Can this problem really be solved using central limit theorem? My friend had this question on a test:
Let $\{X_n\}_{n \in N}$ be a sequence of independent random variables with the same normal distribution $N(0, 2n)$.  Check for the convergence of a sequence $\{Y_n\}_{n \in N}$ : $Y_n = \frac{X_1 + ... X_n}{n}$ and find the limit variable Y.
The supposed solution is "you should use CLT".
I am wondering, is it true? It looks like for each $Y_i$ the definition of $X_i$ changes (it is $N(0, 2i)$), which violates the assumption of CLT, where $X_i$ should always be the same?
 A: Assume the parameter $2n$ is the variance.
Because $Y_n$ is the sum of jointly Normal variables, it, too, has a Normal distribution with mean
$$E[Y_n] = E\left[\frac{1}{n}\sum_{k=1}^n X_k\right] = \frac{1}{n}\sum_{k=1}^n 0 = 0$$
and variance
$$\operatorname{Var}[Y_n] = \operatorname{Var}\left(\frac{1}{n}\sum_{k=1}^n X_k\right) = \frac{1}{n^2} \sum_{k=1}^n \operatorname{Var}(X_k) =  \frac{1}{n^2} \sum_{k=1}^n (2k) = \frac{n+1}{n}.$$
The question thus asks for the limiting distribution of a sequence of zero-mean Normal distributions of variance $1+1/n.$  Intuitively that should be a standard Normal distribution.  This can be proven rigorously by taking the limit of the characteristic functions of $Y_n,$ given by
$$\psi_{Y_n}(t) = \exp\left(-\frac{(n+1)t^2}{2n^2}\right),$$
which converges to $\exp(-t^2/2)$ because $\exp$ is continuous everywhere.  Since the latter is the characteristic function of the standard Normal distribution, we are done.
Notice that the CLT was not needed -- but the mathematical ideas behind a proof of the CLT (namely, examine the limit of characteristic functions) are helpful.

If instead the parameter $2n$ refers to the standard deviation, then the variance is $(2n)^2.$  Using that value in the foregoing analysis would show that
$$\operatorname{Var}(Y_n) = \frac{1}{n^2}\sum_{k=1}^n (2k)^2 = \frac{2(n+1)(2n+1)}{3n} = \frac{4}{3}n +2+\frac{2}{3n},$$
which diverges.  This sequence could therefore have no limiting distribution (intuitively, it becomes more and more spread out as $n$ increases, eventually settling down to nothing like a cloud of dust).
