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Consider two spherical Gaussian distributions in $\mathbb{R}^n$, $A = \mathcal{N}(x, I)$ and $B=\mathcal{N}(y, I)$ where the difference in means is $\delta = y - x$. Let $S \subset \mathbb{R}^n$ be a set with $\mathbb{P}[A \in S] = p$. I'm trying to show a lower bound on $\mathbb{P}[B \in S]$.

My intuition says that the lower bound is achieved when $S$ is a half-space with normal vector $\delta$. More specifically, I'm pretty sure the extremal $S$ is the half-space $S^* = \{z: \frac{\delta}{\| \delta\|} ^T z \le \Phi^{-1}(p) \}$, It's easy to verify that $\mathbb{P}(A \in S^*) = p$ and $\mathbb{P}(B \in S^*) = \Phi(\Phi^{-1}(p) + \|\delta\|)$.

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  • $\begingroup$ In general, the set $A$ that minimizes $\int_A f$ given $\int_A g = p$ is the set $A_p = \{f / g < c(p)\}$ for an appropriate $c(p)$. To see this, simply write $$\begin{aligned}\int_A f - \int_{A_p} f &= \int_{A\setminus A_p} f - \int_{A_p\setminus A}f \\&\ge c(p)\int_{A\setminus A_p} g - c(p) \int_{A_p\setminus A}g \\&= c(p)\int_A g - c(p)\int_{A_p} g \\&= c(p) \times p - c(p) \times p = 0.\end{aligned}$$ $\endgroup$
    – Mateusz Kwaśnicki
    Commented Oct 30, 2018 at 20:05

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Mateusz's comment is correct. Also, it turns out that this result is known in statistics as the Neyman-Pearson lemma.

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