Year: 2017
Paper: 3
Question Number: 13
Course: LFM Stats And Pure
Section: Continuous Uniform Random Variables
The total entry was only very slightly smaller than that of 2016, which was a record entry, but was still over 10% more than 2015. No question was attempted by in excess of 90%, although two were very popular and also five others were attempted by 60% or more. No question was generally avoided with even the least popular one attracting more than 10% of the entry. Less than 10% of candidates attempted more than 7 questions, and, apart from 18 exceptions, those doing so did not achieve very good totals and seemed to be 'casting around' to find things they could do: the 18 exceptions were very strong candidates who were generally achieving close to full marks on all the questions they attempted. The general trend was that those with six attempts fared better than those with more than six.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
The random variable $X$ has mean $\mu$ and
variance $\sigma^2$, and the function ${\rm V}$
is defined, for $-\infty < x < \infty$, by
\[
{\rm V}(x) = \E \big( (X-x)^2\big)
.
\]
Express ${\rm V}(x)$ in terms of $x$, $ \mu$ and $\sigma$.
The random variable $Y$ is defined by $Y={\rm V}(X)$.
Show that
\[
\E(Y) = 2 \sigma^2
%\text{ \ \ and \ \ }
%\Var(Y) = \E(X-\mu)^4 -\sigma^4
.
\tag{$*$}
\]
Now suppose that $X$ is uniformly distributed on the interval $0\le x \le1\,$.
Find ${\rm V}(x)\,$.
Find also the probability density function of $Y\!$ and use it to
verify that $(*)$ holds in this case.\begin{align*}
{\rm V}(x) &= \E \big( (X-x)^2\big) \\
&= \E \l X^2 - 2xX + x^2\r \\
&= \E [ X^2 ]- 2x\E[X] + x^2 \\
&= \sigma^2+\mu^2 - 2x\mu + x^2 \\
&= \sigma^2 + (\mu - x)^2
\end{align*}
\begin{align*}
\E[Y] &= \E[\sigma^2 + (\mu - X)^2] \\
&= \sigma^2 + \E[(\mu - X)^2]\\
&= \sigma^2 + \sigma^2 \\
&= 2\sigma^2
\end{align*}
If $X \sim U(0,1)$ then $V(x) = \frac{1}{12} + (\frac12 - x)^2$.
\begin{align*}
\P(Y \leq y) &= \P(\frac1{12} + (\frac12 - X)^2 \leq y) \\
&= \P((\frac12 -X)^2 \leq y - \frac1{12}) \\
&= \P(|\frac12 -X| \leq \sqrt{y - \frac1{12}}) \\
&= \begin{cases} 1 & \text{if } y - \frac1{12} > \frac14 \\
2 \sqrt{y - \frac1{12}} & \text{if } \frac14 > y - \frac1{12} > 0 \\
\end{cases} \\
&= \begin{cases} 1 & \text{if } y> \frac13 \\
\sqrt{4y - \frac1{3}} & \text{if } \frac13 > y > \frac1{12} \\
\end{cases}
\end{align*}
Therefore $f_Y(y) = \begin{cases} \frac{2}{\sqrt{4y-\frac{1}{3}}} & \text{if } \frac1{12} < y < \frac13 \\
0 & \text{otherwise} \end{cases}$
\begin{align*}
\E[Y] &= \int_{1/12}^{1/3} \frac{2x}{\sqrt{4x-\frac13}} \, dx \\
&= 2\int_{u = 0}^{u=1} \frac{\frac{1}{4}u +\frac1{12}}{\sqrt{u}} \,\frac{1}{4} du \tag{$u = 4x - \frac13, \frac{du}{dx} = 4$}\\
&= \frac{1}{2 \cdot 12}\int_{u = 0}^{u=1} 3\sqrt{u} +\frac{1}{\sqrt{u}} \, du \\
&= \frac{1}{2 \cdot 12} \left [2 u^{3/2} + 2u^{1/2} \right ]_0^1 \\
&= \frac{1}{2 \cdot 12} \cdot 4 \\
&= \frac{2}{12}
\end{align*}
as required
A sixth of the candidates tried this, scoring slightly less well than was done on question 11. Almost everyone found V correctly and the required result for E. Similar success was demonstrated finding V in the uniform case. A lot did not then attempt to find the probability density function, but most who spotted it attempting to calculate the cumulative distribution function of first and then differentiate could do it. It was encouraging that so many correctly found the range of . A variety of methods of integration were used for the final result with varying success.