Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability (resit) / 9 / b

Since $X$ is nonnegative, $F_X(x)=0$ for $x<0$. For $x\ge 0$, use $\frac{d}{dt}{e}^{-t^2/2}=-t{e}^{-t^2/2}$ to obtainThus the cumulative distribution function isIn particular it tends to one as $x\to \infty$, confirming normalization of this unit-scale Rayleigh distribution.

The cumulative distribution function is continuous and strictly increasing for $x>0$, so the median is the unique positive $c$ satisfying $F_X(c)=1/2$. Consequently $e^{-c^2/2}=1/2$, givingThere is no atom at $c$, so $\mathbb{P}(X>c)=1-F_X(c)=1/2$ as required.

Since $X\ge 0$, the function $X\mapsto X^2$ is strictly increasing on its support. For $y\ge 0$,and $F_Y(y)=0$ for $y<0$. Differentiating the cumulative distribution function givesThe density value at the single point $y=0$ is immaterial. Hence $Y$ has an exponential distribution with rate $1/2$. Direct integration by parts, or its standard expected value, yields

By the density formula for expected value,For the standard normal distribution, the integrand in the second moment is even, soComparing these two integrals gives

The standard normal distribution has mean zero and variance one, so $\mathbb{E}[Z^2]=\mathrm{var}(Z)+(\mathbb{E}[Z]{)}^2=1$. Part (iv) therefore givesUsing the second moment from part (iii),