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Question
The discrete random variables \({X_n},{\text{ }}n \in {\mathbb{Z}^ + }\) have probability generating functions given by \({G_n}(t) = \frac{t}{n}\left( {\frac{{{t^n} – 1}}{{t – 1}}} \right)\).
Let \({X_{n – 1}}\) and \({X_{n + 1}}\) be independent.
Use the formula for the sum of a finite geometric series to show that
\[{\text{P}}({X_n} = k) = \left\{ {\begin{array}{*{20}{l}} {\frac{1}{n}}&{{\text{for }}1 \leqslant k \leqslant n} \\ 0&{{\text{otherwise}}} \end{array}.} \right.\]
Find \({\text{E}}({X_n})\).
Find the set of values of \(n\) for which \({\text{E}}({X_{n – 1}} \times {X_{n + 1}}) < 2n\).
Answer/Explanation
Markscheme
using \(\left( {\frac{{{t^n} – 1}}{{t – 1}}} \right) = 1 + t + {t^2} + \ldots {t^{n – 1}}\) M1
\({G_n}(t) = 0 + \frac{t}{n} + \frac{{{t^2}}}{n} + \frac{{{t^3}}}{n} + \ldots \frac{{{t^n}}}{n} + 0 \times {t^{n + 1}} + 0 \times \ldots \) A1A1
Note: A1 for the non-zero terms, A1 for the observation that all other terms are zero.
the statement that the coefficient of \({t^k}\) gives \({\text{P}}({X_n} = k)\) R1
hence \({\text{P}}({X_n} = k) = \left\{ {\begin{array}{*{20}{l}} {\frac{1}{n}}&{{\text{for }}1 \leqslant k \leqslant n} \\ 0&{{\text{otherwise}}} \end{array}} \right.\) AG
[4 marks]
\({\text{E}}({X_n}) = 0 \times 0 + 1 \times \frac{1}{n} + 2 \times \frac{1}{n} + 3 \times \frac{1}{n} + \ldots n \times \frac{1}{n} + (n + 1) \times 0 + \ldots \times 0\) (M1)(A1)
\( = \frac{1}{n} \times \sum\limits_{k = 1}^{k = n} k \)
\( = \frac{1}{n} \times \frac{1}{2}n(n + 1) = \frac{{n + 1}}{2}\) A1
Note: Accept use of \(G'(1)\).
[3 marks]
\({X_{n – 1}}\) and \({X_{n + 1}}\) are independent \( \Rightarrow {\text{E}}({X_{n – 1}} \times {X_{n + 1}}) = {\text{E}}({X_{n – 1}}) \times {\text{E}}({X_{n + 1}})\) M1
\( = \frac{n}{2} \times \frac{{n + 2}}{2}\) A1
required to solve \({n^2} < 6n{\text{ }}({\text{or }}n + 2 < 8)\) M1
solution: \((2 \leqslant ){\text{ }}n < 6\) A1
[4 marks]