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$h(\cdot)$ denotes a strict monotonic increasing transformation such as $\log$.

Another inequality I do not quite get is that

$$\mathsf{P}\left(h(X) \le h(x)\right) \ge \mathsf{P}\left(X \le h(x)\right)$$

Some help would be very much appreciated!

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Which property do you want? The one in the title of your question or the one in the text of your question? –  Dilip Sarwate Aug 23 '12 at 20:23

1 Answer 1

No name that I know for the property in the title, which is a simple consequence of the identity, valid for any strictly increasing function $h$, $$ \{\omega\in\Omega\mid h(X(\omega))\leqslant h(x)\}=\{\omega\in\Omega\mid X(\omega)\leqslant x\}. $$ Note: The inequality in the body cannot be true in general.

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hm ok. for the inequality in the body dont we got something like {ω∈Ω∣h(X(ω))<=h(x)}<={ω∈Ω∣X(ω)<=h(x)}? i mean since x <= h(x) is true the quantatiy {ω∈Ω∣X(ω)<=h(x)} has to be larger or equal to {ω∈Ω∣h(X(ω))<=h(x)}? –  tim Aug 23 '12 at 20:45
    
Did you try this for the function $h=\log_{10}$ mentioned in your post, and $x=10,000,000$? Then $[\log(X)\lt\log(10,000,000)]=[\log(X)\lt7]=[X\lt10,000,000]$ but you suggest to replace it by $[X\lt7]$... –  Did Aug 23 '12 at 20:50
    
oh s*** i turned the <= the wrong way around. it is supposed to be a "=>" instead of "<=". i already edited it. im very sorry about this! –  tim Aug 23 '12 at 20:53
    
Still wrong! Try $h(x)=10^x$. –  Did Aug 23 '12 at 20:55
    
hm well ok than the statement just does not hold. thx alot! –  tim Aug 23 '12 at 20:58

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