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) \\

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.


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