 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2013/m5/b/ii/solution

The two-point distribution gives

M (t) = (e^{t} + e^{- t}) /2 = cosh t

(2 j)! = \prod_{r = 1}^{j} (2 r) (2 r - 1) \geq 2^{j} j!

, so each coefficient satisfies

\frac{t ^{2 j}}{( 2 j )!} \leq \frac{( t ^{2} /2 ) ^{j}}{j !} .

Summing the nonnegative terms proves

M (t) \leq e^{t^{2} /2}

for every real

t

. Part (a) therefore gives

M_{S_{n}} (t) \leq e^{n t^{2} /2} .

This is the sub-Gaussian moment bound for a symmetric random sign.

 New to topics? Read the docs here!