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Consider the random vector $X:=(X_1\dots X_1) \in \mathbb{R}^n, X_1 \sim \mathcal{N}(0,1).$ Notice the identical components, they're identically distributed but not independent.

Now, I was wondering whether we could expect an inequality like the one we expect in the case where $X$ had iid $\mathcal{N}(0,1)$ components, i.e. in this other extreme case, can we expect that for any Lipschitz function $f: \mathbb{R}^n \to \mathbb{R},$ we have for each $t \ge 0:$

$$ P\left[ |f(X) - Ef(X)| \ge t \right] \le 2 exp \left( - \frac{ct^2}{||f||_{Lip}^2} \right), $$

where $c > 0$ is an absolute constant.

P.S. I might add a similar question on what happens to the above probability when $n \to \infty, t $ is fixed, but I'm still thinking about the question.

P.P.S. if you could cite some reference, that'd be very useful too. I do have access to Michel Lédoux's great book, but I find it a bit difficult for myself to quickly navigate through and get the right inequality I want (but that's just me!).

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If $f$ is the sum and all the components are identical then the inequality fails for $t=n$.

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  • $\begingroup$ Thank you for your answer, but do you mind elaborating on it a bit more? For the sum function, the Lipschitz norm is $\sqrt{n}.$ I wonder if your answer is valid for each $n \ge 1,$ or when $n \to \infty ?$ (The limiting case is also quite important for me) I see that $P[ |\sum_{i=1}^{n} X - \sum_{i=1}^{n} EX| \ge t=n ] = P[nX \ge n] =P[X \ge 1] \nrightarrow 0,$ unlike the RHS term $2 exp(- \frac{cn^2}{n}) = 2 exp (- cn)\to 0$ as $n \to \infty.$ So this does prove that the inequality isn't true in the limiting case. Is that what you had in mind? Thanks again! $\endgroup$
    – Learning math
    Commented May 30, 2020 at 19:37
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    $\begingroup$ For every $c$ the inequality fails if $n$ is large enough. $\endgroup$
    – Yuval Peres
    Commented May 31, 2020 at 5:01

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