Maximum Distance

Solution

Answer: \( = \frac{1}{2}. \)

Let \[ X_1, X_2, X_3 \sim \text{Uniform}(0,1) \] \[ X_1, X_2, X_3 \sim \text{Uniform}(0,1) \] and define \[ M = \max(X_1,X_2,X_3), \qquad m = \min(X_1,X_2,X_3). \] The distance between the leftmost and rightmost points is: \[ M - m \] so the quantity we want is \[ \mathbb{E}[M-m] = \mathbb{E}[M] - \mathbb{E}[m]. \] First consider the maximum \(M\): \[ P(M \le x) = P(X_1 \le x, X_2 \le x, X_3 \le x) \] Because the variables are independent, \[ P(M \le x) = x^3, \qquad 0 \le x \le 1. \] Differentiating gives the density \[ f_M(x) = \frac{d}{dx}(x^3) = 3x^2. \] The expected value of the maximum is \[ \mathbb{E}[M] = \int_0^1 x \cdot 3x^2 \, dx \] \[ = 3\int_0^1 x^3 \, dx \] \[ = 3 \left(\frac{1}{4}\right) \] \[ = \frac{3}{4}. \] Now consider the minimum \(m\). It is easier to compute, since we know the symmetry of uniform distribution. The minimum is \[ = \frac{1}{4}. \] Therefore the distance between the two points is: \[ \mathbb{E}[M-m] = \mathbb{E}[M] - \mathbb{E}[m] \] \[ = \frac{3}{4} - \frac{1}{4} \] \[ = \frac{1}{2}. \]