5.1.4 problem 5

a. We have $A=\pi R^2$, so $E\left(A\right)=\mathrm{\pi E}\left(R^2\right)$. We have $E\left(R^2\right)={\int }_0^1{x}^2\ast 1\mathrm{dx}=1/3$, since the PDF of R is always 1. Then $E\left(A\right)=\pi ∕3$.

We have $V\mathrm{ar}\left(A\right)=E\left(A^2\right)-E{\left(A\right)}^2={\pi }^{2}E\left(R^4\right)-{\pi }^2/9$ using linearity. $E\left(R^4\right)={\int }_0^1{x}^4\ast 1\mathrm{dx}=1/5$, so $V\mathrm{ar}\left(A\right)={\pi }^2/5-{\pi }^2/9=4{\pi }^2/45$

b. CDF: $P(A<k)=P(\pi R^2<k)=P(R<\sqrt{k∕\pi })=\sqrt{k∕\pi }$ for $0<k<\pi$ using the CDF of Unif(0,1). The CDF of A is 0 for $k<0$ and $k>\pi$.

PDF: $\frac{d}{\mathrm{dk}}\left(\sqrt{k∕\pi }\right)=\frac{1}{2\sqrt{\mathrm{k\pi }}}$ for $0<k<\pi$ and 0 elsewhere.