Expression for Variance

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  • #1
CAH
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I don't understand a question on finding an expression for the variance of something...


Attempt at solution: also I worked out c as (3/2) previously, which is correct
Var(U) = (3/2)^2 x Var(Xbar) = 9/4n^2 x a^2/18

I'll attach a photo of this too if it's easier to read, my problem is that I thought var(Xbar) = var(x/n) = 1/n^2 x var(x)

... But they have done var(Xbar) = 1/n x var(x) ...?
 

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Answers and Replies

  • #2
statdad
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Think about this:
[tex]
Var\left(\bar X\right) = Var\left(\frac 1 n \sum_{i=1}^n X_i\right) = \frac 1 {N^2} \sum_{i=1}^n Var(X_i)
[/tex]

What happens when you simplify the sum?
 
  • #3
CAH
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Ah I see so it's just sigma^2/n, is this the same for all cases as in, calculating the variance of a sample and estimating a population variance?.. In other words, kinda, will it ever (a level standard) be 1/n^2 x var(x)

Probably a stupid question, just checking
 
  • #4
statdad
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The only time you would have [itex] n^2 [/itex] in the denominator is if you were (for some reason) considering mathematically a single observation [itex] X_1 [/itex] and calculate
[tex]
Var\left( \dfrac{X_1}{n}\right) = \dfrac{Var(X)}{n^2}
[/tex]

"is this the same for all cases as in, calculating the variance of a sample and estimating a population variance?"
I'm not exactly sure what you mean by this, so if my response is off-target that's why.

If you've talked about sampling distributions for the sample mean, the expression [itex] \frac{\sigma^2}n [/itex] is the population variance for that sampling distribution. It will never have denominator [itex] n^2 [/itex], since, as long as the distribution being sampled has a variance, the steps shown above apply.

The (sample) variance of a sample is a different beast. Essentially
* if the population variance is [itex] \sigma^2 [/itex], then the sample variance
[tex]
s^2 = \dfrac 1 {n-1} \sum_{i=1}^n \, \left(x_i - \bar x\right)^2
[/tex]
is an unbiased estimator of the population variance

* If you refer to the variance of the sampling distribution of [itex] \bar x [/itex] - which is given above - then to estimate that you have two options
a) If you have a single sample, use the sample variance [tex] s^2 [/tex] to estimate the sampling distribution's variance by calculating
[tex]
\dfrac{s^2}{n}
[/tex]

b) If you have a large number of samples, all the same sample size, from the same population, then calculate each sample mean and treat those sample means as a new sample. The sample variance of those (call it [itex] s_{\bar x}^2[/itex]) is the estimate of [itex] \dfrac{\sigma^2}{n}[/itex]

So there are several subtleties to wade through, but in none but the one unusual and unrealistic comment I made at the start will [itex] \frac{\sigma^2}{n^2} [/itex] play a role.
 

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