Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 14 / c

Table of contents
- i c
  - Solution i
- ii c
  - Solution ii

i

Solution

The independence of random variables means that fixing

Y = y

leaves the conditional probability density of

X

equal to

f

. Given that value of

Y

,

Z

is the translate

X + y

. Hence

P (z \leq Z \leq z + d z ∣ Y = y) = \int_{z - y}^{z + d z - y} f (x) d x .

At a continuity point of

f

, the expression is

f (z - y) d z + o (d z)

. Since

P (Y = y) = 0

, this conditioning uses a conditional probability density, rather than division by the probability of the single event

Y = y

.

ii

Solution

Use the law of total probability to average the conditional interval probability over

Y

:

P (z \leq Z \leq z + d z) = \int_{- \infty}^{\infty} g (y) \int_{z}^{z + d z} f (v - y) d v d y .

The integrands are nonnegative, so Tonelli's theorem permits interchanging the integrals. Thus

P (z \leq Z \leq z + d z) = \int_{z}^{z + d z} [\int_{- \infty}^{\infty} g (y) f (v - y) d y] d v .

Reading off the probability density gives the convolution of probability densities:

h (z) = \int_{- \infty}^{\infty} g (y) f (z - y) d y .

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