# Does only the Gaussian density provide increasing likelihood ratio?

This question is based on the question which I asked before:

For which pair of probability density functions, $L=f_1(x)/f_0(x)$ is increasing?

however I didnt get any answer other than the Gaussian density and I have the conjecture that perhaps only the Gaussian density provides increasing likelihood ratio.

We know that $y\in(-\infty,\infty)$ for $f_0(y)$ and $f_1(y)$ and both densities are non zero on the domain where they are defined. $L(x)=\frac{f_1(x)}{f_0(x)}$ should be increasing.

I will be very happy to see that my conjecture is incorrect.

Thank you very much in advance.

EDIT: The densities should not be in a partial function. For example:

$$\frac{1}{\mu^+-\mu^-}e^{\frac{-x}{\mu^+}}\quad for\quad x\geq 0$$ $$\frac{1}{\mu^+-\mu^-}e^{\frac{-x}{\mu^-}}\quad for\quad x< 0$$

for $\mu^+>0$ and $\mu^-<0$ is a valid density with increasing $L(x)$

-
When you say "Not a partial function", do you mean "the pdf should be differentiable everywhere"? – Gautam Shenoy Oct 29 '12 at 13:32
@GautamShenoy: I need to make some experiments with the densities that I obtain. After the likelihood ratio is inverted it should be inserted into some complicated equations and I search some solutions for them. The example which I gave has two step functions and when I finally invert the likelihood from that partial function mathematica can not solve the complicated equations. However I guess it means differentiability, althouhgh i didnt check. – Seyhmus Güngören Oct 29 '12 at 14:21

Here is a family of solutions:

Fix any density function $f_0$. For any positive increasing function $g$ integrable with respect to the measure $f_0(x)\,\mathrm dx$, consider the density function $$f_1:x\mapsto f_0(x)g(x)/\mathbb E(g(X_0)),$$ where $X_0$ denotes any random variable with density $f_0$. In other words, $\mathbb E(g(X_0))$ is the integral of $f_0g$ and $f_1$ is the density of any random variable $X_1$ such that, for every bounded measurable test function $\varphi$, $$\mathbb E(\varphi(X_1))=\frac{\mathbb E(\varphi(X_0)g(X_0))}{\mathbb E(g(X_0))}.$$ Then, the pair $(f_0,f_1)$ is a solution.

Naturally, this is barely more than a reformulation of the question, however it shows the diversity of the solutions $f_1$, for any given density function $f_0$.

Assume for concreteness that $f_0:x\mapsto\mathrm e^{-x^2/2}/\sqrt{2\pi}$ is the standard normal density. Then, as already noted, for every $\mu\geqslant0$, the function $g:x\mapsto\mathrm e^{\mu x}$ is admissible. This yields for $f_1$ the normal density with mean $\mu$ and variance $1$.

But many more examples exist, for example $g$ might be $x\mapsto\mathrm e^{cx\cdot|x|^a}$ for some $c\gt0$ and $|a|\lt1$, or the logistic $x\mapsto1/(1+a\mathrm e^{-cx})$ for some $c\gt0$ and $a\gt0$, or the trigonometric inverse $g=\frac\pi2+\arctan$, or the error function $\Phi$, or some shifted versions of these, or, more generally, any positive power of any cumulative distribution function, or...

-
Thank you very much did. Would you mind suggest one couple for my problem. I only want simplest possible such that my non-linear equations will possibly find some solutions in mathematica. The only constraint is that the densities should not be symmetric as in the mean shifted Gaussian case. – Seyhmus Güngören Oct 29 '12 at 14:32
I do not understand what you are asking. Surely you saw that the last paragraph of my answer enumerates a host of examples? (A minor point: of course $f_0$ and $f_1$ cannot both be even, otherwise $f_1/f_0$ would be even as well, which is an odd (if I may say) property for an increasing function.) – Did Oct 29 '12 at 14:38
yes I saw. They are for the function $g$ which is supposed to be multiplied by $f_0$ and then should be normalized to unit area. Did I understand correctly? – Seyhmus Güngören Oct 29 '12 at 14:45
You did.   – Did Oct 29 '12 at 15:05
I am sorry perhaps I was not clear. In my comment I was asking for a probable simple case such that my equations would not explode in mathematica. If I get one of your example, I am very much afraid that $\mathbb{E}(g(X_0))$ would look like horrible. – Seyhmus Güngören Oct 29 '12 at 15:55