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Let $X$ be a random variable with a density $p(x)$ with respect to the Lebesgue measure. We say that $X$ is log concave if $p(x) = \exp(-V(x))dx$ for $V(x)$ a convex function.

Let $X$ be log-concave on $\mathbb{R}^d$. Let $\epsilon\sim \mathsf{Unif}(\{0, 1\}^d)$. I am curious about the distribution of $\langle \epsilon, X\rangle$. In particular, is it log-concave?

I think that conditioned on any particular value of $\epsilon$ occuring this would hold, as the image of log-concave under a linear transformation is log-concave. Moreover, $\epsilon$ of course has incredibly light tails, so I can't rule this out because $\langle \epsilon, X\rangle$ has heavier than sub-exponential tails or something. Randomizing over the $\epsilon$ of course makes that argument fail, and direct arguments I've tried haven't been doing much better. This might be due to inexperience in the area though --- my impression is that these kinds of questions (when they have positive answers) tend to have pretty simple/straightforward proofs (e.g. "it follows from Prékopa–Leindler inequality"). I haven't found one yet though.

Note that I would even be interested in this for "standard" examples of log-concave $X$, say Gaussian, or uniform over a convex body.

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    $\begingroup$ Perhaps explain: notation $\epsilon\gets \{0, 1\}^d$ and "the image of log-concave under a linear transformation is log-concave". $\endgroup$
    – Gerald Edgar
    Feb 9, 2023 at 12:52
  • $\begingroup$ @GeraldEdgar I've adjusted the notation for $\epsilon$. For the image of log-concave, comment, I simply mean that if $X$ is log-concave on $\mathbb{R}^d$, and $A\in\mathsf{Mat}_{m\times d}(\mathbb{R})$ is non-zero, then $AX$ is log-concave. $\endgroup$
    – Mark Schultz-Wu
    Feb 9, 2023 at 20:29
  • $\begingroup$ $Y:=\langle \epsilon,X\rangle=0$ on the event $\{\epsilon=0\}$, and this event has a nonzero probability. So, $Y$ does not have a density. In what sense could it be log concave? Did you want to use $\{-1,1\}^d$ instead of $\{0,1\}^d$? $\endgroup$
    – Iosif Pinelis
    Feb 9, 2023 at 20:32
  • $\begingroup$ $\{-1,1\}^d$ (or small Gaussian, say $\mathcal{N}(0, I_d)$ explicitly) would both be interesting to me as well. Clearly, for $\epsilon$ Gaussian, $\langle X, \epsilon\rangle$ (for $d > 1$) is log concave (one can write the inner product in terms of $\chi^2_{(d)}$ random variables and explicitly compute the resulting density), so if $\epsilon$ is Gaussian I would be more interested in the example of $X$ uniform over a convex body. $\endgroup$
    – Mark Schultz-Wu
    Feb 9, 2023 at 20:46
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    $\begingroup$ In case the intuition helps: you are essentially constructing a mixture of log-concave distributions, and mixtures are often quite far from being log-concave (in some sense, they are the archetypal example of multimodal measures). It is possible that if the centres of each component are very close to one another, relative to their dispersion, then the mixture will be log-concave. I don't actually know of conditions which guarantee that this last claim will hold, but it seems relatively plausible. $\endgroup$
    – πr8
    Feb 9, 2023 at 22:30

1 Answer 1

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$\newcommand\ep\epsilon$According to a comment, the OP actually meant that $\ep$ is a uniformly distributed random element of the set $\{-1,1\}^d$ (rather than $\{0,1\}^d$).

Then, of course, the desired conclusion will not hold in general, even if $d=1$ and $\ep$ is independent of $X$. Indeed, suppose that $X\sim N(a,1)$ for a real $a>1$. Then for the density $p$ of $Y:=\langle\ep,X\rangle=\ep X$ and all real $x$ we have $$p(x)=\frac{f(x-a)+f(x+a)}2,$$ where $f$ is the standard normal density, and for $L(x):=\ln p(x)$ we have $$L''(0)=a^2-1>0.$$ So, $p$ is not log concave.

Here is the graph $\{(x,L(x))\colon|x|\le4\}$ for $a=2$:

Graph of L

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  • $\begingroup$ Does the situation get better if we assume both $\epsilon, X$ are isotropic/mean zero? This is of course not the case for the initial $\epsilon \gets \mathsf{Unif}(\{0,1\}^d)$ I was assuming, but would still be interesting to me. I don't know if this should be a new question though. $\endgroup$
    – Mark Schultz-Wu
    Feb 9, 2023 at 21:43
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    $\begingroup$ @Mark : I would guess that even then the answer would be no, but that guess could be mistaken. I would indeed suggest posting this additional question separately, after giving it some thought and trying some special cases. $\endgroup$
    – Iosif Pinelis
    Feb 9, 2023 at 22:51
  • $\begingroup$ I have a proof of a similar statement here on the other site (as "proof verification" isn't really appropriate for here I think). The proof is that $\langle X, Y\rangle$ for (suitable) gaussian $X$ and uniform $Y$ is log-concave. I expect that it should be required that $Y$ is uniform over a convex set, but I can't seem to see where that is used. $\endgroup$
    – Mark Schultz-Wu
    Feb 14, 2023 at 9:41
  • $\begingroup$ @Mark : I am not really active on math.stackexchange.com. However, I can say that your expression for $p(x)$ (the density of $\langle X, Y\rangle$ at $x$?) is incorrect. You cannot deal with densities as with discrete probabilities. You have to deal with Jacobian determinants here. $\endgroup$
    – Iosif Pinelis
    Feb 14, 2023 at 13:17
  • $\begingroup$ Is the issue with the expression for $\Pr[\langle X,Y\rangle = x\mid Y=y]$, or the overall expression? $\endgroup$
    – Mark Schultz-Wu
    Feb 14, 2023 at 17:42

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