# Comparing degree of dispersion without calculating variance

There are two (price) distributions of the same class, but they differ in parameter values. One distribution has a smaller upper bound and a greater lower bound, so intuitively we know it has a smaller dispersion. Unfortunately, it is very difficult to calculate the variances. Is there another formal way to establish the claim (that one distribution has a smaller dispersion)?

• How do you measure dispersion? Jul 13 at 21:25
• Why can't you calculate variance? Jul 13 at 21:30
• @Michael Greinecker It is a good question. If I am unable to calculate variance (because it is a complicated distribution function), what is a good alternative measure of "dispersion"? I don't necessarily need to use the word "dispersion". I used it only to convey the idea.
Jul 13 at 21:47

The following might help, although whether it's simpler than calculating the variances will depend on the particular functions. Suppose the two distributions are of random variables $$x_1$$ and $$x_2$$. First find the respective means $$\mu_1$$ and $$\mu_2$$. Then replace $$x_1$$ by $$y_1=x_1-\mu_1$$ and $$x_2$$ by $$y_2=x_2-\mu_2$$, with the effect of shifting the distributions so that they both have mean zero while preserving their respective dispersions. Let the cumulative probability distributions after this shifting be:

$$F_{Y_1}(y_1)=P(Y_1\leq y_1)$$ $$F_{Y_2}(y_2)=P(Y_2\leq y_2)$$

Comparing these distributions (the whole distributions, not just the upper and lower bounds), it may (or may not) be found that:

$$\forall y_1,y_2 <0, F_{Y_1}(y_1) = F_{Y_2}(y_2) \implies y_1>y_2$$

and

$$\forall y_1,y_2 >0, F_{Y_1}(y_1) = F_{Y_2}(y_2) \implies y_1

Those conditions, if satisfied, would show that distribution 1 is less dispersed than distribution 2.

• Thanks for the suggestion. Unfortunately for me, even the mean does not have a closed-form solution. I should have also made it clearer, these are two (price) distributions of the same class that differ in only 1 parameter value.