Question

let x have a uniform distribution on the interval [A,B]. Compute V(X)

Answer 1

Answer on Question #63075 – Math – Statistics and Probability

Question

Let $X$ have a uniform distribution on the interval $[A,B]$. Compute $V(X)$.

Solution

Since $X$ is uniformly distributed on $[A,B]$,

$$f(x) = \frac{1}{B - A}, \text{ if } A \leq x \leq B,$$

and

$$f(x) = 0, \text{ if } x < A \text{ or } x > B.$$

The expectation is

$$\mu = \int_{A}^{B} \frac{x dx}{B - A} = \frac{1}{B - A} \left(\frac{x^{2}}{2}\right)_{A}^{B} = \frac{1}{2} \frac{1}{B - A} \left(B^{2} - A^{2}\right) = \frac{A + B}{2}.$$

The variance is

$$\begin{aligned} V(X) &= \int_{-\infty}^{+\infty} (x - \mu)^{2} f(x) dx = \int_{A}^{B} \frac{(x - \mu)^{2} dx}{B - A} = \\ &= \frac{1}{B - A} \int_{A}^{B} (x - \mu)^{2} d(x - \mu) = \frac{1}{B - A} \cdot \left. \frac{(x - \mu)^{3}}{3} \right|_{A}^{B} \\ &= \frac{1}{3(B - A)} \left[ (B - \mu)^{3} - (A - \mu)^{3} \right] \\ &= \frac{1}{3} \frac{1}{B - A} \left[ \left(B - \frac{A + B}{2}\right)^{3} - \left(A - \frac{A + B}{2}\right)^{3} \right] = \frac{(B - A)^{2}}{12}. \end{aligned}$$

Answer: $\frac{(B - A)^{2}}{12}$

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