Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 16 / f

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- Solution f

Solution

The new angular interval has width

π /2

, so its probability density function is

2/ π

there. Its endpoints stay away from the poles of the tangent function, making

Y

bounded and its expected value finite. Integrate over the continuous uniform distribution of angles:

E [Y] = \frac{2 x _{0}}{π} \int_{- π /6}^{π /3} tan θ d θ = \frac{2 x _{0}}{π} [- ln (cos θ)]_{- π /6}^{π /3} .

The two cosines are

3 /2

and

1/2

. Hence

E [Y] = \frac{2 x _{0}}{π} ln 3 = \frac{x _{0} ln 3}{π} .

It has units of length and is positive because the permitted positive angles extend farther than the negative ones.

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