Find \(f(y), E(Y)\) and \(\operatorname{Var}(Y)\). Find \(P(Y>2)\).

Question

The cdf of \(Y\) is given by
\[
F(y)= \begin{cases}0 & y<1 \\ \ln (y) & 1 \leq y \leq e \\ 1 & x>e\end{cases}
\]
Find \(f(y), E(Y)\) and \(\operatorname{Var}(Y)\). Find \(P(Y>2)\).

Answer 1

\begin{equation}
\begin{gathered}
f(y)=\frac{d}{d y} F(y)= \begin{cases}\frac{1}{y} & 1 \leq y \leq e \\
0 & \text { otherwise }\end{cases} \\
E(Y)=\int_{1}^{e} y f(y) d y=\int_{1}^{e} 1 d y=\left.y\right|_{1} ^{e}=e-1 \\
E\left(Y^{2}\right)=\int_{1}^{e} y^{2} f(y) d y=\int_{1}^{e} y d y=\left.\frac{1}{2} y^{2}\right|_{1} ^{e}=\frac{e^{2}-1}{2} \\
\text { So } \operatorname{Var}(Y)=E\left(Y^{2}\right)-E(Y)^{2}=\frac{e^{2}-1}{2}-\left(e^{2}-2 e+1\right)=\frac{-e^{2}+4 e-3}{2} \approx .242 . \\
P(Y>2)=1-P(Y<2)=1-F(2)=1-\ln (2)
\end{gathered}
\end{equation}