Dominance between two functions

Question

Let two functions $f(z)$ and $g(z)$ with $z\in[0,c]$ with $c$ a constant such that $c<b$. I'd like to check whether $f(z)-g(z)>0$. I've tried to set $f(z)$ to its minimal value and $g(z)$ to its maximal one, but nothing emerged intuitively.

Here are both functions: $$f(z)=\int_{\phi(z)}^b(x-{\phi(z)})dH(x)$$

$$g(z)=\int_{\phi(z)}^d(2d-x-{\phi(z)})dH(x)$$ with $x\in[0,b]$ and $d$ a constant such that $c<d<b$ and $H$ a cdf over $[0,b]$.

Recap:

we have that

$X\sim G$ with $supp(H)=[0,b]$

$\phi(z)$ is:

• the inverse function of $t(x)=x-\int_0^xH(t)dt$

• a strictly increasing function on $[0,c]$ with $\phi(z)>z$ for all $z\in[0,c]$.

$0<z<c<d<b$.

$d=E[X]$

and $c=d-\int_0^dH(x)dx$.

My question is:

Would it be sufficient to compute $\lim_{z\mapsto c}\frac{g(z)}{f(z)}$ to see if it equals zero or something less than one to say that function $f$ dominates function $g$, $\forall z\in[0,c]$? Or is it only saying that $f(c)>g(c)$ and nothing else...?

Many thanks for any helpful suggestions!

Answer 1

Note that $f(z)>0$ for $z\in[0,b]$ and $g(z)>0$ for $z\in[d,b]$, which is not helpful...

---

We can write $$f(z)=\int_z^b(x-z)dH(x)=\int_z^d(x-z)dH(x)+\int_d^b(x-z)dH(x)$$ so that

$$f(z)-g(z)=\int_z^d(x-z)dH(x)+\int_d^b(x-z)dH(x)-\int_z^d(2d-x-z)dH(x).$$

Simplifying:

$$f(z)-g(z)=\int_d^b(x-z)dH(x)+2\int_z^d(x-d)dH(x).$$

---

By integration by parts we have:

$$f(z)-g(z)=[(x-z)H(x)]_d^b-\int_d^bH(x)dx+2[(x-d)H(x)]_z^d-2\int_z^dH(x)dx$$

Simplifying:

$$f(z)-g(z)=(b-z)-(d-z)H(d)-\int_d^bH(x)dx-2(z-d)H(z)-2\int_z^dH(x)dx$$

We have $$f(0)-g(0)=b-d\cdot H(d)-\int_d^bH(x)dx-2\int_0^dH(x)dx$$

or

$$f(0)-g(0)=[b-\int_0^bH(x)dx]-d\cdot H(d)-\int_0^dH(x)dx$$

The first term is the mean $d$, so

$$f(0)-g(0)=d-d\cdot H(d)-\int_0^dH(x)dx=d[1-H(d)]-\int_0^dH(x)dx$$

Now $\int_0^dH(x)dx< d\cdot H(d)$, so:

$$f(0)-g(0)>d[1-2H(d)].$$

As long as $d$ is less than the median then $f(0)>g(0)$.

---

Now look at the derivative:

We have $$f'(z)-g'(z)=-\int_z^bdH(x)+\int_z^ddH(x)+2(d-z)=-\int_d^bdH(x)+2(d-z)$$

so that

$$f'(z)-g'(z)=-[1-H(d)]+2(d-z)$$

This is certainly negative if $z\in[d,b]$.

---

The second derivative is

$$f''(z)-g''(z)=-2z<0$$

for all $z\in(0,b)$ so that $f-g$ is strictly concave. Hence it is maximized where the first derivative is zero, i.e. at

$$z^*=d-\frac{1-H(d)}{2}<d.$$

If $z^*>0$ we know that since $f(0)-g(0)>0$, then it must be that $f(z)-g(z)>0$ on $[0,z^*]$

---

Since

$$f(d)-g(d)=\int_b^d(x-d)dH(x)<0.$$

there is no hope of showing that $f(z)-g(z)>0$ for all $z\in[0,c]$ where $c<d$ unless we are allowed to pick $c$ or unless we know $c<z^*$.

Answer 2

Following a question at the end of the solution by @smcc, here is a counter example showing that property $f(z)>g(z)$ cannot be extended to $z \in [0,d]$.

Let us take $c=1,d=2,b=3$.

It suffices to take $dH(x)=h(x)dx$ where:

$h(x):=\begin{cases}\frac{\pi}{6}\cos(\frac{\pi}{3}(x-1.5))&\text{if} \ 0<x<3\\0 & \text{in all other cases}\end{cases}$

See figure below:

Then the curves of $f$ and $g$ are as follows (the plot has been done with Mathematica, using the script below) ; it wouldn't be very difficult to obtain explicit expressions for $f(x)$ and $g(x)$.

Mathematica script:

c = 1; d = 2; b = 3;

h[x_] := (Pi/6)Max[0, Cos[(Pi/3)(x - 1.5)]];

f[z_] := Integrate[(x - z)h[x], {x, z, b}];

g[z_] := Integrate[(2*d - x - z)h[x], {x, z, d}];

Plot[{f[z], g[z]}, {z, 0, b}, PlotRange -> {0, 2.5}]