Prove two random variables are equal almost everywhere

Question

I'm studying for a qualifying exam in probability theory now. Below is a question from an old qual that I've been struggling with.

Problem: Let $\xi,\eta$ be random variables defined on the same probability space such that $1 \le \xi \le \eta$ almost surely. Suppose that all moments of $\eta$ exist and satisfy $$ \mathbb{E} \eta^m \le \mathbb{E}\xi^m + 1, \qquad m \ge 1 $$ Prove $\xi = \eta$ almost surely.

Attempt: Given that $\xi \le \eta$ almost surely, it suffices to show $\xi \ge \eta$ almost surely. This is equivalent to showing $\mathbb{P}(\xi < \eta) = 0$. Fix $\varepsilon > 0$. We aim to demonstrate $\mathbb{P}(\eta > \xi + \varepsilon) = 0$. Since $\eta \ge \xi$ almost surely by hypothesis, $\eta - \xi$ is a nonnegative random variable in $L^m(\Omega,\mathbb{P})$ for all $m \ge 1$. Thus, we may apply Markov's inequality and $$ \mathbb{P}(\eta-\xi>\varepsilon) \le \frac{\mathbb{E}(\eta - \xi)^3}{\varepsilon^3} = \frac{1}{\varepsilon^3} \sum_{k=0}^3 {3 \choose k}(-1)^k\mathbb{E}\eta^{3-k}\xi^k = \frac{1}{\varepsilon^3}\Big( \mathbb{E}\eta^3 - 3 \mathbb{E}\eta^2\xi + 3 \mathbb{E}\eta\xi^2 - \mathbb{E}\xi^3 \Big) $$ Using that $\mathbb{E}\eta^3 \le \mathbb{E}\xi^3 + 1$, we obtain $$ \frac{1}{\varepsilon^3}\Big( \mathbb{E}\eta^3 - 3 \mathbb{E}\eta^2\xi + 3 \mathbb{E}\eta\xi^2 - \mathbb{E}\xi^3 \Big) \le \frac{1}{\varepsilon^3}\Big( 1 - 3 \mathbb{E}\eta^2\xi + 3 \mathbb{E}\eta\xi^2 \Big) = \frac{1}{\varepsilon^3}\Big(1-3\mathbb{E}\xi\eta(\eta-\xi) \Big) $$ If we can show the right-hand side is less or equal to zero, this will imply $\mathbb{P}(\eta-\xi > \varepsilon) = 0$. The right-hand side is less or equal to zero iff $1/3 \le \mathbb{E}\xi\eta(\eta-\xi)$. Since $\eta \ge \xi \ge 1$ a.s., we have the inequality $$ \mathbb{E}\xi\eta(\eta - \xi) \ge \mathbb{E}1^2(\eta - \xi) \ge 0 $$ which is useless.

I'm just not sure how to proceed. Any suggestions are appreciated!

Answer 1

$\newcommand{\E}{\mathbb E}$Not sure if your method would yield fruit if pursued further, but here is an alternative approach using proof by contradiction. Suppose that $\xi \neq \eta$ with positive probability. Then setting $\delta = \eta - \xi$ we know $\delta \geq 0$ and $\mathbb E[\delta] > 0$.

Firstly we have the bound $$ \E[\eta^m] = \E[(\xi + \delta)^m] = \E[\xi^m] + \sum_{i=1}^m \binom{m}{i} \E[\xi^{m - i} \delta^i] \geq \E[\xi^m] + \sum_{i=1}^m \binom{m}{i} \E[\delta^i]$$ where the last inequality follows since $\xi \geq 1$. By Jensen's inequality $\E[\delta^i] \geq \E[\delta]^i$. Thus $$ \sum_{i=1}^m \binom{m}{i} \E[\delta^i] \geq \sum_{i=1}^m \binom{m}{i} \E[\delta]^i = (1 + \E[\delta])^m - 1.$$ Since $\E[\delta]> 0$, there exists $M \geq 1$ such that $$ (1 + \E[\delta])^M - 1 > 1.$$ Thus $$ \E[\eta^M] > \E[\xi^M] + 1$$ which contradicts our assumption.