How do I determine if my variance is significant?

For a given group of data, I have both a mean and variance (already calculated). Looking over the raw spread of data, and its numerous outliers, I expect that variance will be high.

My question is -- now that I have my variance calculated, what do I need to compare it to to determine if the variance is significant (i.e., indicative of highly skewed data)?

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I'm no statistician, but I think you might get better answers if you asked at stats.stackexchange.com instead. – Rahul Nov 28 '10 at 19:37
I did not realize that such a site existed. Could a moderator kindly move the question then? – Raven Dreamer Nov 28 '10 at 19:59
SD [or variance] can be, but is not always related to skewness [lack of symmetry]. it is more related to how far from the center of the distribution some observations tend to be. are you interested in skewness per se - or do you really want to know if the variance is high? – ronaf Nov 29 '10 at 2:21
ps: i don't think a moderator will move your question for you. my understanding is that you have to post it yourself on stats.stackexchange. you can also delete it from this site, if you like - or perhaps just cross-reference it in the new posting. – ronaf Nov 29 '10 at 2:29
@Ronaf -- I'm interested in skewness. And according to this (meta.stats.stackexchange.com/questions/505/…) Jeff already added in a migration path. – Raven Dreamer Nov 29 '10 at 2:31

1 Answer

Taken from this post of mine from a poker forum.

You need to somehow compute the standard deviation for your distribution, and then count how many standard deviations away from the expected value your actual result is.

Let's say we are flipping coins, and each time it lands heads you give me a dollar, and each time it's tails I give you a dollar.

Then let's keep track of how much money I have made or lost. On average I expect to make nothing, but in practice I will most likely lose or win a finite amount. In fact if we flip the coin $N$ times, then I am most likely to win or lose somewhere around the $\sqrt{N}$ dollars. Why is this true? It's actually easy to show.

Let's call my profit/loss $M$. Then $M$ starts at zero, and the average change in $M$ after one or more flips is zero.

Here is the clever trick: let's consider the average change in $M^2$. Well after one flip it's quite clear that this will be 1, since +1 and -1 squared are both equal to 1.

Notice that after one flip the average profit/loss is zero, but in reality I either gained or lost a dollar. $M^2$ however is 1, so you can see it sort of measures the fluctuations in my profit/loss.

Call $M(N)$ my profit/loss after N flips. Then what is $M(N+1)^2$? Since I can either gain or lose 1 dollar, we have either

$M(N+1) = M(N) + 1$ or $M(N+1) = M(N) - 1$

lets square these two expressions and take the average, which I write as $<M(N+1)^2 >$. If you do this, which is quite easy, you will find that:

$<M(N+1)^2>$ = $<M(N)>^2 + 1$

If you think about it, this implies that after $N$ flips, the average of $M^2$ will be $N$; therefore the square root of the average of $M^2$ is $\sqrt{N}$, which is just a fancy way of saying the SD for flipping a coin N times is $\sqrt{N}$.

This shows that the SD for flipping a coin (1:1 odds) is $\sqrt{N}$, and it's simple to modify the above argument to find the SD for weighted coins. Therefore, besides telling us how good/bad you are running, the magnitude of the SD also tells you, in poker, how well on average you are getting your money in. In other words, SD can be used to measure what average chance you have to win the pot when you put all your chips in with cards left to come.

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