Question 6.4: Suppose that some amount of oil is spilled onto the surface ......
Suppose that some amount of oil is spilled onto the surface of an ocean (for instance, due to an oil-tanker accident). The oil spot then spreads away on the sea surface. The radius, R, of the oil spot in kilometers, 24 hours after the accident, has density function
f(r) = \frac{3}{4}\left\{1 − (20 − r)^2 \right\}, \quad 19 ≤ r ≤ 21.Assuming that the area covered by the oil spot is circular, find the density function for the size of this area.
Let Y be the random variable denoting the size of the area. Then, R and Y are related by
Y = πR².
Let also F_R and F_Y denote the distribution functions of R and Y, respectively. Then, we have, for any y ≥ 0,
P(Y\leq y)=P(\pi R^{2}\leq y)=P\left(R^{2}\leq{\frac{y}{\pi}}\right)=P\left(-{\sqrt{\frac{y}{\pi}}}\leq R\leq{\sqrt{\frac{y}{\pi}}}\right)=F_{R}\left({\sqrt{\frac{y}{\pi}}}\right)-F_{R}\left(-{\sqrt{\frac{y}{\pi}}}\right).But the range of the variable R is the interval [19, 21], which means that F_R(r) = 0 for r < 19 and F_R(r) = 1 for r > 21. This implies, in particular, that the last term on the right-hand side above is zero. Taking this into account, we see that we can express the function F_Y as follows:
F_Y(y)=F_R\left(\sqrt{\frac{y}{\pi} } \right) = \begin{cases} 0, \qquad \qquad \qquad \sqrt{\frac{y}{\pi} } \lt 19, \\ F_R\left(\sqrt{\frac{y}{\pi} } \right) , \qquad 19 \leq \sqrt{\frac{y}{\pi} } \leq 21, \\ 1, \qquad \qquad \qquad \sqrt{\frac{y}{\pi} } \gt 21, \end{cases}or, upon simplifying the conditions on the right, we can write this in the form
F_Y(y)=F_R\left(\sqrt{\frac{y}{\pi} } \right) = \begin{cases} 0, \qquad \qquad \qquad y \lt 361 \pi, \\ F_R\left(\sqrt{\frac{y}{\pi} } \right) , \qquad 361 \pi \leq y \leq 441 \pi, \\ 1, \qquad \qquad \qquad y \gt 441 \pi, \end{cases}Differentiating the last expression with respect to y, we find that
f_{Y}(y)=F_{Y}^{\prime}(y)=0,\quad{\mathrm{for~}}y\lt 361\pi{\mathrm{~or~}}y\gt 441\pi,while for y in the range [361π, 441π] we get
f_{Y}(y)=F_{Y}^{\prime}(y)=\left[F_{R}\left({\sqrt{\frac{y}{\pi}}}\right)\right]^{\prime}=F_{R}^{\prime}\left({\sqrt{\frac{y}{\pi}}}\right)\cdot\left({\sqrt{\frac{y}{\pi}}}\right)^{\prime}=f_{R}\left({\sqrt{\frac{y}{\pi}}}\right)\cdot{\frac{1}{2{\sqrt{\pi y}}}}Thus, we obtain finally the required density to be
f_{Y}(y)=\frac{3}{8\sqrt{\pi}}\ y^{-1/2}\left[1-\left(20-\sqrt{\frac{y}{\pi}}\right)^{2}\right],\quad361\pi\leq y\leq441\pi.