Expected value of the smallest of the three numbers

Question

Three numbers are independently and randomly chosen from a uniform distribution between $0$ and $1$. What is the expected value of the smallest of the three numbers?

My answer:

Three numbers $X_1, X_2, X_3$ are independently and randomly chosen from a uniform distribution between $0$ and $1$. We are asked to find the expected value of the smallest of the three numbers, denoted as $M = \min(X_1, X_2, X_3).$

The cumulative distribution function $CDF$ of $M$ is:

$F_M(x) = P(M \leq x) = 1 - P(X_1 > x, X_2 > x, X_3 > x).$

Since the $X_i$ are independent and uniformly distributed, the probability that $X_i > x$ is $1 - x$. Therefore:

$F_M(x) = 1 - (1 - x)^3.$

To find the probability density function $PDF$, we differentiate the CDF:

$f_M(x) = \frac{d}{dx} F_M(x) = \frac{d}{dx} \left( 1 - (1 - x)^3 \right) = 3(1 - x)^2.$

The expected value of $M$ is given by:

$E[M] = \int_0^1 x f_M(x) \, dx = \int_0^1 x \cdot 3(1 - x)^2 \, dx.$

Expanding the integrated:

$E[M] = 3 \int_0^1 x(1 - x)^2 \, dx = 3 \int_0^1 (x - 2x^2 + x^3) \, dx.$

Now, compute each integral:

$\int_0^1 x \, dx = \frac{1}{2}, \quad \int_0^1 x^2 \, dx = \frac{1}{3}, \quad \int_0^1 x^3 \, dx = \frac{1}{4}.$

Thus:

$E[M] = 3 \left( \frac{1}{2} - 2 \cdot \frac{1}{3} + \frac{1}{4} \right) = 3 \left( \frac{6}{12} - \frac{8}{12} + \frac{3}{12} \right) = 3 \cdot \frac{1}{12} = \frac{1}{4}.$

The expected value of the smallest of the three numbers is $\boxed{\frac{1}{4}}.$

Is there a more elegant way to do this?

See this question. All the various solutions apply to your version. — lulu, Oct 13 '24 at 13:45
How are those approaches applicable to this specific question? — Shs Tht, Oct 13 '24 at 14:15
For instance: close the interval by identifying $0$ and $1$: now you are just distributing $4$ points on a circle, the extra one being the point connecting $0$ to $1$. The gaps should be equidistribtuted, hence $\frac 14$. — lulu, Oct 13 '24 at 15:44

Dilip Sarwate · Answer 1 · 2024-10-14T00:17:36.480

Is there a more elegant way to do this?

$F_M(x)$ has value $1-(1-x)^3$ for $x\in [0,1]$ and value $1$ for $x>1$. For nonnegative random variables, \begin{align} E[M] &= \int_0^\infty [1-F_M(x)] \, \mathrm dx\\ &= \int_0^1 (1-x)^3 \, \mathrm dx\\ &= \int_0^1 y^3 \, \mathrm dy & \text{change of variable }y=1-x\\ &= \frac 14. \end{align} Intuitively, if one breaks a stick of unit length at three different randomly chosen points, one should expect that the average (a.k.a. expected) length of each of the four pieces is $\frac 14$. Thus, the expected length of the "leftmost piece" is $\frac 14$.

Expected value of the smallest of the three numbers

1 Answers1