# Is this probability inequality true?

This question may be simple, though I'm not managing to find an answer. Let $$X$$ and $$Y$$ be two dependent random vectors in in $$\mathbb{R}^d$$, with joint probability density $$\mu(x,y)$$ (with respect to the Lebesgue measure). For any subset $$A \subset \mathbb{R}^d$$ and vector $$t \in \mathbb{R}^d$$, define $$A+t=\{x+t=(x_1+t_1, \ldots, x_d+t_d): x \in A\}.$$ Is it true that $$P(X-Y \in A, Y \in B) \leq \sup_{t \in B}P(X \in A+t)$$ where $$A$$ and $$B$$ are measurable proper subset of $$\mathbb{R}^d$$? The inequality is trivially true if $$X$$ and $$Y$$ are independent: $$P(X-Y \in A, Y \in B) =\int_B \left[\int_{A+y} \mu(x|y)dx\right] \mu(y)dy\\ =\int_B \left[\int_{A+y} \mu(x)dx\right] \mu(y)dy\\ =\int_B P(X\in A+y)\mu(y)dy\\ \leq \sup_{y \in B}P(X\in A+y)$$ where $$\mu(x|y)$$, $$\mu(x)$$ and $$\mu(y)$$ are the conditional density of $$X$$ given $$Y=y$$, the marginal density of $$X$$ and the marginal density of $$Y$$, respectively. What about the case where $$X$$ and $$Y$$ are dependent (i.e. $$\mu(x|y)\neq \mu(x)$$)?

EDIT 1 Here is my attempt:

$$\mathbb{P}(X-Y \in A, Y\in B)= \int_{B}\int_{A+y}\mu(x,y)dxdy\\ \leq \sup_{t\in B}\int_{B}\int_{A+t}\mu(x,y)dxdy\\ =\sup_{t\in B}\int_{A+t}\int_B\mu(x,y)dydx\\ \leq \sup _{t\in B}\int_{A+t}\mu(x)dx\\ =\sup _{t\in B}\mathbb{P}(X \in A+t)$$

but I have doubts about the second and third lines, I'm not sure they're correct. I pass from the third to the fourth line by using the fact that $$\mu(x,y)$$ is nonnegative and $$\int_B\mu(x,y)dy \leq \int_{\mathbb{R}^d}\mu(x,y)dy=\mu(x).$$

EDIT 2 Are there conditions (different from independence) under which the inequaliy holds true? In the somewhat patological case $$X=Y$$ with probability one, as highlighted in the answer below, the inequality may not be true. But what if $$X \neq Y$$ with probability 1? Are there conditions under which, in such an istance, the inequality is satisfied?

## 1 Answer

This is false:

Let the probability distribution be $$P(X = 0, Y = 0) = P(X = 1, Y = 1) = \frac{1}{2}$$ and let $$A = \{ 0\}$$, $$B = \{ 0, 1\}$$. Then the left-hand side of your inequality is $$1$$ while the right-hand side is $$\frac{1}{2}$$.

If you want your densities to be continuous just convolve with some smooth highly concentrated function (both $$X, Y$$ and $$A, B$$).

Actually, if $$X=Y$$ with probability one and $$A = \{0\}$$, $$B = \mathbb{R}$$ then your inequality is false as long as distribution of $$X$$ is non-constant. For example, if $$X$$ is normal then left-hand side is $$1$$ while right-hand side is zero.