Year: 2016
Paper: 3
Question Number: 13
Course: UFM Statistics
Section: Bivariate data
A substantially larger number of candidates took the paper this year: 14% more than in 2015. However, the mean score was virtually identical to that in 2015. Five questions were very popular, with two being attempted by in excess of 90% of the candidates, but once again, all questions were attempted by significant numbers, with only one dipping under 10% attempting it, and every question was answered perfectly by at least one candidate. Most candidates kept to six sensible attempts, although some did several more scoring weakly overall, except in six outstanding cases that earned very high marks.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Given a random variable $X$ with mean $\mu$ and standard deviation $\sigma$, we define the \textit{kurtosis}, $\kappa$, of $X$ by
\[
\kappa = \frac{ \E\big((X-\mu)^4\big)}{\sigma^4} -3 \,.
\]
Show that the random variable $X-a$, where $a$ is a constant, has the same kurtosis as $X$.
\begin{questionparts}
\item Show by integration that a random variable which
is Normally distributed with mean 0 has kurtosis 0.
\item Let $Y_1, Y_2, \ldots, Y_n$ be $n$ independent, identically distributed, random variables with mean 0, and let $T = \sum\limits_{r=1}^n Y_r$. Show that
\[
\E(T^4) = \sum_{r=1}^n \E(Y_r^4) +
6 \sum_{r=1}^{n-1} \sum_{s=r+1}^{n} \E(Y^2_s)
\E(Y^2_r)
\,.
\]
\item Let $X_1$, $X_2$, $\ldots$\,, $X_n$ be $n$ independent, identically distributed, random variables each with kurtosis $\kappa$. Show that the kurtosis of their sum is $\dfrac\kappa n\,$.
\end{questionparts}\begin{align*}
&&\kappa_{X-a} &= \frac{\mathbb{E}\left(\left(X-a-(\mu-a)\right)^4\right)}{\sigma_{X-a}^4}-3 \\
&&&= \frac{\mathbb{E}\left(\left(X-\mu\right)^4\right)}{\sigma_X^4}-3\\
&&&= \kappa_X
\end{align*}
\begin{questionparts}
\item $\,$
\begin{align*}
&& \kappa &= \frac{\mathbb{E}((X-\mu)^4)}{\sigma^4} - 3 \\
&&&= \frac{\mathbb{E}((\mu+\sigma Z-\mu)^4)}{\sigma^4} - 3 \\
&&&= \frac{\mathbb{E}((\sigma Z)^4)}{\sigma^4} - 3 \\
&&&= \mathbb{E}(Z^4)-3\\
&&&= \int_{-\infty}^{\infty} x^4\frac{1}{\sqrt{2\pi}} \exp \left ( - \frac12x^2 \right)\d x -3 \\
&&&= \left [\frac{1}{\sqrt{2\pi}}x^{3} \cdot \left ( -\exp \left ( - \frac12x^2 \right)\right) \right]_{-\infty}^{\infty} + \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty 3x^2 \exp \left ( - \frac12x^2 \right) \d x - 3 \\
&&&= 0 + 3 \textrm{Var}(Z) - 3 =0
\end{align*}
\item $\,$
\begin{align*}
&& \mathbb{E}(T^4) &= \mathbb{E} \left [\left ( \sum\limits_{r=1}^n Y_r\right)^4\right] \\
&&&= \mathbb{E} \left [ \sum_{r=1}^n Y_r^4+\sum_{i\neq j} 4Y_iY_j^3+\sum_{i\neq j} 6Y_i^2Y_j^2+\sum_{i\neq j \neq k} 12Y_iY_jY_k^2 +\sum_{i\neq j\neq k \neq l}24 Y_iY_jY_kY_l\right] \\
&&&= \sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+\sum_{i\neq j} \mathbb{E} \left [ 4Y_iY_j^3\right]+\sum_{i\neq j} \mathbb{E} \left [ 6Y_i^2Y_j^2\right]+\sum_{i\neq j \neq k} \mathbb{E} \left [ 12Y_iY_jY_k^2\right] +\sum_{i\neq j\neq k \neq l} \mathbb{E} \left [ 24 Y_iY_jY_kY_l\right] \\
&&&= \sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+4\sum_{i\neq j} \mathbb{E} \left [ Y_i]\mathbb{E}[Y_j^3\right]+6\sum_{i\neq j} \mathbb{E} \left [ Y_i^2]\mathbb{E}[Y_j^2\right]+12\sum_{i\neq j \neq k} \mathbb{E} \left [ Y_i]\mathbb{E}[Y_j]\mathbb{E}[Y_k^2\right] +24\sum_{i\neq j\neq k \neq l} \mathbb{E} \left [ Y_i]\mathbb{E}[Y_j]\mathbb{E}[Y_k]\mathbb{E}[Y_l\right] \\
&&&= \sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+6\sum_{i\neq j} \mathbb{E} \left [ Y_i^2]\mathbb{E}[Y_j^2\right]
\end{align*}
\item Without loss of generality, we may assume they all have mean zero. Therefore we can consider the sitatuion as in the previous case with $T$ and $Y_i$s. Note that $\mathbb{E}(Y_i^4) = \sigma^4(\kappa + 3)$ and $\textrm{Var}(T) = n \sigma^2$
\begin{align*}
&& \kappa_T &= \frac{\mathbb{E}(T^4)}{(\textrm{Var}(T))^2} - 3 \\
&&&= \frac{\sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+6\sum_{i\neq j} \mathbb{E} \left [ Y_i^2\right]\mathbb{E}\left[Y_j^2\right]}{n^2\sigma^4}-3 \\
&&&= \frac{n\sigma^4(\kappa+3)+6\binom{n}{2}\sigma^4}{n^2\sigma^4} -3\\
&&&= \frac{\kappa}{n} + \frac{3n + \frac{6n(n-1)}{2}}{n^2} - 3 \\
&&&= \frac{\kappa}{n} + \frac{3n^2}{n^2}-3 \\
&&&= \frac{\kappa}{n}
\end{align*}
\end{questionparts}
This was the least popular question being attempted by only half the number attempting question 12, and with slightly less success than for question 11. Most candidates picked up a few marks at the start of the question and then a small number used integration by parts in (i), but others attempted this unsuccessfully trying integration by change of variable. The multiplication of 4π was surprisingly badly done, and the expectation of a constant being zero was similarly surprisingly common.