Variance of standard deviation

Just wondering about the variance of standard deviation, does standard deviation of a certain number of events have the same standard deviation applied to it ?

I would assume so but interested to hear from someone wiser than me.

And how is it calculated??

I'm not sure, but I think you're misunderstanding the concept here.

Standard Deviation squared IS Variance.

If what you're asking is how to calculate STD DEV over multiple trials it's pretty easy. The formula is (SQRT of # of trials) * STD DEV. So if it was 100 trials with a STD DEV of 5, then the result would be a STD DEV of 50.

Quote: Walkinshaw30t

Just wondering about the variance of standard deviation, does standard deviation of a certain number of events have the same standard deviation applied to it ?

usually there is a suggestion you read up on it.

First step might be the terms, you are using two of them in your question. You seem to have a grip on variance. The standard deviation is the square root of the variance. I guess it must be true that the variance of standard deviations has a standard deviation, that brings a smile. Once we get this standard deviation of whatever, we get meaning out of a range of numbers.

Clearly the number of events, say the number of bets, does not always produce the same SD. You must be aware that various games have different variances associated with them. Thus the SDs must be different.

I am no math whiz so I can relate to someone trying to comprehend this. The problem for us is that the concept is fairly simple, but the math is not. Here are the hieroglyphics they throw at you.



There are calculators as below. This one shows the mean; in Craps you know that should be 7.

I inserted this range of results:

2,3,3,4,4,4,5,5,5,5,6,6,6,6,6,7,7,7,7,7,7,8,8,8,8,8,9,9,9,9,10,10,10,11,11,12

which Craps players know is a perfect distribution of 36 throws. The mean was indeed calculated to be 7. edit:the SD of Craps is supposed to be close to 1, but I think this is different because there is a range of 36.

For gamblers, for the most part you try to understand the concept, but you rely on the mathematicians to arrive at the facts.

http://www.calculator.net/standard-deviation-calculator.html

Thanks for reply-
The intent of the question is to find out if an outcome that is more than 2 standard deviations away from expected result in a certain number of events happens less than 5% of the time, in a group of 100 trials of that same number of events can I rely on the probability that 5 of those times will be over 2 standard deviations from a average or because of low occurance of being more than 2 standard deviations is there a high variance to the number of times that the results will be more than 2 standard deviations.

Eg betting on a single number in roulette has a low probability of hitting and therefore has a much higher variance than an even chance bet.

I hope I make some kinda sense!

Cheers

in most things 100 trials is not enough, so I don't think this would be any different. If in 100 trials as you describe, a player had 10 outcomes outside of 2 standard deviations on the 'bad side' and never on the 'good side' it would prove nothing, is my guess.

I have no math to show you on that.

Yes I agree. It would be great to hear the Wizard give his input on this matter

Quote: odiousgambit

I guess it must be true that the variance of standard deviations has a standard deviation

The is a conceptual misunderstanding here.

Standard deviation (or it's square the variance) is a property of the statistical ensemble itself. For example in roulette for a given bet, the ensemble consists of all possible spin result with their subsequent payoffs. In Baccarat it consists of all possible combinations of the next 6 cards in the shoe.
If you know the whole ensemble, you can *calculate* the standard deviation. This is a simple task in roulette (there are only 37 or 38 possible outcomes), but is more difficult in card games.

So the standard deviation is as fixed as the rules of the game. The standard deviation of the standard deviation is exactly *zero*, as for any other constant.


Now let's look at a different scenario: Imagine some new game is served, but you don't know the rules of the game at all. You still want to know it's standard deviation. Other than asking the provider or their inventors - since you can't calculate the standard deviation from the ensemble (as you don't know any rules), you will never know the standard deviation. So how do you handle this problem ? You can still place bets and observe the outcomes of the bets. Do it multiple times. From your results you can get *estimates* of the standard deviation by some clever formulas. But these estimates are very different from the (ensemble) standard deviation, as the estimates *do* fluctuate (every different series of results gives you a different estimate).

Back to the original question: There is a non-zero standard deviation of estimates of standard deviations. These depends of course depend on the ensemble standard deviation, but also on the estimator details - mainly the number of tries.

Quote: MangoJ

There is a non-zero standard deviation of estimates of standard deviations.

Yes, but look closely at what he is asking. It seems that once you know the S.D. of what you are playing, it is only natural to wonder how much you can depend on it. Let's say the 2.5% chance of the worst outcomes [say being outside 3 S.D.s on the wrong end] would be important to the player; he decides he can survive that occurring, but not having 5% land there. How much can he count on it?

I must quickly bow to superior knowledge but it seems to me the question is valid. Would it be true that anything that deviates should have a standard deviation? Otherwise aren't you saying he can absolutely count on his luck not being worse than that bell-shaped curve indicates?

Yes thats exactly what Im looking to answer ^^

Quote: odiousgambit

Yes, but look closely at what he is asking. It seems that once you know the S.D. of what you are playing, it is only natural to wonder how much you can depend on it. Let's say the 2.5% chance of the worst outcomes [say being outside 3 S.D.s on the wrong end] would be important to the player; he decides he can survive that occurring, but not having 5% land there. How much can he count on it?

I must quickly bow to superior knowledge but it seems to me the question is valid. Would it be true that anything that deviates should have a standard deviation? Otherwise aren't you saying he can absolutely count on his luck not being worse than that bell-shaped curve indicates?

If anyone can give some input on this it would be much appreciated...

If you could restate your original question in different words, then maybe someone can help you.

Your question "does standard deviation of a certain number of events have the same standard deviation applied to it ?" doesn't make much sense.

The standard deviation of an event is a fixed quantity (at least in the frequentist statistics view), so is the standard deviation of any number of events.
Hence the standard deviation of the "standard deviation of a certain number of events" is zero, no ?

I know its old but I still would be very interested to hear what any maths experts would have to say regarding this

I know its old but I still would be very interested to hear what any maths experts would have to say regarding this

I have perused the thread but am still unclear on the question you wish answered. Before I try to give you a "mathy" explanation can you please try to give me a concrete, real-world example?

My gut reaction mathy answer:
Your goal is to estimate the sd. Hence you want to know how to estimate variance of the sd so you can construct a confidence interval around your sd estimate. Is that right?

Quote: endermike

I have perused the thread but am still unclear on the question you wish answered. Before I try to give you a "mathy" explanation can you please try to give me a concrete, real-world example?

My gut reaction mathy answer:
Your goal is to estimate the sd. Hence you want to know how to estimate variance of the sd so you can construct a confidence interval around your sd estimate. Is that right?

Yes that is correct.

Here are some good references:

http://graphpad.com/support/faqid/1381/

http://www.stat.purdue.edu/~tlzhang/stat511/chapter7_4.pdf

http://en.wikipedia.org/wiki/Variance#Distribution_of_the_sample_variance

http://en.wikipedia.org/wiki/Standard_deviation#Confidence_interval_of_a_sampled_standard_deviation

If they don't clear things up post again.

Thanks mate ill have a read

SD hurts people, it hurts the children and the
elderly the most. Just say no to standard
deviation.

Quote: Walkinshaw30t

Just wondering about the variance of standard deviation, does standard deviation of a certain number of events have the same standard deviation applied to it ?

I would assume so but interested to hear from someone wiser than me.

I don't know about "wiser" but I'll add, in good faith, what I can.

For any random population of outcomes (from dice to shoe sizes) the Central Limit Theorem predicts a convergence of the Probability Density Function (PDF) to be the familiar Gaussian Normal (bell) curve. This assumes that the individual outcomes are not correlated (e.g. I don't just use 100 measurements of my own shoe size, but select them randomly from a diverse set of unrelated strangers). The Normal distribution, if it applies to your problem is attractive for the following reasons:

It can be totally described by only two parameters, the mean and the variance (the standard deviation is not a "third" parameter because it is merely the square root of the variance).

It is defined by only the above two parameters, a wide range of specific questions can be answered simply by comparing the area contained between any two x-axis points. For example, there is a (rounded) 68% chance that any future sample will lie (on the x-axis) between the X1 (mean - SD) and X2 (mean + SD).

I think you probably already knew this, but I mentioned any as background. Now the water gets deeper.

The cherished "bell curve" is a poor fit for many valid random distributions. For example, the probability distribution of a single die being a [1,2,3,4,5,6] is a uniform distribution. When plotted, it looks like a rectangle with constant height and certainly a finite x-range [1-6]. There are other cases (like shoe size) for which the bell curve (say with a mean of 8 and standard deviation of 2 will actually predict a small number of negative shoe sizes, because it extends infinitely on both ends. That's another example of how free-wheeling use of the bell curve can result in silly predictions.

A bell curve will be different depending on whether it is based on an entire population, or a sample of that population. Sampling (think polling) in general is a cheap way to estimate the mean and variance of a much larger population. You can literally make very accurate predictions (say, within 1%) about a population of millions just by sampling a few thousand, so the appeal of this technique is obvious. The only problem is that you must know the limitations of these extrapolations before you go on the air with them. [You'll hear newscasters say things like "this poll is accurate within 3 percentage points"] Actually it's more complicated than that--the correct disclaimer would be ("there is a 95% chance that ... this poll is accurate within 3 percentage points".

One last thing you should know about "sampled populations" is the best estimate of the "true mean" is simply the average of your samples. It ain't perfect, but it's the best you can do with a finite number of samples. However, the best estimate of the true variance is (n)/(n-1) times the formula for standard deviation (the expected value of the squared deviation from each point (Xi) from the mean, or V = E[(X - u)^2] or, if you prefer, V = E(X^2) - [E(X)]^2 where "X" is the vector of your x values: X = [x1, x2, x3 ... xn].

Why are is the formula for sampled estimate the mean the same as the formula for the actual mean, yet not so for standard deviation? Probably too far off track for this topic, but "foxy" in its own right.

OK, I'm getting off topic. Sorry. Let me paraphrase what I think you're asking, and you tell me how close I am to your question. Yes there is such a thing as a standard deviation of a standard deviation, but not for the simple bell curves of the type which best apply for you. IOW, if we consider the standard deviation to be the "first moment" of a distribution, there are also "second moments" and "third moments" and so on. But these "higher moments" are only non-zero for more complex curves than the bell curve. [The fourth standardized moment is sometimes called the kurtosis.] There are also a whole family of "skewed" curves which represent certain populations better than the bell curve.

To summarize, the worst question you can ask of probability theory is "is it possible" because the answer is almost always yes, but usually too loose to have any practical value for making "real life" decisions. Example: Is it possible that all the air molecules in my room will accidentally end up in one corner (just by chance) and I'll suffocate? Yes, it is possible (people have even computed the remote odds of this happening)! But seriously, is that any reason for most of us to care?

Instead, ask how probable something is because therein lies the path to practical wisdom (making the best bet, or finding the "optimal" course of action in daily life. Hope this helps.

Thanks dblanch256.
Yes that answers what I wanted and gives me a further understanding overall.
I appreciate you taking the time to explain it to me.
Cheers mate

Does this answer the practical question?

A random variable (your win/loss on one bet with stake 1) has a mean of M and a standard deviation S. If it were Normal-Gaussian, it would mean that the probability of getting a value higher than two S above M (M+2S) is equal to 2.28%. But one bet is not Gaussian distributed.

If you bet a stake of 100, then your mean is 100M and your SD is 100S. Same conclusions about the probabilities. Still not Gaussian.

Now if you repeatedly bet 1, a hundred independent times, then your overall win/loss has a mean of 100M but a standard deviation of only 10S. Also, the repetition kicks in the theorem that says it finally resembles a Gaussian distribution. Hence you CAN say that the probability of reaching 100M+20S is 2.28%

Example: roulette, red/black.
M = -2/38 = -0.0526 s = 0.2233
100 trials:
100m = -5.263 ; 20s = 4.466
There is but 2.28% chance that your final result is above -0.797
(and only 0.95% of not being negative).

I think you used "variance" as simply meaning "change" in the original post. Beware! it has a very specific meaning in probability theory.

I saw a homeless guy holding up a sign that
said 'Victim of Standard Deviation'. It's insidious.

The variance of the standard deviation is zero. The standard deviation is theoretical, not observed, so every comparable trial gives the same standard deviation.

Seriously?? You're saying that in order to calculate the standard deviation you have to KNOW the standard deviation??? Your formula says:

Standard Deviation = Sqrt(# trials) x Standard Deviation !!!

This works out to Standard Deviation/Standard Deviation = Sqrt(# trials), OR Sqrt(#trials) = 1 !!!!!

Is it true that the 2nd law of Thermodynamics can be used for profit in casino games like roulette?

Quote: JohnB

Is it true that the 2nd law of Thermodynamics can be used for profit in casino games like roulette?
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No, but if you could find a way to circumvent the 2nd Law, you'd be a billionaire.

Ha ha, thx

Seriously: the constant movement between disequilibria promises opportunities. Since it is known in which direction the movement is, it should be possible to take advantage of this knowledge. Is it a matter of stack or tablelimits?

Quote: JohnB

Seriously: the constant movement between disequilibria promises opportunities. Since it is known in which direction the movement is, it should be possible to take advantage of this knowledge. Is it a matter of stack or tablelimits?
link to original post

Taking advantage of movement between disequilibria would seem to require dependent trials.