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It's known that for a random vector $(X_1,\dots,X_n)\in \mathbb{R}^n$ with a log-concave distribution, any subvector has a long-concave distribution. I'm wondering if there are any results about its converse, in particular, when $(X_1,\dots,X_n)$ is isotropic, $X_i$'s are identically distributed (but not necessarily independent) and each $X_i$ is log-concave.

Is this true when $n=2$? That is,

Suppose that $X$ and $Y$ are identically distributed subject to some log-concave distribution with $E[X]=E[Y]=0$, $E[X^2]=E[Y^2]=1$ and $E[XY]=0$. Is the joint distribution of $(X,Y)$ log-concave?

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No. Let $X$ be, say, a standard normal random variable, $Z$ an independent random variable with $P[Z=1] = P[Z=-1] = 1/2$, and $Y=XZ$. Then $X$ and $Y$ are uncorrelated standard normal (in particular log-concave) random variables, but the distribution of $(X,Y)$ is not log-concave.

This is of course also a counterexample to the false theorem one sometimes hears stated that "uncorrelated normal random variables are independent." (The theorem becomes true if "jointly" is inserted in the right place.)

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  • $\begingroup$ In this case the joint distribution of $X$ and $Y$ is only supported on the line $y=x$ in $\mathbb{R}^2$. Isn't it a normal distribution along the line so the joint distribution is log-concave? $\endgroup$
    – Y L
    Commented Feb 13, 2012 at 18:29
  • $\begingroup$ The joint distribution is supported on the union of the line $y=x$ and the line $y=-x$. $\endgroup$
    – Mark Meckes
    Commented Feb 13, 2012 at 19:56

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