# Prove two random variables are equal almost everywhere

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!

$$\newcommand{\E}{\mathbb E}$$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.
• Thank you! This is excellent and makes perfect sense. I believe there is a typo though. In particular, I think you mean there exists $M$ such that $(1 + \mathbb{E}[\delta])^M - 1 > 1$'', and so on. I appreciate it! Aug 12, 2020 at 1:42