Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 40 / 1 / b / ii

Table of contents
- Solution ii

Solution

Integration of the Laplace distribution probability density function gives

F (x) = {\frac{1}{2} e^{(x - 1) /2}, 1 - \frac{1}{2} e^{- (x - 1) /2}, x \leq 1, x \geq 1.

Solving these two equations for

x

gives

F^{- 1} (u) = {1 + 2 lo g (2 u), 1 - 2 lo g (2 (1 - u)), 0 < u \leq \frac{1}{2}, \frac{1}{2} < u < 1.

Generate one uniform

U \in (0, 1)

and return this piecewise expression. The branches agree at

U = 1/2

and the inverse transform sampling proof establishes the target law without any rejection step.

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