# Probability inequality

Let $X,Y,Z$ be non-negative independent r.v. I should find for which $u\geq 0$ holds $$Zu\geq Y - X(1+Z)$$ with probability one. Clearly, if $P(Z = 0)>0$ then the value of $u$ doesn't matter and we should have $X\geq Y$ with probability one.

The question is the following: what if $P(Z = 0) = 0$? Could we divide an inequality by $Z$ and continue working?

If yes I expect an answer like $$u\geq \alpha$$ for $$\alpha = \inf\left(a:P\left[\frac{Y-X}{Z}-X\leq a \right]= 1\right)$$ with possibly $\alpha = \infty$.

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It seems that your we should have $X\ge Z$ with probability one should read we should have $X\ge Y$ with probability one. And that, in the formula for $\alpha$, $+X$ should read $-X$. – Did Jun 15 '11 at 16:00
@Didier, thanks, I'll fix it. Nice to see you feedback in my questions a.s. – Ilya Jun 15 '11 at 17:47

Since $$P[Zu \ge Y - X(1 + Z)] = P[Zu \ge Y - X(1 + Z),Z > 0] + P[Zu \ge Y - X(1 + Z),Z = 0],$$ $P(Z=0)=0$ yields $$P[Zu \ge Y - X(1 + Z)] = P[Zu \ge Y - X(1 + Z),Z > 0] = P\bigg[u \ge \frac{{Y - X}}{Z} - X,Z > 0\bigg].$$

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Is this an answer to the question asked? – Did Jun 17 '11 at 14:01

Recall that the support $S(T)$ of a real valued random variable $T$ is the set of real numbers $t$ such that the events $[|T-t|\le\varepsilon]$ have positive probability for every positive real number $\varepsilon$.

If $Y\le X$ almost surely, that is, if $\sup S(Y)\le\inf S(X)$, then every nonnegative real number $u$ works since $Y-X(1+Z)\le Y-X\le0$ and $Z\ge0$ almost surely.

Otherwise, that is, if $\sup S(Y)>\inf S(X)$, the real number $u$ works if and only if $u\ge u_*$ with $$u_*=\frac{\sup S(Y)-\inf S(X)}{\inf S(Z)}-\inf S(X).$$ To prove this, one optimizes first in $Y$, then in $X$ and finally in $Z$. These three operations can be performed separately because $X$, $Y$ and $Z$ are independent.

In particular, $u_*$ is finite if and only if the ratio defining $u_*$ is finite if and only if the numerator of this ratio is finite and its denominator is positive, that is, if and only if $S(Y)$ is bounded above by a finite $y$ and $S(Z)$ is bounded below by a positive $z$, that is, $[Y\le y]$ and $[Z\ge z]$ both have full probability. Otherwise, $u_*=+\infty$, which means that no value of $u$ is suitable.

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Let's define $x_{\inf} = \max(x| \Pr(X \ge x) = 1)$ and $x_{\sup} = \min(x| \Pr(X \le x) = 1)$ so $\Pr(x_{\inf} \le X \le x_{\sup})=1$, and similarly for $Y$ and $Z$.

The inequality becomes
$$Zu \ge y_{\sup} - x_{\inf}(1+Z) \text{ i.e. } Z \ge \frac{y_{\sup}-x_{\inf}}{u+x_{\inf}}$$

(assuming the denomoinator is positive), which requires either $y_{\sup}\le x_{\inf}$ , or $z_{\inf}$ is at least the right hand side implying a bound for $u$. So you have the cases:

• If $y_{\sup} \le x_{\inf}$ then any value of $u$ will do.

• If $y_{\sup} \gt x_{\inf}$ and $z_{\inf} = 0$ then no value of $u$ will do.

• If $y_{\sup} \gt x_{\inf}$ and $z_{\inf} \gt 0$ then you need $u \ge \dfrac{y_{\sup} - x_{\inf}}{z_{\inf}} - x_{\inf}$

which I think is more or less what you have said.

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The definitions of x-inf and x-sup are incorrect. – Did Jun 17 '11 at 14:00
@Didier: Thank you - I have changed to $0$s to $1$s – Henry Jun 17 '11 at 19:16