Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics III / 6

For a random variable with a continuous probability distribution and probability density function $f_X$, the mean and variance areA finite mean requires $\int |x|f_X(x)dx<\infty$. A finite variance additionally requires the second moment to be finite; without that condition the nonnegative integral for the variance may be infinite.

Expand the square and use linearity of expectation, together with $\mathbb{E}[X]=\mu$ and $\mathbb{E}[1]=1$:This proof applies whenever the second moment is finite, and does not require the random variable to be continuous.

Normalization of the probability density function and the specified mean give two equations:Dividing the second by the first gives $2b/3=1$. ConsequentlyBoth are positive, as required for the stated probability density function.

The second moment isUsing the mean $\mathbb{E}[X]=1$, the variance is therefore

Since $Y=Z^2/2$ is nonnegative, its cumulative distribution function is zero for $y<0$. For $y\ge 0$,The last equality uses the symmetry of the standard normal distribution. Thus

Differentiate the cumulative distribution function for $y>0$, using ${\Phi }^{\prime }(z)=(2\pi {)}^{-1/2}{e}^{-z^2/2}$ and $d\sqrt{2y}/dy=1/\sqrt{2y}$. The probability density function isIts value at zero can be assigned arbitrarily without changing any probability. Although this probability density function diverges as $y\downarrow 0$, $F_Y(0)=0$ and there is no point mass at zero. Equivalently, $Y$ has a gamma distribution with shape $1/2$ and rate one.

The probability density function must integrate to one, soAlso, linearity of expectation and the mean and variance of the standard normal distribution giveUsing the probability density function to compute this expected value givesThese are the values $\Gamma (1/2)$ and $\Gamma (3/2)$ of the gamma function.