 Solution

ID: past-exam-of-the-stochastic-differential-equations-and-brownian-motion-in-complex-analysis-course-of-the-university-of-oxford/2009/a/1/b/solution

For

λ > 1

, use

x / λ \geq 1

on the tail:

\int_{λ}^{\infty} e^{- x^{2} /2} d x \leq \frac{1}{λ} \int_{λ}^{\infty} x e^{- x^{2} /2} d x = \frac{e ^{- λ^{2} /2}}{λ} \leq e^{- λ^{2} /2} .

This proves the requested Gaussian tail bound.

The subsequent printed maximum inequality is false as stated. For

n = 1

its integrand never reaches one, and its right-hand side is

\frac{2}{2 π} \int_{0}^{\infty} e^{- x^{2} /2} d x = 1.

But for two independent absolute standard normal variables, using their common distribution function

F (x) = 2Φ (x) - 1

φ

E M_{1} = 4 \int_{0}^{\infty} x (2Φ (x) - 1) φ (x) d x = 8 \int_{0}^{\infty} φ (x)^{2} d x = \frac{2}{π} > 1.

The middle step is integration by parts with

φ^{'} = - x φ

. Thus no proof of the displayed source inequality can hold for all its stated indices.

E M_{n} = \int_{0}^{\infty} [1 - (2Φ (x) - 1)^{2^{n}}] d x .

A valid union bound gives

P (M_{n} > x) \leq min (1, \frac{2 ^{n + 1}}{2 π} \int_{x}^{\infty} e^{- u^{2} /2} d u) .

In particular, for

x > 1

this is at most

min (1, C_{n} e^{- x^{2} /2})

, where

C_{n} = 2^{n + 1} / 2 π

. This corrected estimate proves the intended summability in part (c), without silently replacing the source formula.

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