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2023 Paper 3 Q11
D: 1500.0 B: 1500.0
poisson distribution expectation series manipulation conditional probability probability generating functions exponential series discrete random variables

Show that \[\sum_{k=1}^{\infty} \frac{k+1}{k!}\, x^k = (x+1)\mathrm{e}^x - 1\,.\] In the remainder of this question, \(n\) is a fixed positive integer.

  1. Random variable \(Y\) has a Poisson distribution with mean \(n\). One observation of \(Y\) is taken. Random variable \(D\) is defined as follows. If the observed value of \(Y\) is zero then \(D = 0\). If the observed value of \(Y\) is \(k\), where \(k \geqslant 1\), then a fair \(k\)-sided die (with sides numbered \(1\) to \(k\)) is rolled once and \(D\) is the number shown on the die.
    1. Write down \(\mathrm{P}(D = 0)\).
    2. Show, from the definition of the expectation of a random variable, that \[\mathrm{E}(D) = \sum_{d=1}^{\infty} \left[ d \sum_{k=d}^{\infty} \left( \frac{1}{k} \cdot \frac{n^k}{k!}\, \mathrm{e}^{-n} \right) \right].\] Show further that \[\mathrm{E}(D) = \sum_{k=1}^{\infty} \left( \frac{1}{k} \cdot \frac{n^k}{k!}\, \mathrm{e}^{-n} \sum_{d=1}^{k} d \right).\]
    3. Show that \(\mathrm{E}(D) = \frac{1}{2}(n + 1 - \mathrm{e}^{-n})\).
  2. Random variables \(X_1, X_2, \ldots, X_n\) all have Poisson distributions. For each \(k \in \{1, 2, \ldots, n\}\), the mean of \(X_k\) is \(k\). A fair \(n\)-sided die, with sides numbered \(1\) to \(n\), is rolled. When \(k\) is the number shown, one observation of \(X_k\) is recorded. Let \(Z\) be the number recorded.
    1. Find \(\mathrm{P}(Z = 0)\).
    2. Show that \(\mathrm{E}(Z) > \mathrm{E}(D)\).

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2018 Paper 3 Q8
D: 1700.0 B: 1516.0
integration Taylor series fractional part substitution partial fractions series manipulation periodic function floor function

In this question, you should ignore issues of convergence.

  1. Let \[ I = \int_0^1 \frac{\f(x^{-1}) } {1+x} \, \d x \,, \] where \(\f(x)\) is a function for which the integral exists. Show that \[ I = \sum_{n=1}^\infty \int_n^{n+1} \frac{\f(y) } {y(1+y)}\, \d y \] and deduce that, if \(\f(x) = \f(x+1)\) for all \(x\), then \[ I= \int_0^1 \frac{\f(x)} {1+x} \, \d x \,. \]
  2. The {\em fractional part}, \(\{x\}\), of a real number \(x\) is defined to be \(x-\lfloor x\rfloor\) where \(\lfloor x \rfloor\) is the largest integer less than or equal to \(x\). For example \(\{3.2\} = 0.2\) and \(\{3\}=0\,\). Use the result of part (i) to evaluate \[ \displaystyle \int _0^1 \frac { \{x^{-1}\}}{1+x}\, \d x \text{ and } \displaystyle \int _0^1 \frac { \{2x^{-1}\}}{1+x}\, \d x \,. \]
  3. (Bonus) Use the same method to evaluate \[ \int_0^1 \frac {x\{x^{-1}\}}{1-x^2} \, \d x \,. \]
  4. (Bonus - harder) Use the same method to evaluate \[ \int_0^1 \frac {x^2\{x^{-1}\}}{1-x^2} \, \d x \,. \]

Solution:

  1. \begin{align*} && I &= \int_0^1 \frac{f(x^{-1})}{1+x} \d x \\ u = x^{-1}, \d u = -x^{-2} \d x: &&&= \int_{\infty}^1 \frac{f(u)}{1+\frac{1}{u}} \frac{-1}{u^2} \d u \\ &&&= \int_1^{\infty} \frac{f(u)}{u(1+u)} \d u \\ &&&= \sum_{n=1}^{\infty} \int_n^{n+1} \frac{f(u)}{u(u+1)} \d u \\ \\ \text{if} f(x) = f(x+1)\, \forall x && &=\sum_{n=1}^{\infty} \int_{0}^1 \frac{f(x+n)}{(x+n)(x+n+1)} \d x \\ &&&= \sum_{n=1}^\infty \int_0^1 \frac{f(x)}{(x+n)(x+n+1)} \d x \\ &&&= \int_0^1 f(x) \l \sum_{n=1}^{\infty} \frac{1}{(x+n)(x+n+1)}\r \d x \\ &&&= \int_0^1 f(x) \l \sum_{n=1}^{\infty} \l \frac{1}{x+n} - \frac{1}{x+n+1} \r\r \d x \\ &&&= \int_0^1 f(x) \l \frac{1}{x+1} \r \d x \\ &&&= \int_0^1\frac{f(x)}{x+1} \d x \\ \end{align*}
  2. Since the fractional part is periodic with period \(1\), we can say \begin{align*} && \int_0^1 \frac{\{ x^{-1} \} }{1+x} \d x &= \int_0^1 \frac{\{ x\}}{x+1} \d x \\ &&&= \int_0^1 \frac{x}{x+1} \d x \\ &&&= \int_0^1 1-\frac{1}{x+1} \d x \\ &&&= [x - \ln (1+x) ]_0^1 \\ &&&= 1 - \ln 2 \end{align*} \begin{align*} && \int_0^1 \frac{\{ 2x^{-1}\}}{1+x} \d x &= \int_0^1 \frac{\{2x\}}{x+1} \d x \\ &&&= \int_0^{1/2} \frac{2x}{x+1} \d x +\int_{1/2}^{1} \frac{2x-1}{x+1} \d x \\ &&&= 2\int_0^1 \frac{x}{x+1} \d x + \int_{1/2}^1 \frac{-1}{x+1} \d x \\ &&&= 2 - 2\ln 2 - \l \ln 2 - \ln \tfrac32 \r \\ &&&= 2 - 4 \ln 2 + \ln 3 \\ &&&= 2 + \ln \tfrac {3}{16} \end{align*}
  3. \begin{align*} && \int_0^1 \frac{x \{ x^{-1} \} }{1-x^2} \d x &= \frac12 \l \int_0^1 \frac{ \{ x^{-1} \}}{1-x} - \frac{\{x^{-1} \}}{1+x} \d x\r \end{align*} Consider for \(f\) periodic with period \(1\) \begin{align*} \int_0^1 \frac{f(x^{-1})}{1-x} \d x &= \int_1^{\infty} \frac{f(u)}{u(u-1)} \d u \\ &= \sum_{n=1}^{\infty}\int_{n}^{n+1} \frac{f(u)}{u(u-1)} \d u \\ &= \sum_{n=1}^{\infty}\int_{0}^{1} \frac{f(u)}{(u+n)(u+n-1)} \d u \\ &= \int_{0}^{1} \sum_{n=1}^{\infty}\frac{f(u)}{(u+n)(u+n-1)} \d u \\ &= \int_{0}^{1} f(u) \sum_{n=1}^{\infty}\l\frac{1}{u+n-1} - \frac{1}{u+n} \r\d u \\ &= \int_0^1 \frac{f(u)}{u} \d u \end{align*} So we have \begin{align*} && \int_0^1 \frac{x \{ x^{-1} \} }{1-x^2} \d x &= \frac12 \l \int_0^1 \frac{ \{ x^{-1} \}}{1-x} - \frac{\{x^{-1} \}}{1+x} \d x \r \\ &&&= \frac12 \int_0^1 \frac{\{ x \}}{x} \d x - \frac12 (1 - \ln 2) \\ &&&= \frac12 - \frac12 + \frac12 \ln 2 \\ &&&= \frac12 \ln 2 \end{align*}

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2012 Paper 3 Q4
D: 1700.0 B: 1500.0
Taylor series exponential series summation series manipulation partial fractions logarithm e constant factorial

  1. Show that \[ \sum_{n=1} ^\infty \frac{n+1}{n!} = 2\e - 1 \] and \[ \sum _{n=1}^\infty \frac {(n+1)^2}{n!} = 5\e-1\,. \] Sum the series $\displaystyle \sum _{n=1}^\infty \frac {(2n-1)^3}{n!} \,.$
  2. Sum the series $\displaystyle \sum_{n=0}^\infty \frac{(n^2+1)2^{-n}}{(n+1)(n+2)}\,$, giving your answer in terms of natural logarithms.

Solution:

  1. \begin{align*} \sum_{n=1}^{\infty} \frac{n+1}{n!} &= \sum_{n=1}^\infty \left ( \frac{1}{(n-1)!} + \frac{1}{n!} \right) \\ &= \sum_{n=0}^\infty \frac{1}{n!} + \sum_{n=1}^\infty \frac{1}{n!} \\ &= \sum_{n=0}^\infty \frac{1}{n!} + \sum_{n=0}^\infty \frac{1}{n!} - 1 \\ &= e + e - 1 \\ &= 2e-1 \end{align*} \begin{align*} \sum_{n=1}^{\infty} \frac{(n+1)^2}{n!} &= \sum_{n=1}^{\infty} \frac{n(n-1) + 3n + 1}{n!} \\ &= \sum_{n=2}^{\infty} \frac{1}{(n-2)!} + 3 \sum_{n=1}^\infty \frac1{(n-1)!} + \sum_{n=1}^\infty \frac{1}{n!} \\ &= \sum_{n=0}^{\infty} \frac{1}{n!} + 3 \sum_{n=0}^\infty \frac1{n!} + \sum_{n=0}^\infty \frac{1}{n!} -1 \\ &= 5e-1 \end{align*} \begin{align*} \sum_{n=1}^\infty \frac{(2n-1)^3}{n!} &= \sum_{n=1}^\infty \frac{8n^3-12n^2+6n-1}{n!} \\ &= \sum_{n=1}^\infty \frac{8n(n-1)(n-2)+12n^2-10n-1}{n!} \\ &= \sum_{n=1}^\infty \frac{8n(n-1)(n-2)+12n(n-1)+2n-1}{n!} \\ &= 8 e+12e+2e-(e-1) \\ &=21e+1 \end{align*}
  2. \begin{align*} \frac{n^2+1}{(n+1)(n+2)} &= \frac{n^2+3n+2-3n-1}{(n+1)(n+2)}\\ &= 1 - \frac{3n+1}{(n+1)(n+2)} \\ &= 1 + \frac{2}{n+1} - \frac{5}{n+2} \\ -\log(1-x) &= \sum_{n=1}^\infty \frac1{n}x^{n} \\ \log(2) &= \sum_{n=1}^\infty \frac{2^{-n}}{n} \\ \sum_{n=0}^\infty \frac{(n^2+1)2^{-n}}{(n+1)(n+2)} &= \sum_{n=0}^{\infty} 2^{-n} + 2 \sum_{n=0}^\infty \frac{2^{-n}}{n+1}-5 \sum_{n=0}^\infty \frac{2^{-n}}{n+2} \\ &= 2 + 2\log2-5 \sum_{n=2}^\infty \frac{2^{-n+2}}{n} \\ &= 2 + 2 \log 2 - 5 \left (2\log 2 - 2 \right) \\ &= 12-8\log2 \end{align*}

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1999 Paper 3 Q14
D: 1700.0 B: 1487.9
probability Poisson distribution expected value geometric distribution summation generating functions conditional expectation series manipulation

In the basic version of Horizons (H1) the player has a maximum of \(n\) turns, where \(n \ge 1\). At each turn, she has a probability \(p\) of success, where \(0 < p < 1\). If her first success is at the \(r\)th turn, where \(1 \le r \le n\), she collects \(r\) pounds and then withdraws from the game. Otherwise, her winnings are nil. Show that in H1, her expected winnings are $$ p^{-1}\left[1+nq^{n+1}-(n+1)q^n\right]\quad\hbox{pounds}, $$ where \(q=1-p\). The rules of H2 are the same as those of H1, except that \(n\) is randomly selected from a Poisson distribution with parameter \(\lambda\). If \(n=0\) her winnings are nil. Otherwise she plays H1 with the selected \(n\). Show that in H2, her expected winnings are $$ {1 \over p}{\left(1-{\e^{-{\lambda}p}}\right)} -{{\lambda}q}{\e^{-{\lambda}p}} \quad\hbox{pounds}. $$

Solution: \begin{align*} && \E[H1] &= \sum_{r=1}^n r \cdot \mathbb{P}(\text{first success on }r\text{th turn}) \\ &&&= \sum_{r=1}^n r \cdot q^{r-1}p \\ &&&= p\sum_{r=1}^n r q^{r-1} \\ \\ && \frac{1-x^{n+1}}{1-x} &= \sum_{r=0}^n x^r \\ \Rightarrow && \sum_{r=1}^n r x^{r-1} &= \frac{-(n+1)x^n(1-x) +(1-x^{n+1})}{(1-x)^2} \\ &&&= \frac{1-(n+1)x^n+nx^{n+1}}{(1-x)^2} \\ \\ && \E[H1] &= p\sum_{r=1}^n r q^{r-1} \\ &&&= p\frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} \\ &&&= p^{-1}(1-(n+1)q^{n} + nq^{n+1}) \end{align*} Not that if \(n =0\) , the formula for \(\E[H1] = 0\). So \begin{align*} && \E[H2] &= \E[\E[H1|n=N]] \\ &&&= p^{-1}\E \left [ 1-(N+1)q^{N} + Nq^{N+1}\right] \\ &&&= p^{-1}\E \left [ 1-((1-q)N+1)q^{N} \right] \\ &&&= p^{-1}\left (1 - p\E[Nq^N] - G_{Po(\lambda)}(q) \right) \\ &&&= p^{-1}(1-e^{-\lambda(1-q)}) - \E[Nq^N] \\ &&&= p^{-1}(1-e^{-\lambda(1-q)}) - q\lambda e^{-\lambda(1-q)} \\ &&&= p^{-1}(1-e^{-\lambda p}) - q\lambda e^{-\lambda p} \end{align*}

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1989 Paper 3 Q9
D: 1700.0 B: 1516.0
sequences sum to infinity power series arithmetico-geometric series logarithmic series binomial series series manipulation

Obtain the sum to infinity of each of the following series.

  1. \(1{\displaystyle +\frac{2}{2}+\frac{3}{2^{2}}+\frac{4}{2^{3}}+\cdots+\frac{r}{2^{r-1}}+\cdots;}\)
  2. \(1{\displaystyle +\frac{1}{2}\times\frac{1}{2}+\frac{1}{3}\times\frac{1}{2^{2}}+\cdots+\frac{1}{r}\times\frac{1}{2^{r-1}}+\cdots;}\)
  3. \({\displaystyle \dfrac{1\times3}{2!}\times\frac{1}{3}+\frac{1\times3\times5}{3!}\frac{1}{3^{2}}+\cdots+\frac{1\times3\times\cdots\times(2k-1)}{k!}\times\frac{1}{3^{k-1}}+\cdots.}\)
[Questions of convergence need not be considered.]

Solution:

  1. \begin{align*} && \frac1{1-x} &= \sum_{r=0}^{\infty} x^r \\ \underbrace{\Rightarrow}_{\frac{\d}{\d x}} && \frac{1}{(1-x)^2} &= \sum_{r=0}^\infty rx^{r-1} \\ \underbrace{\Rightarrow}_{x = \frac12} && 4 &= \sum_{r=0}^{\infty} \frac{r}{2^{r-1}} \end{align*}
  2. \begin{align*} && \frac1{1-x} &= \sum_{r=1}^{\infty} x^{r-1} \\ \underbrace{\Rightarrow}_{\int} && -\ln (1-x) &= \sum_{r=1}^{\infty} \frac1{r} x^r \\ \underbrace{\Rightarrow}_{x = \frac12} && \ln 2 &= \sum_{r=1}^{\infty} \frac1{r } \times \frac{1}{ 2^{r}} \\ \Rightarrow && 2 \ln 2 &= \sum_{r=1}^{\infty} \frac1{r } \times \frac{1}{ 2^{r-1}} \\ \end{align*}
  3. \begin{align*} && (1-x)^{-1/2} &= 1 + \frac{(-\tfrac12)}{1!} (-x) +\frac{(-\tfrac12)(-\tfrac32)}{2!}(-x)^2 + \cdots \\ &&&= \sum_{r=0}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \cdot (2r-1)}{2^rr!} x^r \\ \underbrace{\Rightarrow}_{x = \frac23} && \sqrt{3} &= \sum_{r=0}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \cdot (2r-1)}{r!} \frac1{3^r} \\ &&&= 1 + \frac{1}{1!} \frac23 + \frac13 \sum_{r=2}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \cdot (2r-1)}{r!} \frac1{3^{r-1}} \\ \Rightarrow && 3\sqrt{3}-5 &= \sum_{r=2}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \cdot (2r-1)}{r!} \frac1{3^{r-1}} \\ \end{align*}

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