| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moment generating functions |
| Type | Use MGF to find moments |
| Difficulty | Standard +0.3 This is a straightforward application of MGF techniques: differentiating the given MGF to find moments (routine calculus), then using the independence property that MGFs multiply. Both parts follow standard procedures taught in S4 with no novel insight required, making it slightly easier than average. |
| Spec | 5.02i Poisson distribution: random events model5.02m Poisson: mean = variance = lambda |
| Answer | Marks | Guidance |
|---|---|---|
| Mean \(= \lambda\), Variance \(= \lambda\) | M1 M1 M1 A1 A1 A1 A1 | For \(M_X(t) = e^{\lambda(e^t - 1)}\): \(M_X'(t) = \lambda e^t e^{\lambda(e^t-1)}\), so \(E(X) = M_X'(0) = \lambda e^0 \cdot e^0 = \lambda\). \(M_X''(t) = \lambda e^t e^{\lambda(e^t-1)} + \lambda e^t \cdot \lambda e^t e^{\lambda(e^t-1)} = \lambda e^t e^{\lambda(e^t-1)}(1 + \lambda e^t)\). So \(E(X^2) = M_X''(0) = \lambda(1 + \lambda)\). Therefore \(\text{Var}(X) = \lambda(1 + \lambda) - \lambda^2 = \lambda\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(M_Y(t) = e^{5\lambda(e^t - 1)}\) | M1 M1 A1 | For independent random variables, \(M_Y(t) = [M_X(t)]^5 = [e^{\lambda(e^t-1)}]^5 = e^{5\lambda(e^t-1)}\) |
**(i) Use MGF to verify mean and variance of Poisson($\lambda$)**
| Mean $= \lambda$, Variance $= \lambda$ | M1 M1 M1 A1 A1 A1 A1 | For $M_X(t) = e^{\lambda(e^t - 1)}$: $M_X'(t) = \lambda e^t e^{\lambda(e^t-1)}$, so $E(X) = M_X'(0) = \lambda e^0 \cdot e^0 = \lambda$. $M_X''(t) = \lambda e^t e^{\lambda(e^t-1)} + \lambda e^t \cdot \lambda e^t e^{\lambda(e^t-1)} = \lambda e^t e^{\lambda(e^t-1)}(1 + \lambda e^t)$. So $E(X^2) = M_X''(0) = \lambda(1 + \lambda)$. Therefore $\text{Var}(X) = \lambda(1 + \lambda) - \lambda^2 = \lambda$ |
**(ii) Find MGF of $Y$ where $Y$ is sum of 5 independent Poisson observations**
| $M_Y(t) = e^{5\lambda(e^t - 1)}$ | M1 M1 A1 | For independent random variables, $M_Y(t) = [M_X(t)]^5 = [e^{\lambda(e^t-1)}]^5 = e^{5\lambda(e^t-1)}$ |5 The random variable $X$ has a Poisson distribution with mean $\lambda$. It is given that the moment generating function of $X$ is $e ^ { \lambda \left( e ^ { t } - 1 \right) }$.\\
(i) Use the moment generating function to verify that the mean of $X$ is $\lambda$, and to show that the variance of $X$ is also $\lambda$.\\
(ii) Five independent observations of $X$ are added to produce a new variable $Y$. Find the moment generating function of $Y$, simplifying your answer.
\hfill \mbox{\textit{OCR S4 2015 Q5 [9]}}