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 1


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.


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