Probability — Question 9

Question 9 – Solution

Step 1: Normalize the density. $$\begin{align*} 1&=c\int_1^\infty x^{-3}\,dx\\ &=c\left[-\frac{1}{2x^2}\right]_1^\infty =\frac{c}{2}. \end{align*}$$ Thus $$\boxed{c=2}.$$

Step 2: Test the first moment. $$\begin{align*} E[X] &=\int_1^\infty x\frac{2}{x^3}\,dx =2\int_1^\infty x^{-2}\,dx\\ &=2\left[-\frac1x\right]_1^\infty =2. \end{align*}$$ Therefore, the mean is finite.

Step 3: Test the second moment. $$\begin{align*} E[X^2] &=\int_1^\infty x^2\frac{2}{x^3}\,dx =2\int_1^\infty\frac{dx}{x}. \end{align*}$$ For an upper cutoff $b$, $$2\int_1^b\frac{dx}{x}=2\ln b\longrightarrow\infty.$$ Thus the second moment diverges. $$\boxed{E[X]=2,\qquad E[X^2]=\infty}$$ Consequently, $\operatorname{Var}(X)$ is also infinite.