Answer
See explanation
Work Step by Step
(a)
$\begin{align*}
\int_1^\infty f_Y(y)\ dy &= \int_1^\infty \frac{2}{y^3}\ dy \\
&= 2\int_1^\infty y^{-3}\ dy \\
&= 2\left( \frac{y^{-2}}{-2}\ \right\vert_1^\infty \\
&= \left( -\frac{1}{y^2}\ \right\vert_1^\infty \\
&= -\frac{1}{\infty^2} - \left(-\frac{1}{1^2}\right) \\
&= 0 + 1\\
\color{blue}{\int_1^\infty f_Y(y)\ dy}\ &\color{blue}{= 1.}
\end{align*}$
(b)
$\begin{align*}
E(Y) &= \int_1^\infty y\cdot f_Y(y)\ dy \\
&= \int_1^\infty y\cdot \frac{2}{y^3}\ dy \\
&= 2\int_1^\infty y^{-2}\ dy \\
&= 2\left( \frac{y^{-1}}{-1}\ \right\vert_1^\infty \\
&= 2\left( -\frac{1}{y}\ \right\vert_1^\infty \\
&= 2\left(-\frac{1}{\infty} - \left(-\frac{1}{1}\right)\right) \\
&= 2(0 + 1) \\
\color{blue}{E(Y)}\ &\color{blue}{= 2.}
\end{align*}$
(c)
$\begin{align*}
E(Y^2) &= \int_1^\infty y^2\cdot f_Y(y)\ dy \\
&= \int_1^\infty y^2\cdot \frac{2}{y^3}\ dy \\
&= 2\int_1^\infty y^{-1}\ dy \\
&= 2\left( \ln y\ \right\vert_1^\infty \\
&= 2\left(\ln \infty - \ln 1\right) \\
&= 2(\infty - 0) \\
\color{blue}{E(Y^2)}\ &\color{blue}{= \infty.}
\end{align*}$
Thus,
$\begin{align*}
\text{Var}(Y) &= E(Y^2) - E(Y)^2 \\
&= \infty - 2^2 \\
&= \infty - 4 \\
\color{blue}{\text{Var}(Y)}\ &\color{blue}{= \infty,}
\end{align*}$
so that $\text{Var}(Y)$ is not finite.
