Standard Deviation in terms of Probability Function

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Homework Help Overview

The discussion revolves around understanding the concept of standard deviation in relation to a probability function, particularly focusing on the definitions and calculations involving the mean (μ) and variance (σ²) in a statistical context.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between different notations and definitions of mean and variance, questioning the use of limits and the implications of notation on understanding the problem.

Discussion Status

Some participants have provided clarifications regarding the definitions of μ and σ², while others express confusion about the implications of these definitions and how to proceed with the problem. There is an ongoing exploration of the correct interpretation of the expressions given in the homework statement.

Contextual Notes

Participants note the potential confusion arising from the use of different symbols (N and n) and the notation used in the problem statement. There is also mention of the need to clarify the meaning of μ in the context of the distribution versus measured values.

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.
P4fFhXl.jpg


Homework Equations


I really don't know what to do here.


The Attempt at a Solution

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

3waloSA.jpg
 
There's a 1/N outside the sum.
 
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:

3waloSA.jpg

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?
 
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?
 
  • #10
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.
P4fFhXl.jpg


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:
  • #11
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.
 
  • #12
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'.
 
  • #13
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,
xrR9jUA.jpg
 
  • #14
falranger said:
Ok I realize this. But I'm still not sure how to move on. For one thing,
xrR9jUA.jpg

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
This is where I am currently stuck at:
CR7J0NY.jpg


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

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