2
$\begingroup$

I'm trying to solve an exercise about first order stochastic dominance, and there is this one question that I'm not able to answer.

Here are the hypothesis of the exercise:

Let Y ∼ F and Y' ∼ G, that means Y (resp. Y') has a probability law given by the distribution function F (resp. G).

Assume that Y ≥ 0 and Y' ≥ 0 and E(Y) + E(Y')< ∞.

Here comes the question:

-Show that if the survival function of Y first order dominates over the survival function of Y', then if Y and Y' are independent, we have P(YY') ≥ 1/2.

I don't see how to use that Y and Y' are in this case independant.

Maybe I should set X=Y-Y' and then use the convolution?

Any help, would be appreciated, thanks !

Alex

$\endgroup$
4
  • $\begingroup$ Could you clarify what "stochastically dominates at the order one F" means? Do you mean first-order stochastic dominance of $Y^\prime$ over $Y$? Perhaps you mean the opposite? $\endgroup$
    – whuber
    Apr 20, 2020 at 15:22
  • $\begingroup$ @whuber, yes thanks Y' over Y, I edited my question $\endgroup$
    – Alex
    Apr 20, 2020 at 15:28
  • 1
    $\begingroup$ Suppose $F$ (the distribution of $Y$) is a Uniform$(0,1)$ distribution and $G$ (the distribution of $Y^\prime$) is a Uniform$(1,2)$ distribution. It seems to me that $Y^\prime$ dominates $Y$ but $\Pr(Y\ge Y^\prime)=0.$ $\endgroup$
    – whuber
    Apr 20, 2020 at 15:29
  • $\begingroup$ That's my bad, I tried to rephrase the question but seems like I shouldn't have ! The hypothesis was that the survival function of Y first order dominates over the survival function of Y'. Is it plausible this way? $\endgroup$
    – Alex
    Apr 20, 2020 at 15:33

1 Answer 1

3
$\begingroup$

This is an interesting and instructive question because, in considering it, we obtain no less than five distinct characterizations of stochastic dominance. This provides good material for developing an accurate intuition about the concept.


Let's assume nothing about $F$ or $G$ except that the survival function of $Y$ is never less than that of $Y^\prime;$ that is, for all real numbers $y,$

$$1 - F(y) = \Pr(Y \gt y) \ge \Pr(Y^\prime \gt y) = 1 - G(y).$$

[That's the first characterization of stochastic dominance.] The rules of arithmetic imply this is equivalent to

$$G(y) \ge F(y)\tag{*}$$

for all $y.$ [Second characterization.] Geometrically, the definition means the graph of $F$ lies to the right of the graph of $G$ [third characterization] and result $(*)$ says this is the same as the graph of $G$ lying above the graph of $F$ [fourth characterization]. See the first figure below.

Compute the chance that $Y\ge Y^\prime$ in terms of the expectation of the indicator of this event:

$$\eqalign{ \Pr(Y\ge Y^\prime) &= E\left[\mathcal{I}(Y^\prime \le Y)\right] \\ &= E_Y\left[E_{Y^\prime}\left[Y^\prime \le Y\right]\right] \\ & = E_Y\left[{\Pr}_{Y^\prime}(Y^\prime \le Y \right)] \\ & = E_Y\left[G(Y)\right] \\ & = \int_\mathbb{R} G(y)\,\mathrm{d}F(y). }$$

Evaluate this integral in terms of the new variable $p = F(y)$ (which therefore ranges from $0$ to $1.$) Invert this relation by defining $F^{-1}:[0,1]\to\mathbb{R}$ via

$$F^{-1}(p) = \inf\left\{y\mid F(y) \ge p\right\}.$$

Because $F$ has limits at the left (it is "cadlag"),

$$F\left(F^{-1}(p)\right) \ge p.$$

Here is a sketch of the setup:

Figure showing the graphs of F and G

Substituting $y = F^{-1}(p)$ yields

$$ \int_\mathbb{R} G(y)\,\mathrm{d}F(y) = \int_0^1 G\left(F^{-1}(p)\right)\,\mathrm{d}p \ge \int_0^1 F\left(F^{-1}(p)\right)\,\mathrm{d}p \ge \int_0^1 p\,\mathrm{d}p = \frac{1}{2},$$

QED.

After the substitution, the integral expresses the area beneath the graph of the function that sends $p$ to $G(F^{-1}(p));$ namely, it takes a "round trip" in the first figure from $p$ to $F^{-1}(p) = y$ to $G(y):$

Figure 2

The point of this exercise is that stochastic dominance $(*)$ is equivalent to this graph lying on or above the diagonal (dotted) line [fifth characterization]. The chance that the dominant variable $Y$ equals or exceeds the other variable $Y^\prime$ is the area under this graph. This renders the result immediately obvious:

Because the graph of $p\to G(F^{-1}(p))$ lies above the diagonal, the area beneath it exceeds the area beneath the diagonal, which is $1/2.$

Notice that no assumptions about positive support or existence of expectations were needed: this is a fully general result.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks a lot @whuber for this very clear explanation, it definitely solved my problem ! I appreciate the graphs too :) $\endgroup$
    – Alex
    Apr 22, 2020 at 13:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.