# Proving upper bound for truncated difference

Let $$X$$ and $$Y$$ be real valued random variables. And define a truncation operator as:

\begin{align} X(\tau) = (|X| \wedge \tau) \; \text{sign}(X), \quad \tau > 0 \end{align}

Now, I am not sure how to show the inequality:

\begin{aligned} & \mathbb{E}\left[X Y\right]-\mathbb{E}\left[X(\tau) Y(\tau)\right] \\ \leq & \mathbb{E}\left[\left|XY\right|\left(\mathbb{I}\left\{\left|X\right| \geq \tau\right\}+\mathbb{I}\left\{\left|Y\right| \geq \tau\right\}\right)\right]\end{aligned}

As in this question, the idea is directly comparing $$f(X, Y) = XY - X(\tau)Y(\tau)$$ and $$g(X, Y) = |XY|(I(|X| \geq \tau) + I(|Y| \geq \tau))$$ on different regions of $$\Omega$$.

On the region $$[|X| < \tau, |Y| < \tau]$$, $$f(X, Y) = 0 \leq g(X, Y) = 0$$.

On the region $$[|X| < \tau, |Y| \geq \tau]$$, $$f(X, Y) = X(Y - Y(\tau)) = X(Y - \tau\operatorname{sign}(Y))$$, $$g(X, Y) = |XY|$$. To see $$f(X, Y) \leq g(X, Y)$$ on this region, note that if $$|Y| \geq \tau$$, then $$|Y - \tau\operatorname{sign}(Y)| \leq |Y|$$.

On the region $$[|X| \geq \tau, |Y| \geq \tau]$$, $$f(X, Y) = XY - \tau\operatorname{sign}(X)\tau\operatorname{sign}(Y)$$, $$g(X, Y) = 2|XY|$$. It follows by $$|\tau\operatorname{sign}(X)| = \tau \leq |X|$$ and $$|\tau\operatorname{sign}(Y)| = \tau \leq |Y|$$ on this region that \begin{align} f(X, Y) \leq |f(X, Y)| \leq |XY| + |\tau\operatorname{sign}(X)| |\tau\operatorname{sign}(Y)| \leq |XY| + |XY| = g(X, Y). \end{align}

Now you should be able to finish the comparison on the remaining $$1$$ region.

• +1. But IMHO it would be clearer by first noting you can assume $X$ and $Y$ are non-negative random variables without any loss of generality (proof: $E[XY]\le E[|XY|]=E[|X||Y|]$ allows you to replace each variable with its absolute value) and thereby eliminate all uses of signum and absolute value.
– whuber
Apr 11, 2023 at 19:47
• @whuber How about the second term "$E[X(\tau)Y(\tau)]$" in the left-hand side? Apr 11, 2023 at 22:08
• Good point: I failed to appreciate the effect of the signum in the definition of $X(\tau).$
– whuber
Apr 11, 2023 at 22:11
• Thanks for this @Zhanxiong, but I think what you wrote for the 3rd region is wrong. For $[|X| \geq \tau, |Y| \geq \tau]$ then $f(X,Y) = XY - (\text{sign}(X)\text{sign}(Y) \tau^2)$. But then I think for this region we can say $|XY - (\text{sign}(X)\text{sign}(Y) \tau^2)| \leq |XY|$? Apr 12, 2023 at 8:08
• Additionally, I think that all of what you said holds with the inequality $|f(X,Y)| \leq g(X,Y)$, so we can have a stronger inequality. Apr 12, 2023 at 8:13