Let $X_i$ be a sequence of independent and identically distributed random variables with $\mu = 0$ and $\sigma= 1$. The central limit theorem guarantees that $$\frac{X_1+X_2+..+X_n}{\sqrt n} \rightarrow Z \sim \mathcal{N}(0,1) \ \ \text{converges weakly}$$ I am now trying to show that we generally cannot say that this also holds for convergence in probability . I have tried various things and it seems like the general approach, as suggested in the answer to this question, is to define

$$Y_n:=\frac{S_{2n}}{\sqrt{2n}}-\frac{S_n}{\sqrt n}$$ and say it converges to $0$ in probability, assuming that $\frac{S_n}{\sqrt n}$ converges in probability to $Z$ ($S_n$ denotes the partial sums of the random variable series as usual). Since we can write $Y_n$ as a sum of two independent random variables, who converge to two normal distributions, we could show that $Y_n$ converges to a gaussian with mean $0$ as well and we would have a contradiction.

The prolem I have here is that I do not quite understand how we would finish that argument formally. Lets say we have $Y_n = Y^1_n + Y^2_n$,we know that $Y^1_n$ and $Y^2_n$ converge to normal distributed random variables $Y^1$,$Y^2$ (in distribution) and that they are independent. How do I derive the limit in distribution of $Y_n$?

I was thinking that one might use the fact that for the respective characteristic functions $\phi_{Y^1_n} $, $\phi_{Y^2_n}$ we know that $\phi_{Y_n}=\phi_{Y^1_n+Y^2_n}= \phi_{Y^1_n}\phi_{Y^2_n} \rightarrow \phi_{Y^1}\phi_{Y^2}$ for $n\rightarrow \infty$. This should work becuase convergence in distribution (i.e weak convergence of the distribution functions) implies convergence of the characteristic functions. Then I could just multiply the characteristic functions and I would see that I end up with the characteristic function, which is just another gaussian. Is this correct? Any help would be appreciated!


1 Answer 1


If $S_n/\sqrt{n}$ converged in probability to some random variable $Z$, $Z$ would have to be independent of each $X_j$. But the tail $\sigma$-field is trivial...


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