Year: 2014
Paper: 3
Question Number: 12
Course: UFM Statistics
Section: Cumulative distribution functions
A 10% increase in the number of candidates and the popularity of all questions ensured that all questions had a good number of attempts, though the first two questions were very much the most popular. Every question received at least one absolutely correct solution. In most cases when candidates submitted more than six solutions, the extra ones were rarely substantial attempts. Five sixths gave in at least six attempts.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
The random variable $X$ has probability density function
$f(x)$ (which you may assume is differentiable) and cumulative distribution function $F(x)$ where $-\infty < x < \infty $. The random variable $Y$ is defined by $Y= \e^X$. You may assume throughout this question that $X$ and $Y$ have unique modes.
\begin{questionparts}
\item Find the median value $y_m$ of $Y$ in terms of the median value $x_m$ of $X$.
\item Show that the probability density function of $Y$ is $f(\ln y)/y$, and deduce that the mode $\lambda$ of $Y$ satisfies $\f'(\ln \lambda) = \f(\ln \lambda)$.
\item Suppose now that $X \sim {\rm N} (\mu,\sigma^2)$, so that
\[ f(x) = \frac{1}{\sigma \sqrt{2\pi}\,} \e^{-(x-\mu)^2/(2\sigma^2)} \,. \]
Explain why \[\frac{1}{\sigma \sqrt{2\pi}\,} \int_{-\infty}^{\infty}\e^{-(x-\mu-\sigma^2)^2/(2\sigma^2)} \d x = 1 \] and hence show
that $ \E(Y) = \e ^{\mu+\frac12\sigma^2}$.
\item Show that, when $X \sim {\rm N} (\mu,\sigma^2)$,
\[
\lambda < y_m < \E(Y)\,.
\]
\end{questionparts}\begin{questionparts}
\item \begin{align*}
&& \frac12 &= \mathbb{P}(X \leq x_m) \\
\Leftrightarrow && \frac12 &= \mathbb{P}(e^X \leq e^{x_m} = y_m)
\end{align*}
Therefore the median is $y_m = e^{x_m}$
\item \begin{align*}
&& \mathbb{P}(Y \leq y) &= \mathbb{P}(e^X \leq y) \\
&&&= \mathbb{P}(X \leq \ln y) \\
&&&= F(\ln y) \\
\Rightarrow && f_Y(y) &= f(\ln y)/y \\
\\
&& f'_Y(y) &= \frac{f'(\ln y) - f(\ln y)}{y^2}
\end{align*}
Therefore since the mode satisfies $f'_Y = 0$ we must have $f'(\ln \lambda ) = f(\ln \lambda)$
\item This is the integral of the pdf of $N(\mu + \sigma^2, \sigma^2)$ and therefore is clearly $1$.
\begin{align*}
&& \E[Y] &= \int_{-\infty}^{\infty} e^x \cdot \frac{1}{\sqrt{2\pi \sigma^2}} e^{-(x-\mu)^2/(2\sigma^2)} \d x \\
&&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (x - (x-\mu)^2/(2\sigma^2)) \d x\\
&&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp ((2x \sigma^2- (x-\mu)^2)/(2\sigma^2)) \d x\\
&&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (-(x-\mu-\sigma^2)^2+2\mu \sigma^2-\sigma^4)/(2\sigma^2)) \d x\\
&&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (-(x-\mu+\sigma^2)^2)/(2\sigma^2)+\mu +\frac12\sigma^2) \d x\\
&&&= \e^{\mu +\frac12\sigma^2}\frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (-(x-\mu-\sigma^2)^2)/(2\sigma^2)) \d x\\
&&&= \e^{\mu +\frac12\sigma^2}
\end{align*}
\item Notice that $y_m = e^\mu < e^{\mu + \tfrac12 \sigma^2} = \E[Y]$, so it suffices to prove that $\lambda < e^{\mu}$
Notice that $f'(x) - f(x) = f(x)[-(x-\mu)/\sigma^2 - 1]$ and therefore $\ln y - \mu = -\sigma^2$ so $\lambda = e^{\mu - \sigma^2}$ which is clearly less than $e^{\mu}$ as required.
\end{questionparts}
Less than 8% tried this, scoring just over a quarter of the marks. Very few got the question totally correct, but a number got it mostly right. Nearly all managed the median of Y, but the probability density function of Y caused some to stumble. However the mode result, apart from some poor differentiation, was mostly alright. The explanation in part (iii) eluded some candidates who were otherwise strong. Applying the mode result in part (iv) to obtain the required value surprisingly tripped up some merely through inaccurate differentiation. As one would hope, anyone that got as far as part (iv) spotted that the median of X was the expected value.