Standard Deviation Question

[Troll_Inc]

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Coin flipping has finite variance.

Poker's variance has infinite possibilities.

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Having infinite possibilities doesn't imply infinite variance. And the distribution of the outcome of a single hand in NLH has finite variance unless you assume two players have infinite bankrolls.
Thus the outcome of n hands will converge to a normal distribution with large n. But this convergence is very much slower than most people suspect it to be. I.e. BB/100 is by no means normal, BB/1000 isn't, BB/1m probably comes close.

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Same problem here.....starts talking about variance then stumbles into the mean (average).

[sixhigh]

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starts talking about variance then stumbles into the mean (average).

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wtf? care to elaborate?

[sixhigh]

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A much shorter list are the things that do converge to the normal.


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Btw this statement is wrong.

[Troll_Inc]

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A much shorter list are the things that do converge to the normal.


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Btw this statement is wrong.

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I'm so right it's laughable.

I'll tell you what. For evey non-normal daily distribution I name you have to name a normal one. When you give up, I'll name an extra 30. LOL.

Non normal 1. The weight of items in my kitchen. Totally non-normal. I have tons of small items and very few large.

Your turn.

[sixhigh]

You obviously don't understand what the phrase 'converges to a normal distribution' means. The weights of items from a given set don't converge or whatsoever. And nobody ever claimed that every distribution is normal.

[Troll_Inc]

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You obviously don't understand what the phrase 'converges to a normal distribution' means. The weights of items from a given set don't converge or whatsoever. And nobody ever claimed that every distribution is normal.

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You don't understand the difference between mean and variance as evidenced by your post above. You start talking about variance and then even in the same paragraph all the sudden start talking about means.

I also didn't say that everything is non-normal. There are just way more non-normal distributions than normal. I already proved my argument apparently as you can't name a single normal distribution and already want to change your argument.

[sixhigh]

Also you're obviously not familiar with the phrase 'by no means'.

And I don't know which argument you want to 'prove' here - that most distributions aren't normal? That's a pretty obvious fact. That most distributions don't converge to a normal distribution?

The CLT states sth. like: The normalized sum of independent identically distributed random variables converges to a normally distributed random variable.

[Troll_Inc]

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Also you're obviously not familiar with the phrase 'by no means'.

And I don't know which argument you want to 'prove' here - that most distributions aren't normal? That's a pretty obvious fact. That most distributions don't converge to a normal distribution?

The CLT states sth. like: The normalized sum of independent identically distributed random variables converges to a normally distributed random variable.

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The CLT doesn't rule my kitchen!

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i'm glad it's not just me who has no idea what troll is talking about. possibly trolling? but who would troll a stats forum? :/

btw troll the CLT does rule your kitchen

[jason1990]

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eh, doesn't everything converge to the normal?

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No.

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okay, what doesn't?

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Here is an example. An object is located at the origin of the number line. There are n other objects with the same mass distributed uniformly on the interval [-n,n]. Let us estimate the net gravitational force on the object at the origin, assuming n is large.

Let X_j denote the location of the j-th object. Then the net force on the center object is proportional to

F_n = \sum_1^n sgn(X_j)/|X_j|^2.

For simplicity, let us take the constant of proportionality to be 1. We can write X_j = nY_j, where Y_j are iid uniform [-1,1]. Then F_n = n^{-2}\sum_1^n Z_j, where

Z_j = \sgn(Y_j)/|Y_j|^2.

Note that E|Z_j| = infinity, so these variables do not even have a first moment, let alone a second moment. Hence, we cannot apply the central limit theorem. As it turns out, as n goes to infinity, the distribution of F_n converges to a distribution with characteristic function e^{-c|t|^{1/2}}, for some constant c. For comparison, the characteristic function of the standard normal is e^{-0.5|t|^2}.