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2022 Paper 3 Q11
D: 1500.0 B: 1500.0
binomial distribution mean deviation binomial coefficient combinatorial identity expected value symmetry fair coin

A fair coin is tossed \(N\) times and the random variable \(X\) records the number of heads. The mean deviation, \(\delta\), of \(X\) is defined by \[ \delta = \mathrm{E}\big(|X - \mu|\big) \] where \(\mu\) is the mean of \(X\).

  1. Let \(N = 2n\) where \(n\) is a positive integer.
    1. Show that \(\mathrm{P}(X \leqslant n-1) = \frac{1}{2}\big(1 - \mathrm{P}(X=n)\big)\).
    2. Show that \[ \delta = \sum_{r=0}^{n-1}(n-r)\binom{2n}{r}\frac{1}{2^{2n-1}}\,. \]
    3. Show that for \(r > 0\), \[ r\binom{2n}{r} = 2n\binom{2n-1}{r-1}\,. \] Hence show that \[ \delta = \frac{n}{2^{2n}}\binom{2n}{n}\,. \]
  2. Find a similar expression for \(\delta\) in the case \(N = 2n+1\).

Solution:

  1. When \(N = 2n+1\), \(\mu = n +\frac12\) and so \begin{align*} && \delta &= \E[|X-\mu|] \\ &&&= \sum_{i=0}^n (n + \tfrac12 - i) \frac{1}{2^{2n+1}} \binom{2n+1}{i} + \sum_{i=n+1}^{2n+1} (i-n - \tfrac12) \frac{1}{2^{2n+1}} \binom{2n+1}{i} \\ &&&= 2\sum_{i=0}^n (n + \tfrac12 - i) \frac{1}{2^{2n+1}} \binom{2n+1}{i} \\ &&&= \frac{(2n +1)}{2^{2n+1}}\sum_{i=0}^n \binom{2n+1}i - \frac{2}{2^{2n+1}}\sum_{i=0}^n i \binom{2n+1}{i} \\ &&&= \frac{(2n +1)}{2^{2n+1}}\sum_{i=0}^n \binom{2n+1}i - \frac{2}{2^{2n+1}}\sum_{i=1}^n (2n+1) \binom{2n}{i-1} \\ &&&= \frac{(2n +1)}{2^{2n+1}}2^{2n} - \frac{2(2n+1)}{2^{2n+1}}\sum_{i=0}^{n-1} \binom{2n}{i} \\ &&&= \frac{(2n +1)}{2} - \frac{2(2n+1)}{2^{2n+1}} \frac12\left (2^{2n} - \binom{2n}{n} \right) \\ &&&= \frac{2n+1}{2^{2n+1}}\binom{2n}{n} \end{align*}

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