Convergence of a ratio of singular values Let $X$ be an $n\times p$ matrix with $p>n$, then the number of singular values of $X$ is $n$.
Consider the situation where we increase the dimension $p$ by attaching arbitrary random vector to $X$, i.e., a new $n\times (p+1)$ matrix is defined as $X=[X, y]$ with a new vector $y$.
This procedure continues until $p\rightarrow\infty$.
Based on my simulation study, the condition number $d_1/d_n$ converges a constant as $p\rightarrow \infty$, where $d_1>\dots>d_n$ are singular values of $X$.
Also, I can see $[\prod_{i=1}^n d_i/d_n]^{1/n}$ converges to a constant as $p\rightarrow \infty$ in my simulation.
Can I prove these?
 A: 
Let $X$ denote a $M \times N$ random matrix whose entries are i.i.d. random variables with mean $0$ and variance $\sigma^2 < \infty$. Let
  $$
Y_ N = N^{-1}XX^T,
$$
  and let $\lambda_1, \lambda_2, \ldots, \lambda_M$ be the eigenvalues of $Y_N$. Consider the random measure
  $$
\mu_M(A) = \frac{1}{M}\#\{\lambda_j \in A\},\, \, A \subset \mathbb{R}.
$$
  Theorem.(Marcenko-Pastur) Assume that $M, N \rightarrow \infty$ so that the ratio $M/N \rightarrow \lambda \in (0,+\infty)$. Then $\mu_M \rightarrow \mu$ (in distribution), where
  $$
\mu(A) =\begin{cases} (1-\frac{1}{\lambda}) \mathbf{1}_{0\in A} + \nu(A),& \text{if } \lambda >1\\
\nu(A),& \text{if } 0\leq \lambda \leq 1,
\end{cases}
$$
  and
  $$d\nu(x) = \frac{1}{2\pi \sigma^2 } \frac{\sqrt{(\lambda_{+} - x)(x - \lambda_{-})}}{\lambda x} \,\mathbf{1}_{[\lambda_{-}, \lambda_{+}]}\, dx$$
  with
  $$ \lambda_{\pm} = \sigma^2(1 \pm \sqrt{\lambda})^2. \, $$

which is to say that in your case, if the random variables are i.i.d (each time you append a column), and let's say they are zero mean (you can always re-scale) and variance $\sigma^2$ where $M = n$ and $N = p$, you let $p \rightarrow\infty$ at fixed $n$ (i.e. $\lambda \to 0$), then the singular values of $\frac{1}{\sqrt{p}}X$ all go to $1$. 
I hope this can help.
