STEP III 2007 question 12 solution - The Student Room
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STEP III 2007 question 12 solution

TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 2007 question 12 solution

\displaystyle E(N) = \sum_{k=1}^{2n-1} \frac{k}{2n-1} = \frac{1}{2n-1}\frac{1}{2}2n(2n-1) = n


\displaystyle E(N^2) = \sum_{k=1}^{2n-1} \frac{k^2}{2n-1} = \frac{1}{2n-1}\frac{1}{6}(2n-1)(2n)(4n-1) = \frac{n}{3}(4n-1)

To find the expectation of Y we condition on the value of N:

\displaystyle E(Y) = \sum_{k=1}^{2n-1} E(Y | N=k) P(N=k) = \sum_{k=1}^{2n-1} \frac{k \mu}{2n-1} = n\mu


\displaystyle E(YN) = \sum_{k=1}^{2n-1} E(YN | N=k)P(N=k) = \sum_{k=1}^{2n-1} \frac{k^2 \mu}{2n-1} = \frac{n \mu}{3}(4n-1)

\displaystyle \text{Cov}(Y,N) = E(YN) - E(Y)E(N) = \frac{n \mu}{3}(4n-1) - n^2 \mu = \frac{n \mu}{3}(n-1)

\displaystyle E((\sum_{i=1}^{k} X_i)^2) = E((\sum_{i=1}^{k} X_i^2) + (\sum_{i \neq j} X_iX_j))\\ = k(\sigma^2 + \mu^2) + (k^2-k)\mu^2 = k\sigma^2 + k^2 \mu^2

\displaystyle E(Y^2) = \frac{1}{2n-1} \sum_{k=1}^{2n-1} k\sigma^2 + k^2 \mu^2\\ = n\sigma^2 + \frac{1}{3}(n)(4n-1)\mu^2\\

\displaystyle \text{Var}(Y) = E(Y^2) - E(Y)^2\\ = n\sigma^2 + \frac{1}{3}(n)(4n-1)\mu^2 - n^2\mu^2\\ = n\sigma^2 + \frac{n}{3}(n-1)\mu^2

Solution by SsEe.

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