Slutsky's theorem If I have a set of $N$ i.i.d. random variables $X$ with sample mean $\bar{X}=\frac{1}{N}\sum_i^N X_i$, does Slutsky's theorem http://en.wikipedia.org/wiki/Slutsky%27s_theorem imply that
$$
    E\left[\frac{X_i X_i}{\bar{X}}\right] = \frac{E[X_i^2]}{E[\bar{X}]}.
$$
I'm not particularly familiar with the notation used in probability theory so the wiki is not clear to me.  If the above is not true, are there certain certain constraints I can place on $X$ which make it true?
 A: As it is evident in the link you provided, Slutsky's lemma is concerned with probability limits and convergence in distribution, not with expected values. Moreover, the equality stated in the question is not correct, because (apart from the other issues raised in the comments), the numerator is not independent from the denominator, and so the expected value does not distribute. And, by construction, you cannot make it independent by "placing restrictions on $X$".
But the equality does hold "asymptotically". 
Assume that $E(X_i^2)$ exists and is finite. Then $E(X_i)$ also exists and is finite. Make the additional assumpion that $E(X_i) =\mu \neq 0$.
Let's apply Slutsky's lemma: $X_i^2$ trivially converges in distribution to itself (it does not depend on $N$), while $\bar X$ converges in probability to the non-zero mean $\mu$, which is a constant. Then Slutsky's lemma says that 
$$\frac{X_i ^2}{\bar{X}} \xrightarrow{d} \frac{X_i^2 }{E(\bar X)} \equiv Z$$
The expected value of the limiting random variable will be
$$E(Z) = E\left(\frac{X_i^2 }{E(\bar X)}\right) = \frac{E[X_i^2]}{E[\bar{X}]}$$
where here we "distributed" the expected value because the denominator is a constant.
But, the moments of the limiting distribution are not necessarily equal to the limit of the moments of the finite sample distribution. We need an additional condition for that, which is (for our particular case)
$$\exists \;\delta>0:\; E\left(\left|\frac{X_i ^2}{\bar{X}}\right|^{1+\delta}\right) < M<\infty \;\;\forall N$$
Then, what we can say is that
$$\lim_{N\rightarrow \infty}E\left[\frac{X_i^2 }{\bar{X}}\right] = E(Z)  = \frac{E[X_i^2]}{E[\bar{X}]}$$
Another way to approach the matter, more intuitively but less formally, is to think that as $N$ grows very large, $X_i$ and $\bar X$ tend to become independent, since a single  $X_i$ contributes relatively less and less to $\bar X$. So at the limit, we can distribute the expected value.
