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Suppose we draw a sample $x$ from a population $X$, this sample has $n$ random variables $(x_1, x_2, x_3... x_n)$, the sample mean is $\bar{x}$, and it's variance is $v(x)$, the whole population is $X$, has a mean is $\mu$ , variance $V(X) = \sigma^2$,

say this specific sample mean $\bar{x}$ follows a sampling distribution $W$, the variance of this $\bar{x}$ is $V(\bar{x})$.

and the variance of this specific sample is $V(x)$,

are $V(x)$ and $V(\bar{x})$ exactly the same?

we know that $V(\bar{x}) = \sigma^2 /n$

but is $V(x)$ also equal to $\sigma^2 /n$?

please notice $V(X)$ and $V(x)$ and $V(\bar{x})$ has three total different meaning:

$V(X)$ is the real variance of the whole population, $V(x)$ is the variance of one specific sample. $V(\bar{x})$ is the sample mean, when we change the sample, this sample mean changes too, and this changed sample mean follows a specific distribution W

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  • $\begingroup$ It is not obvious what you mean by $V(\bar x)$. On the other hand, $V(\bar X)$ could be meaningful and would be $\frac1n V(X)$ if it is finite and you are sampling with replacement or from a continuous distribution $\endgroup$
    – Henry
    Commented Nov 3, 2020 at 1:34
  • $\begingroup$ @Henry 𝑋¯ bar is the mean of the whole population which is a fixed number, it will never be changed (assume this population is static), 𝑉(x¯) means ,as we changing the sample, each time we draw a different size of the sample from this poplulation, these sample mean varies, each sample will have a different mean, this V(x¯) is dynamic , it will follow a distribution $\endgroup$
    – Ernie Sender
    Commented Nov 3, 2020 at 1:41
  • $\begingroup$ See that $Var(x)$ is a matrix as $x$ is a vector. $\endgroup$
    – Dayne
    Commented Nov 3, 2020 at 2:15
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    $\begingroup$ Not in my notation. I use $\bar X$ to mean $\frac{1}{n} \sum\limits_1^n X_i$ and since each $X_i$ is a random variable so too is $\bar X$. Meanwhile $\bar x$ is a particular observation of $\bar X$ $\endgroup$
    – Henry
    Commented Nov 3, 2020 at 2:16
  • $\begingroup$ I think the edit changed the question incorrectly. This was not what the original question intended to ask. The original question asked (rather poorly) what the variance of the sample variance was. This question as currently written simply asks what is the variance of one specific sample, which is obvious and simply equal to the sample variance $\endgroup$
    – astel
    Commented Nov 3, 2020 at 16:04

2 Answers 2

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say this specific sample mean $\bar{x}$ follows a sampling distribution $W$, the variance of this $\bar{x}$ is $V(\bar{x})$.

and the variance of this specific sample is $V(x)$,

are $V(x)$ and $V(\bar{x})$ exactly the same?

No $V(\bar{x})=V(x)/n$

we know that $V(\bar{x}) = \sigma^2 /n$

This part is false, as you stated before that $V(X) = \sigma^2$ is the population variance (by your notation).

but is $V(x)$ also equal to $\sigma^2 /n$?

No again, $V(x)$ is itself, an estimator of $\sigma^2$.

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  • $\begingroup$ σ is the variance of the whole population, it's not necessary to be the variance of a specific sample $\endgroup$
    – Ernie Sender
    Commented Nov 3, 2020 at 1:26
  • $\begingroup$ $\sigma$ is the population standard deviation; $\sigma^2$ is the population variance; $\sigma/\sqrt{n}$ is the standard error of the sample mean $\bar X =\frac1n\sum X_i$ [i.e, $SD(\bar X)$]; and $S/\sqrt{n}$ is the [estimated] standard error of the sample mean (where the word estimated is often omitted). $\endgroup$
    – BruceET
    Commented Nov 3, 2020 at 6:38
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What you are looking for is the variance of the sample variance, commonly denoted V(S^2) not V(x) as you have it. There are two full derivations of its formula here (one assumes normality the other does not): https://math.stackexchange.com/questions/72975/variance-of-sample-variance

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  • $\begingroup$ Which is exactly what I said in my answer... $\endgroup$
    – astel
    Commented Nov 3, 2020 at 15:57
  • $\begingroup$ Oops, misread you. I still don't see how you got to that, nowhere it's mentioned by OP $\endgroup$
    – Firebug
    Commented Nov 3, 2020 at 16:14
  • $\begingroup$ Again, it was what the original post was asking (poorly) but the edit (by someone other than the OP) changed the meaning of the question entirely. I think OP needs to clarify $\endgroup$
    – astel
    Commented Nov 3, 2020 at 16:30

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