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?
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.
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.
