Standard Deviation after N trials

[QuadsOverQuads]

Thanks to all posters for their replies.

I think the critical idea I needed here was that VAR(A+B) = VAR(A) + VAR(B).

Therefore, VAR(nA) = n VAR(A), and StdDev(nA) = SQRT(n VAR(A)). So, the StdDev increases in proportion to SQRT(n) as n increases.

That solved the problem for me [img]/images/graemlins/smile.gif[/img]


q/q

[peterchi]

[ QUOTE ]
Thanks to all posters for their replies.

I think the critical idea I needed here was that VAR(A+B) = VAR(A) + VAR(B).

Therefore, VAR(nA) = n VAR(A), and StdDev(nA) = SQRT(n VAR(A)). So, the StdDev increases in proportion to SQRT(n) as n increases.

That solved the problem for me [img]/images/graemlins/smile.gif[/img]


q/q

[/ QUOTE ]
Just to be a little nit-picky,

Var(A + B) = Var(A) + Var(B) + 2 Cov(A, B)

When A and B are independent, Cov(A, B) = 0 and then your equation is fine. But you have to know that they are; you can't just assume it if it's unwarranted. Here they clearly are though.

Also, be careful when you say Var(nA). It's not Var(nA) that you're going after. If that were the case, you have to square constants when you bring them out of the Variance quantity. So Var(nA) = n^2 x Var(A). But that isn't what you wanted anyways.

I know that might not make intuitive sense. I'm not sure how better to explain it, either. Just realize that nA is not your quantity of interest. You don't want the variance of "nA," that is, the number of spins times your outcome. Rather, you have n random variables that are identical and independent, each with some variance that you've found already, and you can simply sum these variances to find the variance of n spins.

[QuadsOverQuads]

Thanks for the additional info.

I ran into the concept of covariance when looking at the multinomial approach, but I wasn't clear how it applied. I think I'm catching on a bit now.

Anyway, the 'n' thing above was just a bit of shorthand for "the variance of n spins is equal to n times the variance for a single spin". All points well-taken, nonetheless [img]/images/graemlins/smile.gif[/img]

q/q

[jogsxyz]

[ QUOTE ]
Suppose I have a slot machine that pays off according to the following table:

1% : $10
2% : $5
3% : $2
5% : $1
89% : nothing

I understand how to compute the average payout and standard deviation for a single spin.

<u>But:</u> how do I determine the expected standard deviation over an arbitrary number of trials (ie: "N spins")?


[/ QUOTE ]

That's only $29. No machine is that much of a ripoff.
Calculate the sd of one trial.
The sd of n trials is the square root of n times the sd of one trial.

[DiceyPlay]

Let x be the the return of a single spin.
E(x) is the expected return on any spin.
Let sd(x) be the standard deviation of 1 spin.
Let x-bar be the average return on n spins.
E(x-bar) = E(x).
sd(x-bar) = sd/SQRT(n).

Notice as n approaches infinity sd(x-bar) approaches 0 and your observed average return approaches E(x).

Hope that's clear - it's correct.

[peterchi]

[ QUOTE ]
Let x be the the return of a single spin.
E(x) is the expected return on any spin.
Let sd(x) be the standard deviation of 1 spin.
Let x-bar be the average return on n spins.
E(x-bar) = E(x).
sd(x-bar) = sd/SQRT(n).

Notice as n approaches infinity sd(x-bar) approaches 0 and your observed average return approaches E(x).

Hope that's clear - it's correct.

[/ QUOTE ]
I debated whether to point this out, as you're definitely right. And part of me did think this was what he was really asking.

To try and clarify for you OP, as you might be confused about the two opposing answers now...

Before this post, we were all talking about the variance of the sum of n random variables, i.e. n spins, that are identical and independent. This variance will equal the sum of the independent variances as we said.

But here, DiceyPlay is talking about the variance (or standard deviation) of x-bar. This is a different quantity than the sum. But it is related... as we know from basic math, the average is the sum/n.

So now here's where my earlier point about Var(nA) actually matters. Let T = the sum of the Xs. So Var(T) = nVar(X) as we said before.

But what's x-bar? Well x-bar is the same as T/n. So Var(T/n) = (1/n^2)Var(T) = (1/n^2)(nVar(X)) = (1/n)Var(X) -&gt; stdev(T/n) = stdev(X)/sqrt(n)

Which is exactly what DiceyPlay has.

So which answer is "correct?"
Again,
Var(T) = nVar(X)
Var(T/n) = (1/n)Var(X)
Where T is the sum of n spins, T/n = X-bar, each spin is X (each identical and independent, so we can call them all the same)

If you want the variance of the sum of n trials, then it's the first one. If you want the variance of the mean after n trials, it's the second one.

And yes, notice that one goes to infinity and the other goes to zero, as n gets large.

[Beavis68]

[ QUOTE ]
"the squareroot of the variance for one spin times the numbers of spins in your sample is the SD for that number of spin I think"

I interpreted this as:
(STDEV of 1 spin) * n = (STDEV of n spins)

That's wrong and it should be:
(VAR of 1 spin) * n = (VAR of n spins)

[/ QUOTE ]

Yeah, I didnt say it very clearly, I was confused myself when I reread it.

[Beavis68]

[ QUOTE ]
[ QUOTE ]
"the squareroot of the variance for one spin times the numbers of spins in your sample is the SD for that number of spin I think"

I interpreted this as:
(STDEV of 1 spin) * n = (STDEV of n spins)

That's wrong and it should be:
(VAR of 1 spin) * n = (VAR of n spins)

[/ QUOTE ]



[/ QUOTE ]Yeah, I didnt say it very clearly, I was confused myself when I reread it.

I was trying to say "take the variance for one spin, multiply it by N, and then take the squareroot of that number for N SD.

[gaijinuronin]

I had the same problem at one moment.
A very similar problem has the following answer:
http://www.mathhelpforum.com/math-help/a...eviation-4.html

[jogsxyz]

VAR(X)=E(X²)+[E(X)]²

No need for sampling to calculate the VAR and sd.
The distribution was given.