Distribution of "normalised" Gaussian random variables Let $X_1, \dots, X_n$ be independent normally distributed random variables. What is the distribution of:
$$
Y_i = \frac{X_i}{\mathrm{stdDev}(X_1, \dots, X_n)},
$$
where $\mathrm{stdDev}(X_1, \dots, X_n)$ is the standard deviation of the sample? I came across this in a simulation, where the simulated random variables were "normalised" before being used, but no statistical analysis was provided.
 A: Not intended as an answer ... but more a comment that is too long for the comment box ...
Updated for OP's change of sample variance to sample standard deviation
To get an idea of the difficulty of the problem ... consider the simplest possible form this question can take, namely:


*

*a sample of size $n = 2$, where ...

*$X_1$ and $X_2$  are random draws from a common standard Normal parent. 


Then, using the $(n-1)$ version of sample variance, and defining sample standard deviation as the square root of the latter, ... the problem is to find the distribution of: 
$$Y = \frac{\sqrt{2}\, X_1}{\big|X_1-X_2\big|} \quad \quad \text{where } X_i \sim N(0,1) $$
This does not appear to be easy at all ... never mind solving for general $n \geq2$. 
Monte Carlo simulation of the pdf (for different sample sizes $n$)
What will the general $n$ solution look like?  The following diagram constructs the empirical Monte Carlo pdf of your ratio $Y$, for samples of size $n = 2, 3, 5$ and $25$. Each plot compares the:


*

*empirical Monte Carlo pdf [squiggly blue curve] to

*a standard Normal pdf (dashed red curve)



There is perhaps some solace in that, by the time $n = 25$, the distribution appears to be well-approximated by a standard Normal (provided the parent is standard Normal). But the small sample sizes are tricky. 


*

*If you are interested in general Normal distributions (i.e. with non-zero means), then there is, of course, the additional complication of asymmetry to the plots. 

