Total Variation distance is often too strong to be useful.

While reading lecture notes on Stein's method, there's one example that I cannot prove myself.

For iid random variables $X_1, ..., X_n$ where $\Pr(X_i=1) = \Pr(X_i=-1) = 1/2$, define $S_n=\frac{1}{\sqrt{n}}\sum_{i=1}^n X_i$. Then $S_n$ converges to normal distribution due to Central Limit Theorem. Still, $d_{TV}(S_n,Z)=1$ for all $n$ where $d_{TV}$ is the total variation distance and $Z\sim N(0,1)$.

It's an example of 'total variation distance is often too strong to be useful', but I don't know how to prove it. Any hint or suggestion?

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HINT Try to show the more general fact that the total variation distance between a continuous distribution and a discrete distribution is $1$. – Srivatsan Dec 19 '11 at 18:39

1 Answer

What's the probability that $S_n$ is an integer multiple of $1/\sqrt{n}$?

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Let $\mathcal{E}$ be the event that $S_n$ is an integer multiple of $1/\sqrt{n}$. This event corresponds to finitely many points, and finitely many points in a continuous distribution amount to zero. I guess this is what you meant, right? – Federico Magallanez Dec 19 '11 at 22:39
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