# A theorem on dominance of random variables

I wrote previously about a dominance relation amongst $$\mathbb{N}$$ valued random variables. I’ve realised a nice characterisation of the relationship, which is what this post is about.

We’ll change the setting slightly to $$\mathbb{R}$$ valued random variables, as proving this theorem works more nicely in this context, but the result will still hold for $$\mathbb{N}$$ as a subset of it.

Definition: Let $$X, Y$$ be real valued random variables. Say $$x \preceq y$$ if $$\forall t. P(X \geq t) \leq P(Y \geq t)$$.

Theorem: $$X \preceq Y$$ iff for every monotone function $$h : \mathbb{R} \to \mathbb{R}$$, $$E(h(X)) \leq E(h(y))$$.

Proof of this will need a version of a lemma from the last post:

Lemma: Let $$X$$ be a real-valued random variable. Then $$E(X) = \int\limits_{-\infty}^\infty P(X \geq t) dt$$.

I’m going to skip proving this for now. I have proofs, but they’re a little fiddly and I got lost in the details when trying to write this up pre-lunch (it’s easy to prove given some stronger continuity assumptions if you use integration by parts). So IOU one proof.

Proof of theorem:

If $$X \preceq Y$$ and $$h$$ is monotone, then $$E(h(X)) = \int\limits_{-\infty}^\infty P(h(X) \geq t) dt$$ and similarly for $$y$$.

But $$h$$ is monotone, so $$H_t = \{x : h(x) \geq t\}$$ is either $$[y, \infty)$$ or $$(y, \infty)$$ for some $$y$$. The latter can be written as $$\bigcup [y_n, \infty)$$ for some sequence $$y_n$$. We know that $$P(X \geq y) \leq P(Y \geq y)$$, so we know that $$P(X \in H_t) \leq P(Y \in H_t)$$ due to this characterisation. But these probabilities are respectively $$P(h(X) \geq t)$$ and $$P(h(Y) \geq t)$$.

So \begin{align*} E(h(X)) & = \int\limits_{-\infty}^\infty P(h(X) \geq t) dt \\ & \leq \int\limits_{-\infty}^\infty P(h(Y) \geq t) dt \\ & = E(Y) \\ \end{align*}

as desired.

Now for the converse:

Let $$t \in \mathbb{R}$$ with $$P(X \geq t) > P(Y \geq t)$$. Let $$h(x) = 0$$ if $$x < t$$ and $$h(x) = 1$$ if $$x \geq t$$. Then $$h$$ is monotone increasing and $$E(h(X)) = P(X \geq t) > P(Y \geq t) = E(h(Y))$$ as desired.

QED

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