Suppose I have a probability density function $\pi(x_1, \ldots, x_n)$, which is the density of a vector-valued random variable $X$ in $\mathbb{R}^n$. Assume that $\pi$ is strongly log-concave, i.e., the function $E(x_1, \ldots, x_n) = - \log \pi(x_1, \ldots, x_n)$ is strongly convex. This implies that $\pi$ possesses a unique global maximum $$ (x_1^\star, \ldots, x_n^\star) = \operatorname{argmax}_{(x_1, \ldots, x_n)} \pi(x_1, \ldots, x_n). $$
Now, suppose that I look at the marginal density for any coordinate, e.g., $$ \pi(x_k) = \int_{\mathbb{R}^{n-1}} \pi(x_1, \ldots, x_n) dx_1 \cdots dx_{k-1} dx_{k+1} \cdots dx_n, $$ for any $1 \leq k \leq n$. We then know that the marginal density is also log-concave and possesses a unique maximizer (e.g., see here). Let the mode of this marginal be denoted by $$ x^\dagger_k = \operatorname{argmax}_{x_k} \pi(x_k). $$
Question: Under the assumption that the joint density $\pi(x_1, \ldots, x_n)$ is strongly log-concave, does the mode of the joint necessarily agree with the mode of the marginals? By "agree", I am meaning whether $$ x_k^\star = x_k^\dagger $$ for each $k$?
I will note that without the assumption of log-concavity, the answer to this question is no, as discussed here. I am asking whether assuming log-concavity makes this statement true.
