0
$\begingroup$

I am thinking of the property in probability of inequality. In particular, we assume \begin{equation} P[\zeta>a]\leq b, \end{equation} 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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – MAN-MADE Sep 1 '17 at 5:11
  • $\begingroup$ Thanks. How about the condition of locally non-decreasing transformation? That is, $g$ is non-decreasing on $[0,+\infty]$ and $\zeta$ is continuous on $R$. $\endgroup$ – aaronyxt Sep 1 '17 at 5:15
  • $\begingroup$ I think then you have to truncate $\zeta$ in that domain too. $\endgroup$ – MAN-MADE Sep 1 '17 at 5:18
  • 1
    $\begingroup$ 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 $$P(\zeta^2>a^2)>P(\zeta>a)$$ $\endgroup$ – Did Sep 1 '17 at 6:48
  • $\begingroup$ @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? $\endgroup$ – Did Sep 1 '17 at 6:50
1
$\begingroup$

$$\zeta = \begin{cases} 1 & w.p. 0.5 \\ -1 & w.p. 0.5\end{cases} $$

Let $a=0.1$, $P(\zeta > 0.1) \leq 0.6$ is a true statements.

However, $P(\zeta^2 > 0.1^2) =1 > 0.6$

$\endgroup$
  • $\begingroup$ Thanks. I lack the condition of $\zeta$ being continuous. In fact, I am thinking of monotonic non-decreasing locally density transformation. $\endgroup$ – aaronyxt Sep 1 '17 at 5:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.