Question

Find the mean and the variance of the distribution that has the cdf $$ F(x)=\left\\{\begin{array}{ll} 0 & x<0 \\ \frac{x}{8} & 0 \leq x<2 \\ \frac{x^{2}}{16} & 2 \leq x<4 \\ 1 & 4 \leq x \end{array}\right. $$

Short answer

The mean (expected value) of the distribution is \(\frac{5}{3}\) and the variance is \(\frac{1}{9}\).

Step-by-step solution

Find the PDF

The first step to finding the mean and variance of a distribution given the CDF is to find the PDF. The PDF can be found by taking the derivative of the CDF with respect to \(x\). Hence, Depending on the value of \(x\), the derivative of \(F(x)\) is different. For \(0\leq x <2\), \(f(x)= F'(x) = \frac{1}{8} \)For \(2\leq x <4\), \(f(x)= F'(x) = \frac{x}{8}\)

Compute Expected Value

Next, we compute the expected value, or the mean, by integrating the product of \(x\) and the PDF over all \(x\). This gives us\(E[X] = \int_{0}^{2}x \frac{1}{8} dx + \int_{2}^{4}x \frac{x}{8} dx = \frac{1}{32}x^{2} \Big|_{0}^{2} + \frac{1}{24}x^{3} \Big|_{2}^{4} = \frac{5}{3}\)

Compute Second Moment

To compute variance, we also need the second moment \(E[X^2]\). It's computed by\(E[X^2] = \int_{0}^{2}x^{2} \frac{1}{8} dx + \int_{2}^{4}x^{2} \frac{x}{8} dx =\frac{1}{96}x^{3} \Big|_{0}^{2} + \frac{1}{32}x^{4} \Big|_{2}^{4} = \frac{8}{3}\)

Compute Variance

Finally, variance is calculated via \(Var(X) = E[X^2] - (E[X])^2\). Substituting the previously calculated values, we get:\(Var(X) = \frac{8}{3} - \left(\frac{5}{3}\right)^{2} = \frac{1}{9}\)