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2005 Paper 1 Q14
D: 1516.0 B: 1513.9
probability continuous probability distributions mixed distribution expectation variance median exponential distribution substitution absolute value Gaussian integral

The random variable \(X\) can take the value \(X=-1\), and also any value in the range \(0\le X <\infty\,\). The distribution of \(X\) is given by \[ \P(X=-1) =m \,, \ \ \ \ \ \ \ \P(0\le X\le x) = k(1-\e^{-x})\,, \] for any non-negative number \(x\), where \(k\) and \(m\) are constants, and \(m <\frac12\,\).

  1. Find \(k\) in terms of \(m\).
  2. Show that \(\E(X)= 1-2m\,\).
  3. Find, in terms of \(m\), \(\var (X)\) and the median value of \(X\).
  4. Given that \[ \int_0^\infty y^2 \e^{-y^2} \d y = \tfrac14 \sqrt{ \pi}\;,\] find \(\E\big(\vert X \vert^{\frac12}\big)\,\) in terms of \(m\).

Solution:

  1. We must have the total probability summing to \(1\), therefore \(1 =m + k\) (as \(x \to \infty\)) therefore \(k = 1-m\).
  2. \(\,\) \begin{align*} && \E[X] &= \mathbb{P}(X=-1) \cdot (-1) + \int_0^{\infty} kx e^{-x} \d x \\ &&&= -m + (1-m) = 1-2m \end{align*}
  3. \(\,\) \begin{align*} && \var[X] &= \E[X^2]-\E[X]^2 \\ &&&= \mathbb{P}(X=-1)\cdot(-1)^2 + (1-m)\int_0^{\infty} x^2e^{-x} \d x - (1-2m)^2 \\ &&&= m + (1-m)(1+1^2) - (1-2m)^2 \\ &&&= 3-4m - 1+4m -4m^2 \\ &&&= 2(1-m^2) \end{align*} To find the median \(q\), we need \begin{align*} && \frac12 &= \mathbb{P}(X \leq q) \\ &&&= m + (1-m)(1-e^{-q}) \\ \Rightarrow && e^{-q} &= 1-\frac{\frac12-m}{1-m} \\ &&&= \frac{1-m - \frac12+m}{1-m} \\ &&&= \frac{1}{2(1-m)} \\ \Rightarrow && q &= \ln 2(1-m) \end{align*}
  4. \(\,\) \begin{align*} && \E\left [|X|^{\frac12}\right] &= \mathbb{P}(X=-1) \cdot 1 + \int_0^{\infty} \sqrt{x} (1-m)e^{-x} \d x \\ &&&= m + (1-m)\int_0^\infty \sqrt{x} e^{-x} \d x \\ u^2 = x, \d x = 2u \d u : &&&= m + (1-m) \int_{u=0}^{u=\infty} u e^{-u^2} \cdot 2u \d u \\ &&&= m + 2(1-m) \int_0^{\infty} u^2 e^{-u^2} \d u \\ &&&= m + (1-m)\frac{\sqrt{\pi}}2 \end{align*}

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1996 Paper 2 Q14
D: 1600.0 B: 1500.0
central limit theorem uniform distribution moments expectation binomial theorem inequality mixed distribution normal approximation

The random variable \(X\) is uniformly distributed on \([0,1]\). A new random variable \(Y\) is defined by the rule \[ Y=\begin{cases} 1/4 & \mbox{ if }X\leqslant1/4,\\ X & \mbox{ if }1/4\leqslant X\leqslant3/4\\ 3/4 & \mbox{ if }X\geqslant3/4. \end{cases} \] Find \({\mathrm E}(Y^{n})\) for all integers \(n\geqslant 1\). Show that \({\mathrm E}(Y)={\mathrm E}(X)\) and that \[{\mathrm E}(X^{2})-{\mathrm E}(Y^{2})=\frac{1}{24}.\] By using the fact that \(4^{n}=(3+1)^{n}\), or otherwise, show that \({\mathrm E}(X^{n}) > {\mathrm E}(Y^{n})\) for \(n\geqslant 2\). Suppose that \(Y_{1}\), \(Y_{2}\), \dots are independent random variables each having the same distribution as \(Y\). Find, to a good approximation, \(K\) such that \[{\rm P}(Y_{1}+Y_{2}+\cdots+Y_{240000} < K)=3/4.\]

Solution: \begin{align*} && \E[Y^n] &= \frac14 \cdot \frac1{4^n} + \frac14 \cdot \frac{3^n}{4^n} + \frac12 \int_{1/4}^{3/4}2 y^n \d y \\ &&&= \frac{3^n+1}{4^{n+1}} + \left [ \frac{y^{n+1}}{n+1} \right]_{1/4}^{3/4} \\ &&&= \frac{3^n+1}{4^{n+1}} + \frac{3^{n+1}-1}{(n+1)4^{n+1}} \end{align*} \begin{align*} && \E[Y] &= \frac{3+1}{16} + \frac{9-1}{2 \cdot 16} \\ &&&= \frac{1}{4} + \frac{1}{4} = \frac12 = \E[X] \end{align*} \begin{align*} && \E[X^2] &= \int_0^1 x^2 \d x = \frac13 \\ && \E[Y^2] &= \frac{9+1}{64} + \frac{27-1}{3 \cdot 64} = \frac{56}{3 \cdot 64} = \frac{7}{24} \\ \Rightarrow && \E[X^2] - \E[Y^2] &= \frac13 - \frac{7}{24} = \frac{1}{24} \end{align*} \begin{align*} && \E[X^n] &= \frac{1}{n+1} \\ && \E[Y^n] &= \frac{1}{n+1} \frac{1}{4^{n+1}}\left ( (n+1)(3^n+1)+3^{n+1}-1 \right) \\ &&&= \frac{1}{n+1} \frac{1}{4^{n+1}}\left ( 3^{n+1} + (n+1)3^n +n \right) \\ \\ && (3+1)^{n+1} &= 3^{n+1} + (n+1)3^n + \cdots + (n+1) \cdot 3 + 1 \\ &&&> 3^{n+1} + (n+1)3^n + n + 1 \end{align*} if \(n \geq 2\) Notice that by the central limit theorem: \begin{align*} &&\frac{1}{240\,000} \sum_{i=1}^{240\,000} Y_i &\sim N \left ( \frac12, \frac{1}{24 \cdot 240\,000}\right) \\ \Rightarrow && \mathbb{P}\left (\frac{\frac{1}{240\,000} \sum_{i=1}^{240\,000} Y_i - \frac12}{\frac1{24} \frac{1}{100}} \leq \frac23 \right) &\approx 0.75 \\ \Rightarrow && \mathbb{P} \left ( \sum_i Y_i \leq 240\,000 \cdot \left ( \frac2{3} \frac1{2400}+\frac12 \right) \right ) & \approx 0.75 \\ \Rightarrow && K &= 120\,000 + 66 \\ &&&\approx 120\,066 \end{align*}

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