Standard Deviation in terms of Probability Function

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falranger
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Homework Statement


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.
Typeset derivation: standard deviation sigma squared equals the limit as N goes to infinity of (1/N) times the sum over i from 1 to N of (x_i minus mu) squared, then rewritten as the sum over j from 1 to n of (x_j minus mu) squared times P(x_j), equal to the sum over j of x_j squared P(x_j) minus mu squared.


Homework Equations


I really don't know what to do here.


The Attempt at a Solution

 
But I can't really see how to turn mu^2/n into P(x_j)
 
$$\frac{1}{N}\sum_{i=1}^N \mu^2 = ?$$
... though you seem to be changing between n and N in the sums.
 
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.
 
falranger said:
I did:

Handwritten-style derivation of mean and variance in limit form: mu equals the limit as N goes to infinity of the sum over j of x_j P(x_j), and sigma squared expansions with summation over i and j, ending in sigma squared equals the limit of the sum over j of (x_j - mu) squared times P(x_j).

The expression
[tex]\mu = \lim_{N \to \infty} \frac{1}{N} \sum_{i=1}^N x_i[/tex]
is correct, but the expression
[tex]\mu = \lim_{N \to \infty} \sum_{j=1}^n x_j P(x_j)[/tex] is false. Do you see why?
 
falranger said:
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.
So what is ##\mu## the mean of?
 
falranger said:

Homework Statement


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.
Handwritten-style typeset derivation: standard deviation squared defined as limit as N goes to infinity of (1/N) times sum from i=1 to N of (xi minus mu) squared, rewritten as limit of sum from j=1 to n of (xj minus mu) squared times P(xj), then limit of sum of xj squared P(xj) minus mu squared.


Homework Equations


I really don't know what to do here.


The Attempt at a Solution


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:
Simon Bridge said:
So what is ##\mu## the mean of?

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.
 
falranger said:
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.

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'.
 
Ray Vickson said:
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'.

Ok I realize this. But I'm still not sure how to move on. For one thing,
Typed equation mu equals the sum from j = 1 to n of v_j times F(v_j), followed by the line 'Then what is mu squared?'
 
falranger said:
Ok I realize this. But I'm still not sure how to move on. For one thing,
Typeset equation mu = sum from j=1 to n of v_j F(v_j), followed by the line 'Then what is mu^2?'

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##.
 
This is where I am currently stuck at:
Handwritten-style derivation of standard deviation sigma squared as a limit over N of sums of x_j N P(x_j) minus mu squared, expanded into three successive lines of equations.


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