Verified answer

A chi-square distribution is the distribution of a sum of squared normal r.v.'s. As the d.f. -> infinity, it just means you are summing more and more normal squared r.v.s together.

Now if you take sum of chi-square r.v. which is itself a chi-square r.v. and divide by n, then you get something which converges to a normal r.v. by the CLT. So I guess the answer to your question is yes and no - a chisquare r.v. divided by its degrees of freedom approaches a normal r.v. as df -> infinity but a chi-square r.v. by itself does not.

Note: Vekkus - he is talking about convergence in distribution. A sequence of r.v. X[n] -> Y in distn if their CDF's converge at all continuity points in the usual sense of convergence from calculus (i.e., doesn't have to be uniform or anything).

2) If it does then that means for any epsilon greater than zero there exists M > 0 such that N>=M implies N is a big enough sample size. But I'm not sure which norm you used for chi squared -> normal. More significantly, I don't know how big your epsilon is allowed to be.

1) Superficially I don't think it does, because the graph of the chi squared function isn't there for x < 0