Answer to Question #87367 in Statistics and Probability for Shivam Nishad

Question #87367
Let X be a random variable with density function
f(X) = {2/x^3, if x≥1
{0, otherwise
Show that E(X) exists and E(X) = 2 but Var(X) does not exist.
1
Expert's answer
2019-04-09T10:41:28-0400
f(x)=2/x3x1,0,otherwisef(x)= \begin{matrix} 2/x^3 & x\geqslant1, \\ 0, & otherwise \end{matrix}

E(X)=1x2x3dx=[2x]1=2E(X)=\displaystyle\int_{1}^\infin x{2 \over x^3}dx=[{-2 \over x}]\begin{matrix} \infin \\ 1 \end{matrix}=2

Var(X)=E(X2)(E(X))2Var(X)=E(X^2)-(E(X))^2

1x22x3dx=2[lnx]1=\displaystyle\int_{1}^\infin x^2{2 \over x^3}dx=2[ln|x|]\begin{matrix} \infin \\ 1 \end{matrix}=\infin

Therefore, Var(X) does not exist.



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