Suppose $X_1,X_2,\ldots, X_n$ are random variables distributed independently as $N(\theta , \sigma^2)$. define $$S^2=\frac{1}{n-1}\sum_{i=1}^{n} (X_i-\bar{X})^2 ,\qquad \bar{X}=\frac{1}{n}\sum_{i=1}^{n} X_i\,.$$

Take $n=10$. How can $\operatorname{var} \left(\dfrac{X_1-\bar{X}}{S}\right)$ be calculated?

  • $\begingroup$ I made a number of corrections to make your post more readily understood. Please check it still says what you intend. ... How does this question arise? $\endgroup$ – Glen_b Aug 22 '15 at 11:42
  • $\begingroup$ This seems a tough one to me. I guess that $E[(X_1-\bar{X})/S]=0$, so the variance to be calculated is $E[(X_1-\bar{X})^2/S^2]$. But then...? $\endgroup$ – StijnDeVuyst Aug 22 '15 at 21:10
  • $\begingroup$ Because it is a normal sample, $S^2$ and $\bar{X}$ are independent, but I don't know if that helps here. $\endgroup$ – StijnDeVuyst Aug 22 '15 at 21:21
  • $\begingroup$ stats.stackexchange.com/questions/181964/… $\endgroup$ – StubbornAtom Jul 23 '18 at 16:33

I think it is possible to arrive at an integral representation of $\text{Var}[\frac{X_1-\bar{X}}{S}]$. First, let us express the sample mean $\bar{X}$ and the sample variance $S^2$ in terms of their counterparts for the observations other than $X_1$: \begin{equation*} \bar{X}_* = \frac{1}{n-1}(X_2+\ldots+ X_n) \quad\text{ and }\quad S_*^2 = \frac{1}{n-2} \sum_{i=2}^n (X_i-\bar{X}_*)^2 \end{equation*} It is not so difficult to prove that (see also here) \begin{equation*} \bar{X} = \frac{1}{n} X_1 + \frac{n-1}{n} \bar{X}_* \quad\text{ and }\quad S^2 = \frac{n-2}{n-1}S_*^2 + \frac{1}{n}(X_1-\bar{X}_*)^2 \end{equation*} We may agree that $E[\frac{X_1}{S}]=E[\frac{\bar{X}}{S}]=0$, so that $E[\frac{X_1-\bar{X}}{S}]=0$ and therefore $\text{Var}[\frac{X_1-\bar{X}}{S}] = E[\frac{(X_1-\bar{X})^2}{S^2}]$. The quantity of which we need the expectation can be rewritten as \begin{align*} \frac{(X_1-\bar{X})^2}{S^2} & = \frac{(X_1 - \frac{1}{n}X_1 - \frac{n-1}{n} \bar{X}_*)^2}{\frac{n-2}{n-1}S^2_* + \frac{1}{n}(X_1-\bar{X}_*)^2}\\ & = \big(\frac{n-1}{n}\big)^2 \frac{(X_1-\bar{X}_*)^2}{\frac{n-2}{n-1}S^2_* + \frac{1}{n}(X_1-\bar{X}_*)^2} \end{align*} The import thing now is that $X_1\sim N(\mu,\sigma^2)$, $\bar{X}_*\sim N(\mu,\frac{1}{n-1}\sigma^2)$ and $\frac{n-2}{\sigma^2}S^2_* \sim \chi^2_{n-2}$ are jointly independent. Define $Y=X_1-\bar{X}_*$, which is $N(0,\frac{n}{n-1}\sigma^2)$ and therefore $\frac{n-1}{n\sigma^2} Y^2 \sim \chi^2_1$. Then \begin{align*} E[\frac{(X_1-\bar{X})^2}{S^2}] & = \big(\frac{n-1}{n}\big)^2 E[\frac{Y^2}{\frac{n-2}{n-1}S^2_* + \frac{1}{n}Y^2}] = \frac{(n-1)^2}{n} E[\frac{\chi_1^2}{\chi_{n-2}^2 + \chi_1^2}]\\ \end{align*} with $\chi_1^2$ and $\chi_{n-2}^2$ still independent. Expanding the expectation operator and using the density $f_{\chi^2_m}(x)=(\frac{x}{2})^{\frac{m}{2}-1}\frac{1}{2\Gamma(m/2)}e^{-\frac{x}{2}}$ of the $\chi^2_m$-distribution we may numerically evaluate \begin{align*} E[\frac{(X_1-\bar{X})^2}{S^2}] & = \frac{(n-1)^2}{n} \int_0^\infty \int_0^\infty \frac{a}{b+a} f_{\chi^2_1}(a) f_{\chi^2_{n-2}}(b) \text{d}a\text{d}b \end{align*} Unfortunately, I see no easy way to do this.

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It seems people want to ask the variance of a t distribution which is $\frac{v}{v-2}$ when degree of freedom is bigger than 2.

But we cannot know the distribution of the $ \left(\dfrac{X_1-\bar{X}}{S}\right)$

Please see discussions of this post Standardizing a Standard normal Variable

It shows $\frac{X_1}{S}-\frac{\bar{X}}{S} $ has an unknown distribution.

Let $Y=\frac{X_1}{S}-\frac{\bar{X}}{S} $

($\frac{\bar{X}}{S}$ follows a $t$ distribution ($\bar{X}$ and $S$ are independent),but not $\frac{X_1}{S}$,since $X_1$ and $S$ are not independent)

We don't know the distribution of $\frac{X_1}{S}$, therefore,we also don't know the distribution of $Y$

If the distribution is unknown, people may not able to use the definition of variance

$Var(Y)=E(Y^2)-[E(Y)]^2=\int_{-\infty}^{\infty}Y^2f(y)dy-[\int_{-\infty}^{\infty}Yf(y)dy]^2$ to calculate the variance since $f(y)$ is unknown. (For discrete case it is the same).

So I think your question is unanswerable. Or you may need to modify your question.

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  • 1
    $\begingroup$ Surely $(X_1-\bar{X})/S$ definitely HAS some well-defined distribution. The fact that we here seem unable to identify it or give a formula for its variance, does not mean that "we cannot know the distribution" or that is "unknown". The OP's question is difficult, but not unanswerable. Secondly, you do not always have to know the complete distribution of a random variable to obtain its variance... $\endgroup$ – StijnDeVuyst Aug 23 '15 at 13:02
  • $\begingroup$ Yes, if we know $E(Y^2)$ and $E(Y)$ already we don't need know the distribution. Then how get this two things? Or by other methods, maybe we need more extra information. $\endgroup$ – Deep North Aug 23 '15 at 13:15
  • $\begingroup$ The link in this question is very helpful. Not knowing an explicit form for the distribution doesn't mean that we can't calculate the variance, however. If it were possible to determine $Var(X_1/S)$ and $Covar(X_1/S,\bar{X}/S)$, then the standard rules for variance would give the answer. It seems at first glance, however, that these might depend on all of the $X_i$. $\endgroup$ – EdM Aug 23 '15 at 15:23
  • $\begingroup$ @DeepNorth Your comment Yes, if we know $E(Y^2)$ and $E(Y)$ already we don't need know the distribution. misses StijnDevuyst's point entirely. If $Y=g(Z)$ where $Z$ has a known distribution, one can find $E(Y^2)$ and $E(Y)$ via the law of the unconscious statistician without needing to know the distribution of $Y$. $\endgroup$ – Dilip Sarwate Aug 23 '15 at 18:18
  • $\begingroup$ Thanks, I see, I think the question is interesting, there are many people doing a standardization of a random variable by $\frac{X_i-\bar{X}}{S}$. Then I am wondering what is the theory base of this practice. $\endgroup$ – Deep North Aug 24 '15 at 0:04

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