# Equivalent definition of stochastic dominance

Note that a distribution function (cadlag etc) $$F$$ is said to be stochastically dominated by a distribution function $$G$$ if $$F(x)\geq G(x)$$ for all $$x \in \mathbb{R}$$. The following result characterizes stochastic dominance equivalently:

Theorem: $$F$$ is stochastically dominated by $$G$$ if and only if for every increasing function $$u$$ $$\mathbb{E}_F[u(x)] \leq \mathbb{E}_G[u(x)].$$

I have seen proofs for this result when $$F$$ and $$G$$ are absolutely continuous (and thus admit densities) using integration by parts. Is there a more general proof that holds for arbitrary distribution functions/measures on the real line?

To be clear, the integral definition implies the CDF one by using $$u(x) = \mathbb{1}\{x \in (z,\infty)\}$$ for all $$z \in \mathbb{R}$$. The converse direction doesn't seem immediately obvious.

• Geometrically this is clear: stochastic dominance implies the graph of $G$ lies to the right of the graph of $F.$ The function $u$ merely distorts the horizontal axis, but does not alter the right-left relation, QED. You can translate this into a formal proof if you like.
– whuber
Commented Apr 12, 2023 at 3:35
• formally, you always have $$\mathbb{E}_F[u(X)] = \int_{\mathbb R} u(x)\ dF(x)$$ where the integral is to be understood as a Lebesgue-Stieltjes integral. Using the characterization of LS integral as a supremum of Daniell integrals (see wiki), which can be computed as Riemann-Stieltjes integrals, I believe the proofs you know for differentiable $F$ and $G$ should extend in a straightforward way (there may be technicalities on the way, but the idea remains the same) Commented Apr 12, 2023 at 9:16

The integration by parts formula still holds for general distribution functions (under appropriate technical conditions). For example, Theorem 18.4 in Probability and Measure by Patrick Billingsley (do not confuse $$F, G$$ in this theorem with $$F, G$$ in your question):

Let $$F$$ and $$G$$ be two nondecreasing, right-continuous functions on an interval $$[a, b]$$. If $$F$$ and $$G$$ have no common points of discontinuity in $$(a, b]$$, then \begin{align} \int_{(a, b]}G(x)dF(x) = F(b)G(b) - F(a)G(a) - \int_{(a, b]}F(x)dG(x). \tag{1} \end{align}

Equation $$(1)$$ is a good start for proving the result of your interest -- if we can extend the integral region $$(a, b]$$ to $$\mathbb{R}$$. To this end, we would need to impose integrability conditions of $$u$$ (of course, for $$(1)$$ to hold, we need to also assume that $$u$$ and $$F$$ / $$G$$ have no common discontinuity in $$\mathbb{R}$$ and $$u$$ is right-continuous): \begin{align} \int_{\mathbb{R}}|u(x)|dF(x) < \infty, \; \int_{\mathbb{R}}|u(x)|dG(x) < \infty. \tag{2} \end{align}

By $$(1)$$, for every $$n$$: \begin{align} & \int_{(-n, n]}u(x)dF(x) = F(n)u(n) - F(-n)u(-n) - \int_{(-n, n]}F(x)du(x). \tag{3} \\ & \int_{(-n, n]}u(x)dG(x) = G(n)u(n) - G(-n)u(-n) - \int_{(-n, n]}G(x)du(x). \tag{4} \end{align} It then follows by $$(3), (4)$$ and $$F(x) \geq G(x)$$ for all $$x \in \mathbb{R}$$ that \begin{align} & \int_{(-n, n]}u(x)dF(x) - \int_{(-n, n]}u(x)dG(x) \\ =& u(n)(F(n) - G(n)) - u(-n)(F(-n) - G(-n)) - \int_{(-n, n]}[F(x) - G(x)]du(x) \\ \leq & u(n)(F(n) - G(n)) - u(-n)(F(-n) - G(-n)). \tag{5} \end{align}

If $$u$$ is non-negative, then the right-hand side of $$(5)$$ is bounded by $$u(n)(F(n) - G(n))$$, which can be rewritten as $$u(n)(1 - G(n)) - u(n)(1 - F(n))$$, which converges to $$0$$ as $$n \to \infty$$ by the integrability of $$F, G$$ and the monotonicity of $$u$$. Similarly, if $$u$$ is non-positive, then the right-hand side of $$(5)$$ is bounded by $$-u(-n)(F(-n) - G(-n))$$ and converges to $$0$$ as $$n \to \infty$$ (see the next paragraph for a more detailed derivation).

If $$u$$ takes both negative and positive values, it follows by the monotonicity of $$u$$ that for sufficiently large $$N$$, $$u(N) > 0$$ and $$u(-N) < 0$$, whence for all $$n > N$$, again by monotonicity of $$u$$: \begin{align} 0 \leq u(n)(1 - F(n)) \leq \int_n^\infty u(x)dF(x), \; \int_{-\infty}^{-n}u(x)dF(x) \leq u(-n)F(-n) \leq 0. \tag{6} \end{align} By condition $$(2)$$ and Lebesgue's dominated convergence theorem (DCT), $$(6)$$ implies that $$u(n)(1 - F(n)) \to 0$$ and $$u(-n)F(-n) \to 0$$ as $$n \to \infty$$. Similarly, $$u(n)(1 - G(n)) \to 0$$ and $$u(-n)G(-n) \to 0$$ as $$n \to \infty$$. Therefore, the right-hand side of $$(5)$$ always converges to $$0$$ as $$n \to \infty$$ for $$u$$ that is nondecreasing and integrable.

Now the result follows by passing $$n \to \infty$$ on both sides of $$(5)$$ (note that condition $$(2)$$ and DCT imply the left-hand side of $$(5)$$ converges to $$E_F[u] - E_G[u]$$).

• Wow I can't believe I missed a remarkably simple theorem on partial integration with Stieltjes integrals (very classic application of Fubini). I haven't checked the details of your answer but it looks roughly correct and in line with how the proof in the absolutely continuous case goes. Can we assume that $u$ is right-continuous without any loss of generality (although I think that this shouldn't really impede the existence of the Stieljes integral with respect to $u$) Commented Apr 13, 2023 at 20:34
• @YashaswiMohanty Yes. Since $u$ is nondecreasing, it is very natural (or it is a convention in probability) to treat it as right-continuous (basically you have the freedom to modify the countably many jumping points of a monotone function to make it is right-continuous). Commented Apr 13, 2023 at 21:19
• I'm missing how DCT implies that the LHS of (5) goes to the expression we need. Don't we need that $\lvert u \mathbb{1}[-n,n] \rvert \leq u$ for that ? Commented Apr 13, 2023 at 21:19
• @YashaswiMohanty Exactly, and $|uI_{(-n, n]}| \leq |u|$ is trivial :). Commented Apr 13, 2023 at 21:20

This is not really a theorem about stochastic dominance: it's a property of areas. It comes down to this lemma, which will be applied in the last two paragraphs:

When $$f:\mathbb R\to\mathbb R$$ is an integrable function with non-zero norm $$|f|=\int |f(x)|\,\mathrm dx \lt \infty$$ and $$\mathcal A$$ is a set of positive measure $$|\mathcal A| = \int_{\mathcal A}\mathrm dx \gt 0$$ on which the values of $$f$$ all exceed some positive number $$\epsilon \gt 0,$$ then there exists an increasing (measurable) function $$u$$ for which the transformed function $$f\circ u$$ has a positive integral, $$\int_\mathbb{R}f(u(x))\,\mathrm dx \gt 0.$$

The idea is to make the image of $$u$$ focus on $$\mathcal A$$ while practically skipping over everything else: the integral is then at least $$\epsilon$$ (the minimum value of $$f$$ on $$\mathcal A$$) times the measure of $$\mathcal A$$ -- plus any negative contributions elsewhere. By limiting the latter we wind up with a positive integral.

In this illustration, the set $$\mathcal A$$ is highlighted in orange along the horizontal axis and the area under $$f$$ over the region $$\mathcal A$$ is shaded.

One such function $$u$$ is obtained by inverting the (strictly) increasing function

$$v(y) = \int_{-\infty}^y \mathcal{I}_\mathcal{A}(x) + \delta(1-\mathcal{I}_\mathcal{A}(x))\,\mathrm dx$$

for a positive $$\delta$$ to be determined. ($$\mathcal I$$ is the indicator function.)

This illustration graphs $$v$$ for $$\delta = 0.05.$$ Its slopes are $$1$$ (orange) and $$0.05$$ (gray).

The Fundamental Theorem of Calculus and the rule of differentiating inverse functions show the inverse $$u=v^{-1}$$ is (a) differentiable with (b) derivative equal to $$1$$ on $$\mathcal A$$ and $$1/\delta$$ elsewhere. Writing $$v(\mathcal A)^\prime$$ for the complement of $$v(\mathcal A)$$ within the image of $$v$$ (which is $$\mathbb R$$ itself), use the standard integral inequalities (Holder's, for instance) and the change of variables formula for integrals to deduce

\begin{aligned} \int f(u(x))\,\mathrm dx &= \int_{v(\mathcal A)} f(u(x))\,\mathrm dx + \int_{v(\mathcal A)^\prime} f(u(x))\frac{|u^\prime(x)|}{|u^\prime(x)|}\,\mathrm dx\\ &\ge \int_{v(\mathcal A)} f(u(x))\,\mathrm dx - \left(\sup_{x\in v(\mathcal A)^\prime} \frac{1}{|u^\prime(x)|}\right)\left|\int f(u(x))|u^\prime(x)|\,\mathrm dx\right|\\ &\ge |\mathcal A|\epsilon - \delta|f|. \end{aligned}

Taking $$\delta = |\mathcal A|\epsilon / (2|f|)$$ produces a strictly positive value, proving the lemma.

This illustration of the graph of $$f\circ u$$ shows how the horizontal axis has been squeezed at all places where $$f\lt \epsilon,$$ thereby giving the entire integral a positive value. Making $$\delta$$ sufficiently close to zero will effectively eliminate the dips in the graph below $$\epsilon.$$

As a corollary, applying the lemma to $$-f$$ shows that when there is a set of positive measure on which $$f$$ has negative values below $$-\epsilon \lt 0,$$ then there is an increasing function $$u$$ for which $$f\circ u$$ has a negative integral.

Consequently, if for all increasing (measurable) functions $$u$$ the integral in the lemma is positive, it follows that the set of places where $$f$$ has a negative value has measure zero.

That's the heart of the matter.

Let's pause to notice two things. The first is technical: in this construction of $$u,$$ $$u^{-1}$$ is also almost everywhere differentiable and therefore continuous and measurable, allowing us to focus on such "nice" functions.

The second is probabilistic: when $$F_X$$ is the distribution function of a random variable $$X$$ -- that is, $$F_X(x)=\Pr(X\le x)$$ -- and $$u$$ is an increasing (measurable) function with an increasing (measurable) inverse $$u^{-1},$$ then the distribution function of $$u^{-1}(X)$$ is

$$F_{u^{-1}(X)}(y) = \Pr(u^{-1}(X)\lt y) = \Pr(X \le u(y)) = F_X(u(y)).$$

That is, $$F_{u^{-1}(X)} = F_X\circ u.$$

Now observe that when $$F$$ and $$G$$ are distinct distribution functions for a random variable $$X$$ and $$u$$ is an increasing (measurable) function,

$$E_G[u^{-1}(X)] - E_F[u^{-1}(X)] = \int F(u(x)) - G(u(x))\,\mathrm dx = \int (F-G)(u(x))\,\mathrm dx.$$

(For the elementary proof see Expectation of a function of a random variable from CDF for instance. It's just an integration by parts.)

## Proof of the theorem

Applying the corollary to the function $$f = F-G$$ (which has a nonzero norm since $$F$$ and $$G$$ are distinct), under the assumption $$f$$ has finite norm, shows that when all such integrals are positive, the set on which $$F-G$$ is negative has measure zero: that is, $$G$$ stochastically dominates $$F,$$ QED.

We can eliminate the finite-norm assumption by noting that $$F-G$$ can have an infinite norm only by diverging at infinity: it cannot have vertical asymptotes. (The values are differences of probabilities, whence they are bounded by $$\pm 1.$$) Consequently we can approximate $$F-G$$ on an expanding sequence of compact sets, such as the intervals $$(-n,n)$$ for $$n=1,2,3,\cdots,$$ and apply a limiting argument. But that should be viewed as a technicality, because the underlying idea remains the same, as expressed in the lemma.

• Can you elaborate why the last inequality "$E_G[u] - E_F[u] = \cdots$" is true? According to the link you provided (and other technical conditions), $E_G[u] = \int_{-\infty}^\infty (1 - G(x))u'(x)dx$. With additional conditions (like $u$ is strictly increasing), it can be written as $\int_a^b [F(u^{-1}(x)) - G(u^{-1}(x))]dx$ (with $a$, $b$ are asymptotes of $u$), which is different from the last equation. Commented Apr 13, 2023 at 19:30
• In addition, if the last equation were true, then OP's question would be immediate (all he asked is to prove $E_G[u] \geq E_F[u]$ under the condition $F \geq G$), why would we need the lemma to get the job done? Commented Apr 13, 2023 at 19:36
• @Zhanxiong I have added some text to elaborate on the argument. I don't understand your second comment, because I don't believe that such as statement was all that was asked.
– whuber
Commented Apr 16, 2023 at 14:12
• The second comment comments the equation before your edit: "$E_G[u(X)] - E_F[u(X)] = \int (F - G)(u(x))dx$". The OP wants to prove $E_G[u(X)] \geq E_F[u(X)]$ if $F \geq G$. So if the pre-editing formula held, then $E_G[u(X)] \geq E_F[u(X)]$ would hold trivially (because the integrand is nonnegative). Commented Apr 16, 2023 at 14:38
• @Zhanxiong That's the other direction of the implication.
– whuber
Commented Apr 16, 2023 at 15:30