OCR S3 2010 January — Question 4 9 marks

Exam BoardOCR
ModuleS3 (Statistics 3)
Year2010
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCumulative distribution functions
TypeCDF of transformed variable
DifficultyStandard +0.8 This S3 question requires understanding of CDF transformations and the relationship between CDFs and PDFs. Part (i) involves careful manipulation of inequalities when transforming variables (non-trivial since Y is a decreasing function of V), requiring students to show a specific form and determine the valid interval. Part (ii) requires finding E(1/Y²) which involves recognizing this equals E((1+V)²) and either integrating or using the PDF of Y cleverly. While systematic, this goes beyond routine S3 exercises and requires solid understanding of transformation techniques.
Spec5.03g Cdf of transformed variables

4 The continuous random variable \(V\) has (cumulative) distribution function given by $$\mathrm { F } ( v ) = \begin{cases} 0 & v < 1 \\ 1 - \frac { 8 } { ( 1 + v ) ^ { 3 } } & v \geqslant 1 \end{cases}$$ The random variable \(Y\) is given by \(Y = \frac { 1 } { 1 + V }\).
  1. Show that the (cumulative) distribution function of \(Y\) is \(8 y ^ { 3 }\), over an interval to be stated, and find the probability density function of \(Y\).
  2. Find \(\mathrm { E } \left( \frac { 1 } { Y ^ { 2 } } \right)\).

4(i)
\(G(y) = P(Y \leq y) = P(1/(1+V) \leq y) = P(V \geq 1/y - 1) = 1 - F(1/y - 1)\)
\[G(y) = \begin{cases} 0 & y \leq 0 \\ 8y^3 & 0 < y \leq 1/2 \\ 1 & y > 1/2 \end{cases}\]
AnswerMarks Guidance
\(8y^3\) obtained correctlyM1, A1 Correct range. Condone omission of \(y \leq 0\)
\(g(y) = \begin{cases} 24y^2 & 0 < y \leq 1/2 \\ 0 & \text{otherwise} \end{cases}\)M1, A1 For \(G(y)\) - Correct answer with range
\(\int 24y^2/y^3 \, dy\) with limits \(= 12\)M1, A1 With attempt at integration
Total: [9]
## 4(i)

$G(y) = P(Y \leq y) = P(1/(1+V) \leq y) = P(V \geq 1/y - 1) = 1 - F(1/y - 1)$

$$G(y) = \begin{cases} 0 & y \leq 0 \\ 8y^3 & 0 < y \leq 1/2 \\ 1 & y > 1/2 \end{cases}$$

| $8y^3$ obtained correctly | M1, A1 | Correct range. Condone omission of $y \leq 0$ |
| --- | --- | --- |
| $g(y) = \begin{cases} 24y^2 & 0 < y \leq 1/2 \\ 0 & \text{otherwise} \end{cases}$ | M1, A1 | For $G(y)$ - Correct answer with range |
| $\int 24y^2/y^3 \, dy$ with limits $= 12$ | M1, A1 | With attempt at integration |

**Total: [9]**

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4 The continuous random variable $V$ has (cumulative) distribution function given by

$$\mathrm { F } ( v ) = \begin{cases} 0 & v < 1 \\ 1 - \frac { 8 } { ( 1 + v ) ^ { 3 } } & v \geqslant 1 \end{cases}$$

The random variable $Y$ is given by $Y = \frac { 1 } { 1 + V }$.\\
(i) Show that the (cumulative) distribution function of $Y$ is $8 y ^ { 3 }$, over an interval to be stated, and find the probability density function of $Y$.\\
(ii) Find $\mathrm { E } \left( \frac { 1 } { Y ^ { 2 } } \right)$.

\hfill \mbox{\textit{OCR S3 2010 Q4 [9]}}