问题及解答

Let $X$ denote checkout time duration with pdf given in Exercise 1 , 

\[
f(x)=\begin{cases}
       .5x, & 0\leqslant x\leqslant 2, \\
       0, & \mbox{otherwise}.
     \end{cases}
\]

• (a) Compute $E(X)$.
• (b) Compute $V(X)$ and $\sigma_X$.
• (c) If the borrower is charged an amount $h(X)=X^2$ when checkout duration is $X$, compute the expected charge $E[h(X)]$.
   

(a)

\[
E(X)=\int_0^2 xf(x)dx=\int_0^2 x\cdot\frac{1}{2}xdx=\frac{1}{2}\cdot\frac{1}{3}x^3\biggr|_0^2=\frac{4}{3}\approx 1.333333
\]

(b)

\[
\begin{split}
V(X)&=E(X^2)-\bigl[E(X)\bigr]^2\\
&=\int_{0}^{2}x^2\cdot f(x)dx-\Bigl(\frac{4}{3}\Bigr)^2\\
&=\int_{0}^{2}\frac{1}{2}x^3dx-\frac{16}{9}\\
&=\frac{1}{2}\cdot\frac{1}{4}x^4\biggr|_{0}^{2}-\frac{16}{9}\\
&=\frac{1}{8}\cdot 2^4-\frac{16}{9}\\
&=2-\frac{16}{9}\\
&=\frac{2}{9}\approx 0.22222222
\end{split}
\]

\[
\sigma_X=\sqrt{V(X)}=\sqrt{\frac{2}{9}}=\frac{\sqrt{2}}{3}\approx 0.47140452
\]

(c)

For $h(X)=X^2$,

\[
\begin{split}
E[h(X)]&=E(X^2)=\int_{0}^{2}x^2\cdot f(x)dx\\
&=\int_{0}^{2}x^2\cdot\frac{1}{2}xdx\\
&=\frac{1}{2}\int_{0}^{2}x^3dx\\
&=\frac{1}{2}\cdot\frac{1}{4}x^4\biggr|_{0}^{2}\\
&=\frac{1}{8}\cdot 16\\
&=2
\end{split}
\]

In fact, we have computed $E(X^2)$ in (b).