Let $X$ be a continuous random variable with probability density function $f (x) = 2 x$ for $0 l e x l e 1$ and $f (x) = 0$ otherwise. Find the expected value $E [X]$ and variance $Var (X)$ .

medium 4 marks Class 12 · Mathematics · Probability

Answer

Expected value = 2/3, Variance = 1/18

The correct answer is Expected value = 2/3, Variance = 1/18 — Thus $E [X] = \frac{2}{3}$ and $Var (X) = \frac{1}{18}$ .

Step-by-step solution

Verify that $f (x)$ is a valid pdf: $in t_{0}^{1} 2 x, d x = 1$ .
Compute the expectation: $E [X] = \int_{0}^{1} x c d o t 2 x, d x = \int_{0}^{1} 2 x^{2}, d x = 2 \cdot \frac{x ^{3}}{3}_{0}^{1} = \frac{2}{3}$ .
Compute $E [X^{2}] = \int_{0}^{1} x^{2} \cdot 2 x, d x = \int_{0}^{1} 2 x^{3}, d x = 2 \cdot \frac{x ^{4}}{4}_{0}^{1} = \frac{1}{2}$ .
Use the variance formula: $Var (X) = E [X^{2}] - (E [X])^{2} = \frac{1}{2} - (\frac{2}{3})^{2} = \frac{1}{2} - \frac{4}{9} = \frac{9}{18} - \frac{8}{18} = \frac{1}{18}$ .
Thus $E [X] = \frac{2}{3}$ and $Var (X) = \frac{1}{18}$ .

Common mistake: Using the wrong integral bounds or forgetting to square

x

E [X^{2}]

Concept tested: Expectation and variance of continuous distribution