Distribution of $\dfrac{X_{i}}{\max X_{i}}$? Suppose we have a sample of standard normal i.i.d. observations $X_{1}, X_{2}, \cdots ,X_{n}$, what can we say about the asymptotic distribution of $V_{i}$ and $W_{i}$, where:
$V_{i}=\dfrac{X_{i}}{\max X_{i}}$ and $ W_{i}=X_{i}-\max X_{i}$?
 A: For the case where there is a fixed maximum and you record any $X_i$ below that maximum then the distribution of $X_i$ will follow a truncated normal distribution.
Of course your case is a bit more complicated, your maximum depends on the values of $X_i$.
Let $\phi(z)$ be the pdf of the standard normal distribution and $\Phi(z)$ is the CDF of the standard normal distribution. Also let $Y=max(X_i)$.
In order to get a maximum of $y$ from $n$ samples you need one sample to be equal to $y$ and the others must be less than or equal to $y$. This is like a binomial distribution; there are $n$ samples which could be equal to $y$ with a pdf of $\phi(\frac{y-\mu}{\sigma})$ and the remaining $n-1$ samples which are less than $y$ with a probability of $\Phi(\frac{y-\mu}{\sigma})^{n-1}$. Therefore the distribution of the maximum is:
$$f(y)= n\ \phi \left(\frac{y-\mu}{\sigma}\right)\ \Phi \left(\frac{y-\mu}{\sigma}\right)^{n-1}$$
Clearly the distribution of $W_i$ will have one value equal to $0$ and the rest are negative. You are interested in the negative values.
As mentioned before $W$ will have a truncated normal distribution with a constant subtracted from it. However, the upper limit to the truncated normal is now a random variable with distribution $f(y)$.
For a given maximum $y$ the distribution of $w$ is:
$g(w)=\frac {\phi ({\frac {w-\mu+y }{\sigma }})}{\Phi ({\frac {y-\mu }{\sigma }})}$    for $w\leq 0$
Considering that $y$ is a random variable we can integrate over all possible $y$ to get the distribution of $w$
$g(w)= \int_{-\infty}^{\infty} \frac{\phi ({\frac {w+y-\mu }{\sigma }})}{\Phi ({\frac {y-\mu }{\sigma }})} n\ \phi \left(\frac{y-\mu}{\sigma}\right)\ \Phi \left(\frac{y-\mu}{\sigma}\right)^{n-1} dy$ $\ \ \ $   for $w\leq 0$
Perhaps from this you can determine the asymptotic distribution of $W$ and apply a similar approach to the distribution of $V$
