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)?

  • $\begingroup$ How do you measure dispersion? $\endgroup$ Commented Jul 13, 2021 at 21:25
  • $\begingroup$ Why can't you calculate variance? $\endgroup$
    – Tortar
    Commented Jul 13, 2021 at 21:30
  • $\begingroup$ @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. $\endgroup$
    – Adam
    Commented Jul 13, 2021 at 21:47

1 Answer 1


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:

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


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

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

  • $\begingroup$ 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. $\endgroup$
    – Adam
    Commented Jul 14, 2021 at 17:32
  • $\begingroup$ Can you collect several sample points? It seems unreasonable that a price distribution is lacking a mean and variance $\endgroup$ Commented Jul 15, 2021 at 13:50

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