Expected value of the CDF of a function of a random variable.

Question

I have pdfs $f(x), g(x)$ with CDFs $F(x), G(x)$. I am looking for conditions relating distributions $f$ and $g$, where

$\int_0^\infty f(x)F(\frac{1}{2}x)dx \geq \int_0^\infty g(x)G(\frac{1}{2}x)dx$

The problem arose in my research as an application of the Hotelling model of spatial competition. There is an incumbent seller located at 0, and a distribution of buyers $f(x)\sim [0,\infty)$. One of the buyers is selected at random from $f(x)$ to be a challenger. The remaining buyers buy from either the incumbent or the seller, whichever is closer. Let $s$ denote the proportion of sales made by the incumbent. I want to know the expected value of $s$.

Translating this to a mathematical problem, this should be $\mathbf{E}[s] = \mathbf{E}[F(\frac{1}{2}x)]=\int_0^\infty f(x)F(\frac{1}{2}x)dx $. (that is the buyer half way between the two is indifferent, so evaluating the CDF at this point give the proportion that buy from the incumbent)

But the real question is this: Suppose that I have the same problem, but with two different distributions, $f(x)$ and $g(y)$. Is there some property of $f(x)$ and $g(y)$ such that $\mathbf{E}[F(\frac{1}{2}x)]\geq \mathbf{E}[G(\frac{1}{2}y)]$, where $x\sim f(x)$ and $y\sim g(y)$?

That is, I want to know some property of the distributions $f$ and $g$ such that I can say "The incumbent can expect a higher share of the sales under $f$ than under $g$$, or other words, that satisfies this inequality:

$\int_0^\infty f(x)F(\frac{1}{2}x) \geq \int_0^\infty g(x)G(\frac{1}{2}x)$

What I tried:

Note that if $g$ is just a shifted version of $f$, that is $g(x)=f(x)+a$ for $a\geq 0$, this works. That is, shifting the distribution doesn't change the distance from the challenger to the buyers, but increases the distance from the incumbent seller to the buyers.

I would like something more general than this. I have been trying to apply first order stochastic dominance. But I can't seem to make it work.

I thought that this answer was relevant, but couldn't figure out how to apply it. I tried rewriting the problem by noting that the distribution of the pivotal buyer is given by $h(x)=2f(2x)$. Then writing down the expected value as $\int_0^\infty 2f(2x)F(x)dx$ and $\int_0^\infty 2g(2x)G(x)dx$. So the question is, what conditions can I impose such that $\int_0^\infty 2f(2x)F(x) dx \geq \int_0^\infty 2g(2x)G(x) dx$? The desired inequality could then be rewritten: $\int_0^\infty f(x)F(\frac{1}{2}x) \geq \int_0^\infty g(x)G(\frac{1}{2}x)$. But couldn't make any progress from there.

Any thoughts on applying first order stochastic dominance to this problem, or any ideas about what property of $f$ and $g$ I could use would be appreciated!

Answer 1

A result: Suppose $F$ and $G$ have support $[0,1]$ and have the same mean. Then, if $F$ is larger than $G$ in the rotation order, with rotation point at least $1/2$, and $F$ or $G$ is convex on $[0,1/2]$, we have

$$\int_0^1 f(x)F\left(\frac{1}{2}x\right)dx \geq \int_0^1 g(x)G\left(\frac{1}{2}x\right)dx.$$

Proof: Consider the case where $F$ is convex on $[0,1/2]$. From the definition of the convex order, we have

$$\int_0^1 f(x)F\left(\frac{1}{2}x\right)dx \geq \int_0^1 g(x)F\left(\frac{1}{2}x\right)dx $$

Since $F$ is larger than $G$ in the rotation order with rotation point at least $1/2$, we have $F(x)\geq G(x)$ for all $x\leq 1/2$. Thus

$$\int_0^1 g(x)F\left(\frac{1}{2}x\right)dx\geq\int_0^1 g(x)G\left(\frac{1}{2}x\right)dx. $$

The other case (where $G$ is convex on $[0,1/2]$) is similar.

---

Definition (Convex Order): The distribution function $F$ is larger than $G$ in the convex order if $$\int_0^1 f(x)\phi(x)dx\geq \int_0^1 g(x)\phi(x)dx$$ for all functions $\phi$ that are convex on the union of the supports of $F$ and $G$ (provided the expectations exist).

(Note that the definition implies that $F$ and $G$ have the same mean. To see this, take the cases $\phi(x)=-x$ and $\phi(x)=x$.)

Definition (Rotation Order): The distribution function $F$ is larger than $G$ in the rotation order if there exists a rotation point $\psi$ such that

$$F(x)\gtreqqless G(x) \iff x\lesseqqgtr\psi $$

Theorem 3.A.44 (Shaked and Shantikumar, "Stochastic Orders"): If $F$ and $G$ have the same mean and $F$ is larger than $G$ in the rotation order, then $F$ is larger than $G$ in the convex order.

---

Note that $F$ is larger than $G$ in the convex order if and only if $G$ second order stochastically dominates $F$ ($G$ is larger than $F$ in the concave order). A (clockwise) rotation of a distribution is a special type of spread of the distribution.

The condition that $F$ or $G$ is convex on $[0,1/2]$ is ensured if one of them is a unimodal distribution with a mode of at least $1/2$.

An example of a family of distributions ordered by rotation about $1/2$ and satisfying the convexity condition is the set of symmetric beta distributions with density $$f(x)=\frac{[x(1-x)]^n}{\beta(n+1,n+1)}$$ where $n\geq 0$. When $n=0$, $f$ is the uniform density and $\mathbb{E}[s]=1/2$. When $n=1$, $\mathbb{E}[s]=7/40=0.175$ and when $n=2$, $\mathbb{E}[s]=57/448=0.127$ (to 3dp).

Answer 2

Your linked answer handles the well-known case that $$ \int_0^\infty f(x)F(x)\,dx=\frac{1}{2} $$ which holds because any random variable $X$ when inserted into its own CDF $F$ yields a r.v. $F(X)$ that is uniformly distributed on $[0,1]\,.$ The proof of this is very easy.

I think your case to calculate $$ \int_0^\infty f(x)F(ax)\,dx\,,\quad a>0 $$ is much harder.

Possible Approach

From

\begin{align} \frac{d}{da}\int_0^\infty f(x)F(ax)\,dx= a\int_0^\infty f(x)f(ax)\,dx \end{align} it looks promising to make a few assumptions on the PDF $f\,.$ A popular distribution on $[0,\infty)$ is the exponential distribution: $$ f(x)=\lambda e^{-\lambda x}\,,\quad f(x)f(ax)=\lambda^2e^{-\lambda(1+a)x}\,. $$ Then $$ a\int_0^\infty f(x)f(ax)\,dx=a\int_0^\infty \lambda^2e^{-\lambda(1+a)x}\,dx=\frac{a\lambda}{1+a}\,. $$ With the known value $1/2$ for $a=1$ we find by integrating $$ \int_0^\infty f(x)F(ax)\,dx=\frac{1}{2}+\int_1^a\frac{b\lambda}{1+b}\,db= \frac{1}{2}+\lambda a-\lambda\log(a+1)-\lambda+\lambda\log 2\,. $$

• I recommend now to apply this approach to another popular distribution on $[0,\infty)$ which is the lognormal distribution.