The point \(P\) lies on the circumference of a circle of unit radius and centre \(O\). The angle, \(\theta\), between \(OP\) and the positive \(x\)-axis is a random variable, uniformly distributed on the interval \(0\le\theta<2\pi\).
The cartesian coordinates of \(P\) with respect to \(O\) are \((X,Y)\).
Find the probability density function for \(X\), and calculate \(\var (X)\).
Show that \(X\) and \(Y\) are uncorrelated and discuss briefly whether they are independent.
The points \(P_i\) (\(i=1\), \(2\), \(\ldots\) , \(n\)) are chosen independently on the circumference of the circle, as in part (i), and have cartesian coordinates \((X_i, Y_i)\).
The point \(\overline P\) has coordinates \((\overline X, \overline Y)\), where \(\overline X =\dfrac1n \sum\limits _{i=1}^n X_i\) and \(\overline Y =\dfrac1n \sum\limits _{i=1}^n Y_i\).
Show that \(\overline X\) and \(\overline Y\) are uncorrelated.
Show that, for large \(n\), \(\displaystyle \P\left(\vert \overline X \vert \le \sqrt{\frac2n}\right)\approx 0.95\,\).
Solution:
\(X = \cos \theta\) \(\theta \sim U(0, 2\pi)\). Noting that \(\mathbb{P}(X \geq t ) = \frac{2}{2\pi}\cos^{-1} t\) so \(f_X(t) = \frac{1}{\pi} \frac{1}{\sqrt{1-x^2}}\)
\begin{align*}
&& \E[X] &= 0 \tag{by symmetry} \\
&& \E[X^2] &= \int_0^{2\pi} \cos^2 \theta \frac{1}{2 \pi} \d \theta \\
&&&= \frac{1}{2} \cdot 2\pi \cdot \frac{1}{2\pi} \\
&&&= \frac12 \\
\Rightarrow & &\var[X] &= \frac12 \\
\\
&& \E[XY] &= \int_0^{2\pi} \cos \theta \sin \theta \frac{1}{2 \pi} \d \theta \\
&&&= \frac{1}{4\pi} \int_0^{2\pi} \sin 2\theta \d \theta \\
&&& =0 = \E[X]\E[Y]
\end{align*}
But note that clearly \(X\) and \(Y\) are not independent, since given \(X\) there are only two possible values of \(Y\).
\(\,\) \begin{align*}
&& \E \left [ XY \right] &= \E \left [ \left ( \frac1n \sum_{i=1}^n X_i \right)\left ( \frac1n \sum_{i=1}^n Y_i\right) \right] \\
&&&= \frac{1}{n^2} \sum_{i=1}^n \sum_{j=1}^n \E [X_i Y_j] \\
&&&= 0 = \E[X] \E[Y]
\end{align*}
Therefore \(X\) and \(Y\) are uncorrelated.
Note that \(\E[X_i] = 0, \var[X_i] = \frac12\) so we can apply the central limit theorem to see that \(X \approx N(0, \frac{1}{2n})\), in particular
\begin{align*}
&& 0.95 &\approx \mathbb{P}(|Z| < 2) \\
&&&= \mathbb{P} \left ( \Big |\frac{X}{\sqrt{\frac{1}{2n}}} \Big | < 2 \right ) \\
&&&= \mathbb{P}\left (|X| < \sqrt{\frac{2}{n}} \right)
\end{align*}