If $X\_1, \ldots , X\_k$ are i.i.d normal random variables with mean $0$ and variance $1$, then is there a way to sample $Y\_1, \ldots , Y\_m$ for $m=\omega(k)$ such that each of the $Y\_i$'s is a normal random variable with mean $0$ and variance $1$ and they are pairwise independent?



How to sample pairwise independent gaussians
Here is the answer I promised in my last comment.


Instead of considering ${\rm N}(0,1)$ variables, we may consider uniform$[0,1)$ variables.
Indeed, if $Z\_i$ are i.i.d. ${\rm N}(0,1)$ variables, then, with $\Phi(\cdot)$ denoting the ${\rm N}(0,1)$ distribution function, $U\_i := \Phi (Z\_i)$ are i.i.d. uniform$[0,1)$ variables. In turn, if $\tilde U\_i$ are pairwise independent uniform$[0,1)$ variables, then $\tilde Z\_i := \Phi^{-1} (\tilde U\_i)$ are pairwise independent ${\rm N}(0,1)$ variables.


The rest of this answer is based on the recent paper "Recycling physical random numbers",
available at [1](http://projecteuclid.org/euclid.ejs/1262617417) or [2](http://www-stat.stanford.edu/~owen/reports/recycle.pdf). Henceforth, we use the same letters as in that paper. Suppose that $U\_1,\ldots,U\_n$ are independent uniform$[0,1)$ variables. Fix $2 \leq m \leq n$, and define $N\_m = {n \choose m}$. Now let $X\_i$, for $i = 1,\ldots,N\_m$, comprise all $N\_m$ distinct sums of the form $U\_{r\_1 } \oplus U\_{r\_2 } \oplus \cdots \oplus U\_{r\_m }$, for $1 \le r\_1 < r\_2 < \cdots < r\_m \le n$. Here $U\_{r\_1 } \oplus U\_{r\_2 } \oplus \cdots \oplus U\_{r\_m }$ is the sum modulo $1$ of the $U\_{r\_i}$, given explicitly by 
$$
U\_{r\_1 } \oplus U\_{r\_2 } \oplus \cdots \oplus U\_{r\_m } = U\_{r\_1 } + U\_{r\_2 } + \cdots + U\_{r\_m } - \left\lfloor {U\_{r\_1 } + U\_{r\_2 } + \cdots + U\_{r\_m } } \right\rfloor ,
$$
where $\left\lfloor \cdot \right\rfloor$ is the floor function. Then, the $X\_i$ are pairwise independent uniform$[0,1)$ variables. In particular, by letting $m=2$, we can efficiently construct $n(n-1)/2$ pairwise independent uniform variables from $n$ independent ones.


Finally, for general purposes it might be worth stating the following simple fact (Proposition 2 in the aforementioned paper). For $N \geq 2$, let $Y\_1,\ldots,Y\_N$ be pairwise independent random variables with common mean $\mu$ and common variance $\sigma^2 < \infty$. Define $\bar Y = \frac{1}{N}\sum\nolimits\_{i = 1}^N {Y\_i }$ and $s^2 = \frac{1}{{N - 1}}\sum\nolimits\_{i = 1}^N {(Y\_i - \bar Y)^2 }$. Then, ${\rm E}(\bar Y) = \mu$, ${\rm Var}(\bar Y) = \sigma^2/N$, and ${\rm E}(s^2) = \sigma^2$. Combined with the previous paragraph, a straightforward implication is that for a square-integrable function $f$ defined on $[0,1)$, we can approximate the integral $\mu = \int\_{[0,1)} {f(x)\,{\rm d}x}$ using a modest number $n$ of independent random inputs. Indeed, note that $n$ independent random inputs can be used to get unbiased Monte Carlo estimates for $\mu$ with the same variance as with $N\_m = {n \choose m}$ independent random inputs.


