<p>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 <span class="math-container">$x_1$</span> and <span class="math-container">$x_2$</span>. First find the respective means <span class="math-container">$\mu_1$</span> and <span class="math-container">$\mu_2$</span>.  Then replace <span class="math-container">$x_1$</span> by <span class="math-container">$y_1=x_1-\mu_1$</span> and <span class="math-container">$x_2$</span> by <span class="math-container">$y_2=x_2-\mu_2$</span>, with the effect of shifting the distributions so that they both have mean zero while preserving their respective dispersions.  Let the cumulative probability distributions <em>after this shifting</em> be:</p>
<p><span class="math-container">$$F_{Y_1}(y_1)=P(Y_1\leq y_1)$$</span>
<span class="math-container">$$F_{Y_2}(y_2)=P(Y_2\leq y_2)$$</span></p>
<p>Comparing these distributions (the whole distributions, not just the upper and lower bounds), it may (or may not) be found that:</p>
<p><span class="math-container">$$\forall y_1,y_2 &lt;0, F_{Y_1}(y_1) = F_{Y_2}(y_2) \implies y_1&gt;y_2$$</span></p>
<p>and</p>
<p><span class="math-container">$$\forall y_1,y_2 &gt;0, F_{Y_1}(y_1) = F_{Y_2}(y_2) \implies y_1&lt;y_2$$</span></p>
<p>Those conditions, if satisfied, would show that distribution 1 is less dispersed than distribution 2.</p>

CC BY-SA 4.0
2021-07-14T17:32:03.733Z
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.
Adam
CC BY-SA 4.0
2021-07-15T13:50:43.240Z
Can you collect several sample points? It seems unreasonable that a price distribution is lacking a mean and variance
RegressForward
CC BY-SA 4.0
2021-07-14T11:01:31.190Z
2021-07-14T11:01:31.190Z
Adam Bailey
<p>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)?</p>

CC BY-SA 4.0
2021-07-13T21:25:23.623Z
How do you measure dispersion?
Michael Greinecker
CC BY-SA 4.0
2021-07-13T21:30:59.337Z
Why can't you calculate variance?
Tortar
CC BY-SA 4.0
2021-07-13T21:47:00.300Z
@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.
Adam
<variance>
Comparing degree of dispersion without calculating variance
Adam
# Comparing degree of dispersion without calculating variance

(Asked by: username_0 on 2021-07-13)

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)?

**Tags:** variance
> **Comments:**
> **username_1**: How do you measure dispersion?  
> **username_2**: Why can't you calculate variance?  
> **username_0**: @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.  

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## 1 answers

### Answer by username_3 on 2021-07-14. Score: 2

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.

> **Comments:**
> **username_0**: 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.  
> **username_4**: Can you collect several sample points? It seems unreasonable that a price distribution is lacking a mean and variance  

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