1

I guess it means this: let $X, Y, Z$ be i.i.d. random variables with distribution $U[0,1]$. What is $E[min(X,Y,Z)]$?

I have not checked the answers, but you can check them yourself: In this question I asked the good people of math.se to check an expression I derived for the maximum of random variables. Maybe it can inspire you. Density function of $\max(X_1,\dots,X_n)$..

Answer 2

I thought I'd write up the solution so that I'd understand it better, and so that others reading this question could understand it as well.

So, the question is: "What is the average of the smaller of three random numbers from 0 to 1?"

What this question actually asks, as noted by Leo in his answer, is this:"If you choose three random numbers in $[0,1]$, what is the expected value of the minimum amongst them?"

For this, consider the three random numbers as random variables $X_1, X_2, X_3$ , each having a uniform distribution in $[0,1]$.This means that $X_1, X_2,$ and $ X_3$ all have equal probability of being any number in $[0,1]$.

Now this has a cumulative distribution function $F(x) = P( X<=x ) = x$, meaning that the probability that a random number between $0$ and $1$ is smaller than $x$, is $x$.

Define $Y =$ "the smallest variable" $= min(X_1, X_2, X_3)$. We wish to find the cumulative distribution function of $Y = P( Y<=x )$ which is the same as $1 - P( Y>x )$. The reason for this will become apparent soon.

Now, if $Y>x$, then $X_1, X_2, X_3 > x$, since $Y = min(X_1,X_2,X_3)$.

So $1 - P( Y>x )$ can be rewritten as $1 - P( X_1>x, X_2>x, X_3>x )$

Since these events are independent of each other, this can be further rewritten as $1-P(X_1>x) \times P(X_2>x) \times P(X_3>x) = 1-[P(X_1>x)]^3 =1-[P(X_1<=x)]^3$

This is because $X_1, X_2$ and $X_3$ all have the same distribution.

$P(X_1<=x)$ is the cumulative probability distribution which I mentioned earlier, equal to $x$.

So, $1-[P(X_1<=x)]^3 = 1-[1-x]^3$.

Now, the probability mass function $f(x)$ = $\dfrac{d}{dx}(1-[1-x]^3)$.

And the Expected value (Average) is given by $E(Y) = \int_0^1 x \times f(x) dx$. Evaluate this integral, and one obtains the answer as $E(Y) = 1/4$.

Answer 3

Although computing distribution functions in this case isn't too bad, there's a way to do this without getting your hands dirty:

Let X, Y and Z be iid Unif[0,1] random variables. We want to find $\mathbb{E}[X|X\leq Y \leq Z]$.

First note that $\mathbb{E}[X|X\leq Y \leq Z] + \mathbb{E}[Y|X\leq Y \leq Z] + \mathbb{E}[Z|X\leq Y \leq Z] = \mathbb{E}[X+Y+Z|X\leq Y \leq Z] = 3/2$.

Then note that $\mathbb{E}[Z|X\leq Y \leq Z] = \mathbb{E}[1-X|X\leq Y \leq Z] = 1-\mathbb{E}[X|X\leq Y \leq Z]$, where the first inequality comes from the fact that the law of 1-X and Z are equal on the event $\{X\leq Y \leq Z\}$ - it is the law of the maximum of 3 independent Unif[0,1] random variables.

Finally, by a similar argument, $\mathbb{E}[Y|X\leq Y \leq Z]=\mathbb{E}[1-Y|X\leq Y \leq Z] = 1-\mathbb{E}[Y|X\leq Y \leq Z]$, as both Y and 1-Y have the law of the second largest of 3 independent Unif[0,1] random variables on $\{X\leq Y \leq Z\}$.

It's then immediate that $\mathbb{E}[Y|X\leq Y \leq Z]=1/2$, and substituting this and the relationship $\mathbb{E}[Z|X\leq Y \leq Z] = 1-\mathbb{E}[X|X\leq Y \leq Z]$ into the first equation gives a linear equation in $\mathbb{E}[X|X\leq Y \leq Z]$, which is readily solved to give $\mathbb{E}[X|X\leq Y \leq Z]=1/4$.