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2023 Paper 2 Q11
D: 1500.0 B: 1500.0
normal distribution binomial distribution variance mixture distribution normal approximation probability limit

  1. \(X_1\) and \(X_2\) are both random variables which take values \(x_1, x_2, \ldots, x_n\), with probabilities \(a_1, a_2, \ldots, a_n\) and \(b_1, b_2, \ldots, b_n\) respectively. The value of random variable \(Y\) is defined to be that of \(X_1\) with probability \(p\) and that of \(X_2\) with probability \(q = 1-p\). If \(X_1\) has mean \(\mu_1\) and variance \(\sigma_1^2\), and \(X_2\) has mean \(\mu_2\) and variance \(\sigma_2^2\), find the mean of \(Y\) and show that the variance of \(Y\) is \(p\sigma_1^2 + q\sigma_2^2 + pq(\mu_1 - \mu_2)^2\).
  2. To find the value of random variable \(B\), a fair coin is tossed and a fair six-sided die is rolled. If the coin shows heads, then \(B = 1\) if the die shows a six and \(B = 0\) otherwise; if the coin shows tails, then \(B = 1\) if the die does not show a six and \(B = 0\) if it does. The value of \(Z_1\) is the sum of \(n\) independent values of \(B\), where \(n\) is large. Show that \(Z_1\) is a Binomial random variable with probability of success \(\frac{1}{2}\). Using a Normal approximation, show that the probability that \(Z_1\) is within \(10\%\) of its mean tends to \(1\) as \(n \longrightarrow \infty\).
  3. To find the value of random variable \(Z_2\), a fair coin is tossed and \(n\) fair six-sided dice are rolled, where \(n\) is large. If the coin shows heads, then the value of \(Z_2\) is the number of dice showing a six; if the coin shows tails, then the value of \(Z_2\) is the number of dice not showing a six. Use part (i) to write down the mean and variance of \(Z_2\). Explain why a Normal distribution with this mean and variance will not be a good approximation to the distribution of \(Z_2\). Show that the probability that \(Z_2\) is within \(10\%\) of its mean tends to \(0\) as \(n \longrightarrow \infty\).

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1990 Paper 1 Q8
D: 1500.0 B: 1516.0
trigonometric identity double angle formula proof by induction infinite product small angle approximation Vieta product limit

Show that \[ \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)=\frac{\sin\alpha}{4\sin\left(\dfrac{\alpha}{4}\right)}\,, \] where \(\alpha\neq k\pi\), \(k\) is an integer. Prove that, for such \(\alpha\), \[ \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right)=\frac{\sin\alpha}{2^{n}\sin\left(\dfrac{\alpha}{2^{n}}\right)}\,, \] where \(n\) is a positive integer. Deduce that \[ \alpha=\frac{\sin\alpha}{\cos\left(\dfrac{\alpha}{2}\right)\cos\left(\dfrac{\alpha}{4}\right)\cos\left(\dfrac{\alpha}{8}\right)\cdots}\,, \] and hence that \[ \frac{\pi}{2}=\frac{1}{\sqrt{\frac{1}{2}}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}}\cdots}\,. \]

Solution: \begin{align*} &&\sin \alpha &= 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \\ &&&= 4 \sin \frac{\alpha}{4} \cos \frac{\alpha}{4} \cos \frac{\alpha}{2} \\ \Rightarrow && \frac{\sin \alpha}{4 \sin \frac{\alpha}{4}} &= \cos \frac{\alpha}{2} \cos \frac{\alpha}{4} \end{align*} We proceed by induction on \(n\). Clearly this is true for \(n = 1\) (as we just established). Assume it is true for \(n=k\). Then: \begin{align*} && \frac{\sin \alpha}{2^n \sin \frac{\alpha}{2^n}} &= \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right) \\ \Rightarrow && \frac{\sin \alpha}{2^{n+1} \sin \frac{\alpha}{2^{n+1}} \cos \frac{\alpha}{2^{n+1}}} &= \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right) \\ \Rightarrow && \frac{\sin \alpha}{2^{n+1} \sin \frac{\alpha}{2^{n+1}} } &= \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right)\cos \left ( \frac{\alpha}{2^{n+1}} \right) \\ \end{align*} Therefore it is true for \(n=k+1\) Therefore since it is true for \(n=1\) and if it is true for \(n=k\) it is also true for \(n=k+1\) by the principle of mathematical induction it is true for all \(n \geq 1\) \begin{align*} \lim_{n \to \infty} \frac{\sin \alpha}{\cos\left(\frac{\alpha}{2}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right)} &= \lim_{n \to \infty} 2^n \sin \frac{\alpha}{2^n} \\ &= \lim_{n \to \infty} \alpha \frac{\sin \frac{\alpha}{2^n}}{\frac{\alpha}{2^n}} \\ &= \alpha \lim_{t \to 0} \frac{\sin t}{t} \\ &= \alpha \end{align*} When \(\alpha = \frac{\pi}{2}\) notice that \(\sin \alpha =1\), \(\cos \frac{\alpha}{2} = \sqrt{\frac12}\) and \(2\cos^2 \frac{\alpha}{2^{n+1}}-1 = \cos \frac{\alpha}{2} \Rightarrow \cos \frac{\alpha}{2^{n+1}} = \sqrt{\frac12 + \cos \frac{\alpha}{2^n}}\) exactly the series we see.

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