 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2013/discrete-mathematics-probability-and-continuous-mathematics/9/b/i/solution

Assume

a < b

. An increasing affine map sends the two endpoints to zero and one when

c a + d = 0

and

c b + d = 1

. Solving gives

c = \frac{1}{b - a}, d = - \frac{a}{b - a} .

Indeed, for

0 \leq t \leq 1

P (\frac{V - a}{b - a} \leq t) = P (V \leq a + (b - a) t) = \frac{a + ( b - a ) t - a}{b - a} = t,

[0, 1]

A decreasing affine map works too, by swapping the endpoints:

c = - \frac{1}{b - a}, d = \frac{b}{b - a} .

These are the two possible choices, since the image interval must have endpoints zero and one. The case

a = b

cannot be transformed into a nondegenerate continuous uniform distribution by deterministic constants.

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