Question

Let X be given by its "distribution" function F(x), such that:

F(x) = 0 if x ≤ 0
F(x) = (x^2)/4 if 0<x≤2
F(x) =1 if x>2

Find E(x), var (x) and std deviation (x).

Answer 1

Answer on Question #43537, Math, Statistics and Probability

Let $X$ be given by its "distribution" function $F(X)$ such that:

$$F(x) = \begin{cases} 0, & \text{if } x \leq 0 \\ \frac{x^2}{4}, & \text{if } 0 < x \leq 2 \\ 1, & \text{if } x > 2 \end{cases}$$

Find $E(X)$, $\operatorname{var}(X)$ and $\operatorname{std}$ deviation $(X)$.

Solution.

By definition,

$$E(X) = \int_{-\infty}^{\infty} x f(x) \, dx$$

where $f(x) = F'(X)$. So,

$$f(X) = \begin{cases} 0, & \text{if } x \notin (0, 2] \\ \frac{x}{2}, & \text{if } x \in (0, 2] \end{cases}$$

Then,

$$E(X) = \int_{-\infty}^{0} x \cdot 0 \, dx + \int_{0}^{2} \frac{x^2}{2} \, dx + \int_{2}^{\infty} x \cdot 0 \, dx = 0 + \frac{2^3}{6} - 0 + 0 = \frac{4}{3}$$

By definition,

$$\begin{aligned} \operatorname{var}(X) &= \int_{-\infty}^{\infty} x^2 f(x) \, dx - \left(E(X)\right)^2 = \int_{-\infty}^{0} x^2 \cdot 0 \, dx + \int_{0}^{2} \frac{x^3}{2} \, dx + \int_{2}^{\infty} x^2 \cdot 0 \, dx - \frac{16}{9} = \\ &= 0 + \frac{2^4}{8} + 0 - \frac{16}{9} = 2 - \frac{16}{9} = \frac{2}{9} \end{aligned}$$

By definition,

$$\operatorname{std} \text{ deviation}(X) = \sqrt{\operatorname{var}(X)}$$

So,

$$\operatorname{std} \text{ deviation}(X) = \frac{\sqrt{2}}{3}$$

Answer: $E(X) = 4/3$, $\operatorname{var}(X) = 2/9$, $\operatorname{std} \text{ deviation}(X) = \sqrt{2}/3$.

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