# Does square of an inequality holding in probability change the probability value?

I am thinking of the property in probability of inequality. In particular, we assume $$P[\zeta>a]\leq b,$$ where $a>0$, $b>0$ and $\zeta\in R$ is a random variable.

Now we would like to consider whether the inequality of $P[\zeta^2>a^2]\leq b$ holds.

In fact, for differential and monotonic transformation, e.g., exponential function, the inequality holds. That is, $P[\exp(\zeta)>\exp(a)]\leq b$.

Can someone give hints for me on this issue? Thanks a lot in advance.

• I think, if $g$ is a continuous inecreasing function in $\mathbb{R}$, $\zeta>a\Leftrightarrow g(\zeta)>g(a)$. Then $P(\zeta>a)= P(g(\zeta)>g(a))$. Then you can make such statement easily. – MAN-MADE Sep 1 '17 at 5:11
• Thanks. How about the condition of locally non-decreasing transformation? That is, $g$ is non-decreasing on $[0,+\infty]$ and $\zeta$ is continuous on $R$. – aaronyxt Sep 1 '17 at 5:15
• I think then you have to truncate $\zeta$ in that domain too. – MAN-MADE Sep 1 '17 at 5:18
• The implication clearly does not hold in general since $$\{\zeta^2>a^2\}=\{\zeta>a\}\cup\{\zeta<-a\}$$ and this union is disjoint hence, unless there are reasons to believe that $$P(\zeta<-a)=0$$ one knows that $$P(\zeta^2>a^2)>P(\zeta>a)$$ – Did Sep 1 '17 at 6:48
• @MANMAID "I think then you have to truncate ζ in that domain too" Sorry but I have no idea what you are talking about. Have you? – Did Sep 1 '17 at 6:50

$$\zeta = \begin{cases} 1 & w.p. 0.5 \\ -1 & w.p. 0.5\end{cases}$$
Let $a=0.1$, $P(\zeta > 0.1) \leq 0.6$ is a true statements.
However, $P(\zeta^2 > 0.1^2) =1 > 0.6$
• Thanks. I lack the condition of $\zeta$ being continuous. In fact, I am thinking of monotonic non-decreasing locally density transformation. – aaronyxt Sep 1 '17 at 5:13