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

A probability density function must be nonnegative and have definite integral one. HereThereforeThis value is positive, so the nonnegativity requirement also holds.

The cosine function is positive on $(0,\pi /2)$ and negative on $(\pi /2,2)$. Thus any nonzero $\alpha$ makes $\alpha \cos x$ negative on an interval of positive length. Taking $\alpha =0$ makes the total definite integral zero. HenceNormalization alone would suggest $\alpha =1/\sin 2$, but that candidate still violates nonnegativity and therefore is not a probability density function.

We need $0<\alpha <2$. Indeed, if $\alpha <0$, the function is negative on $(\alpha ,0)$; if $\alpha =0$, the improper integral diverges at zero; and if $\alpha \ge 2$, its support interval is empty. For $0<\alpha <2$,Setting this equal to one gives the unique admissible valueOn this interval $1/x>0$, so both requirements for a probability density function are met.

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),