# Probability Inequalities – How to Prove Given Inequality in Self-Study

probability-inequalitiesself-study

For any nonnegative random variable $$X$$ independent of $$U$$ where $$U \sim \operatorname{Uniform}(-t,t)$$ and any $$t\ge 0,$$

$$P(X+U\ge t)\le\frac{E(X)}{2t}.$$

Any hints to prove this inequality?

It's convenient to define $$U$$ once and for all to be a uniform variable on the interval $$[-1,1]$$ and simply multiply it by $$t$$ to produce the $$U$$ used in the question.

Two useful, easily proven, but not widely known facts about random variables $$X$$ in general are

1. No matter what the distribution function (CDF) $$F_X$$ of $$X$$ might be, the random variable $$X + tU$$ is absolutely continuous with a density function $$f_{X+tU}(y) = \frac{1}{2t}\left(F_X(y+t) - F_X(y-t)\right).$$ There are many ways to prove this, as explained at https://stats.stackexchange.com/a/43075/919 (which concerns sums of the closely related uniform variable supported on $$[0,1]$$).

2. The expectation of any non-negative random variable $$X$$ with CDF $$F_X$$ equals $$E[X] = \int_0^\infty \left[1 - F_X(x)\right]\,\mathrm{d} x.$$ This is repeatedly demonstrated in many threads here on CV; one is Expectation of a function of a random variable from CDF. They all amount to performing an integration by parts.

With these facts in mind, evaluate the probability in the question as

\begin{aligned} \Pr(X + tU \gt t) &= \int_t^\infty f_{X+tU}(y)\,\mathrm{d}y \\ &= \frac{1}{2t}\int_t^\infty \left(F_X(y+t) - F_X(y-t)\right)\,\mathrm{d}y\\ &= \frac{1}{2t}\int_t^\infty \left( \left[1 - F_X(y-t)\right] - \left[1 - F_X(y+t)\right]\right)\,\mathrm{d}y\\ &= \frac{1}{2t}\left(\int_0^\infty \left[1 - F_X(x)\right]\,\mathrm{d}x - \int_{2t}^\infty \left[1 - F_X(x)\right]\,\mathrm{d}x\right)\\ &= \frac{1}{2t}\left(E[X] - \int_{2t}^\infty \left[1 - F_X(x)\right]\,\mathrm{d}x\right)\\ &\le \frac{1}{2t} E[X]. \end{aligned}

The justifications of these steps are (1) definition of density, (2) fact $$(1)$$ above, (3) algebra, (4) linearity of integration followed by changes of variables, (5) fact $$(2)$$ above, and (6) since $$1-F_X(x)$$ is a probability for all $$x,$$ it is never negative, whence its integral is non-negative.

Finally, $$\Pr(X + tU \ge t) = \Pr(X + tU \gt t)$$ because (as previously noted) the variable $$X + tU$$ is absolutely continuous, QED.

One thing I like about this way of proceeding is the insight it provides into the tightness of the inequality: the amount by which the probability (on the left hand side) falls short of $$E[X]/(2t)$$ (on the right hand side) is proportional to the integral of $$1-F_X$$ from $$2t$$ on up. Thus, for instance, when $$X$$ is bounded and $$2t$$ exceeds this bound, that integral is zero and the inequality becomes an equality.