Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 2 / 3F (Probability)

Table of contents
- a 3F
  - Solution a
- b 3F
  - Solution b
- c 3F
  - Solution c

a

Solution

For a nonnegative random variable

Y

and

a > 0

, the Markov inequality is

P (Y \geq a) \leq \frac{E Y}{a} .

Indeed,

Y \geq a 1_{{Y \geq a}}

pointwise. Taking expectations and dividing by

a

proves the claim.

b

Solution

Completing the square gives

M_{X} (t) = \frac{1}{2 π} \int_{- \infty}^{\infty} e^{t x - x^{2} /2} d x = e^{t^{2} /2} \frac{1}{2 π} \int_{- \infty}^{\infty} e^{- (x - t)^{2} /2} d x = e^{t^{2} /2} .

c

Solution

For any

t > 0

, the exponential Markov bound and part (b) give

P (X \geq v) \leq e^{- t v} M_{X} (t) = e^{- t v + t^{2} /2} .

The minimum occurs at

t = v

, so the Gaussian tail bound is

\int_{v}^{\infty} \frac{e ^{- x^{2} /2}}{2 π} d x \leq e^{- v^{2} /2} .

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