Standard Deviation in terms of Probability Function

1. Jan 27, 2013

falranger

1. The problem statement, all variables and given/known data
I've still yet to learn Latex since I'm pretty good with words equation editor, so here's the question typed out in words.

2. Relevant equations
I really don't know what to do here.

3. The attempt at a solution

2. Jan 27, 2013

Simon Bridge

Have you tried starting by expanding out the brackets in the sum?

3. Jan 27, 2013

falranger

I did:

4. Jan 27, 2013

Simon Bridge

There's a 1/N outside the sum.

5. Jan 27, 2013

falranger

But I can't really see how to turn mu^2/n into P(x_j)

6. Jan 27, 2013

Simon Bridge

$$\frac{1}{N}\sum_{i=1}^N \mu^2 = ?$$
... though you seem to be changing between n and N in the sums.

7. Jan 27, 2013

falranger

Yeah, N is the number of trials, and n is the number of possible outcomes from each trial. and also, considering mu is also a summation, I don't really know how to calculate mu^2.

8. Jan 27, 2013

Ray Vickson

The expression
$$\mu = \lim_{N \to \infty} \frac{1}{N} \sum_{i=1}^N x_i$$
is correct, but the expression
$$\mu = \lim_{N \to \infty} \sum_{j=1}^n x_j P(x_j)$$ is false. Do you see why?

9. Jan 27, 2013

Simon Bridge

So what is $\mu$ the mean of?

10. Jan 27, 2013

Ray Vickson

A lot of your difficulties are, I suspect, a result of using bad notation, so first, let's change it to something sensible. You have a random sample $X_1, X_2, \ldots, X_N$ drawn independently from a distribution $\{ v_j, P(v_j), \: j = 1,2,\ldots n \}$. That is the possible values of each $X_i$ are $v_1, v_2, \ldots, v_n$ with probabilities $P(v_1), P(v_2), \ldots, P(v_n).$ Now by definition, $\mu = \sum_{j=1}^n P(v_j) v_j$ and $\sigma^2 = \sum_{j=1}^n P(v_j)(v_j - \mu)^2.$ Note that there are no N's or limits, or anything like that in these two expressions!

You are being asked to show that μ and σ2 are also given by the limiting expressions wiritten in the question. (To be technical, these are "almost-sure" equalities, but never mind than for now.)

Last edited: Jan 27, 2013
11. Jan 27, 2013

falranger

μ is supposed to be the average value of measured x.

And also, the question is given to us like that, but I understand what you mean since I do believe that's what it should be. And I've been wondering about the N and limit myself since at the end there are no N values in the expression.

12. Jan 27, 2013

Ray Vickson

No, $\mu$ is not supposed to be the average of the measured x; it is supposed to be the mean of the distribution---given by the formula in my previous post.

Yes, there are no N values and limits in the end, because what you are being asked to show is the the large-N limit equals something, and that 'something' is just a number, not a function of N. For example, $\lim_{N \to \infty} 1 + (1/N) = 1,$ and at that point the '1' does not have any N's in it, or any 'lim', or anything like that: it is just the number '1'.

13. Jan 27, 2013

falranger

Ok I realize this. But I'm still not sure how to move on. For one thing,

14. Jan 27, 2013

Ray Vickson

The number $\mu^2$ is just the square of the number $\mu$---no, I am not kidding! It is a number, and has nothing at all to do with $X_1, X_2, \ldots, X_N$.

15. Jan 27, 2013

falranger

This is where I am currently stuck at:

If μ is just a number then I don't see a way to move on.