# Math Help - Order for quasi-convex functions

1. ## Order for quasi-convex functions

Hey,

does anyone know if there exist some conditions on the variance and mean of normally distributed random variables $X,Y$ such that it holds
$Ef(X) \leq Ef(Y)$ for $f$ quasi-convex (but not convex)?

Unfortunatly, I am not very familiar with stochastic orders (convex, increasing convex,..) so far. I just started familiarizing myself with it using the book of Mueller "Comparison Methods for Stochastic Models and Risks". Can anyone recommend other books/papers?

2. ## Re: Order for quasi-convex functions

Hey Juju.

One suggestion I would have is to look at the parts that are increasing and decreasing and use the fact that if f() is monotonic increasing and E[X] < E[Y], then E[f(X)] < E[f(Y)]. Similarly if f is monotonic decreasing, you will have E[f(X)] < E[f(Y)].

However if there are turning points, then one needs to look at a mixture of this above results. If for example f(x) is always greater than 0 for all x, then you should get the above result that E[f(X)] < E[f(y)] for E[X] < E[Y].

3. ## Re: Order for quasi-convex functions

Thank you.

But I think that $f$ monotonically increasing and $EX does not imply $Ef(X) in general (without any conditions on the variance).

Best regards.

4. ## Re: Order for quasi-convex functions

The means and the variances are independent parameters.

If you modify the variance for a normal distribution, the mean is still the same due to symmetry around the mean.

If you need to prove the result, use the symmetry property as a means to do so in combination with the invariance principle for MLE estimators.

Under MLE if an estimator for mu is mu_hat, then the estimator for f(mu) is f(mu_hat).

This proves the result.

5. ## Re: Order for quasi-convex functions

Thank you Chiro.

But the stochastic order $(\leq_{st})$ tells us that
$X\leq_{st} Y,$ if $Ef(X)\leq Ef(Y)$ for all increasing $f.$

In my situation we have normally distributed random variables $X\stackrel{d}{=}\mathcal{N}(\mu_1,\sigma_1^2)$ and $Y\stackrel{d}{=}\mathcal{N}(\mu_2,\sigma_2^2).$ Now it can be shown that
$X\leq_{st} Y,$ if and only if $\mu_1\leq \mu_2$ and $\sigma_1^2=\sigma_2^2.$
Then, we need to have conditions on the variance.

What is wrong in my "thinking"?
Thanks in advance and best regards.

6. ## Re: Order for quasi-convex functions

How did you prove your claim? I used the MLE invariance property (that assumes monotonicity as well).

Maybe you can show us your proof so we can take it from there.

7. ## Re: Order for quasi-convex functions

Thank you very much for your help.

A. Mueller has shown the following statement in his paper "stochastic ordering of multivariate normal distributions":

$X\stackrel{d}{=}\mathcal{N}(\mu_1,\sigma_1^2)$ and $Y\stackrel{d}{=}\mathcal{N}(\mu_2,\sigma_2^2).$ Now it can be shown that
$X\leq_{st} Y,$ if and only if $\mu_1\leq \mu_2$ and $\sigma_1^2=\sigma_2^2.$

Sketch of his proof:
$"\Rightarrow"$ $X\leq_{st}Y$ $\Rightarrow$ $\lim\limits_{t \to -\infty}\frac{f_Y(t)}{f_X(t))}\leq 1$ and $\lim\limits_{t \to +\infty}\frac{f_Y(t)}{f_X(t))}\geq 1$. But this is only possible if $\mu_1\leq \mu_2$ and $\sigma_1^2\leq \sigma_2^2.$

8. ## Re: Order for quasi-convex functions

You don't need to specifically have the probabilities (i.e. the cumulative ones) to be less at every point for the expectations to be less.

The condition you give is a lot stronger than what is required. Your condition is really really strong, and the normal distribution doesn't require that specifically.

The MLE invariance procedure given that the function is monotonic increasing satisfies the constraints you desire.

Note that the invariance principle is for an estimator so if you wanted specific probability proofs, then you might need to use results like the one shown.

9. ## Re: Order for quasi-convex functions

Maybe I do not understand your argumentation in the right way.
But the claim I stated was an "if and only if" condition. Such that it follows:
if $\sigma_1^2\neq \sigma_2^2, \mu_1\leq \mu_2$ it does NOT hold $Ef(X)\leq Ef(Y)$ for all increasing $f.$