# What's the common way to measure the “stability” or “volatility” of a distribution? Variance seems insufficient

for example, in a discrete distribution A

P(X) = 0.999  when x = 0
P(X) = 0.001  when x = 1000

E(X) = 1
Var(X) = 1000-1=999


compared to B

P(X) = 0.6  when x = 0
P(X) = 0.2  when x = 1
P(X) = 0.2/2  when x = 2
P(X) = 0.2/4  when x = 4
P(X) = 0.2/4  when x = 8

E(X) = 1 =A
Var(X) = 0.2+0.4+0.8+3.2-1=3.6 <<A


However B seems more volatile(instable) than A. What is the common correct stats to measure this sort of thing?

Update:

Entropy might be a good candidate, however it doesn't look into the information itself (it only cares about the chance). I feel the time series graph of a fair coin which could score (0 or 1) is less volatile / more stable than a fair coin which could score (0 or 100).

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Entropy might be what you're after. –  paul Apr 25 '12 at 20:26
@paul thank u. did cryptography course 3 years ago. Have to relearn that now... –  colinfang Apr 25 '12 at 20:28
What is your "Var(X)"? It's definitely not the variance! –  whuber Apr 25 '12 at 21:41
@whuber: I suspect you are supposed to read example A as $\Pr (X=0)=0.999$ and $P(X=1000)=0.001$, which would indeed make $var(X)=999$. –  Henry Apr 26 '12 at 0:19
Thank you, @Henry: that works. –  whuber Apr 26 '12 at 13:56
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If you wanted something less dramatic than variance because of your concerns about the impact of a rare extreme outlier, you might try the average absolute deviation from the median. For for example A, you would get $$|0-0|\times 0.999 + |1000-0|\times0.001 =1$$ while for example B $$|0-0|\times 0.6 + |1-0|\times 0.2 + |2-0|\times 0.1 + |4-0|\times 0.05 + |8-0|\times 0.05 = 1.$$
It is not quite a coincidence this measure of dispersion is equal to the expectation for these two examples: this will happen for any random-variable where $\Pr(X \lt 0)=0$ and $\Pr(X = 0) \ge \frac12$.