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I have the following depictions in my lecture notes, which I fail to understand. It talks about functions of random variables and distinguishes between the two cases, when the function of a random variable $g(X)$ is strictly monotonic. $Y,X$ are random variables and $Y=g(X)$:

enter image description here

I do not understand why $P(X\leq x)=P(Y \leq y)$

and the second case: enter image description here

Same question. Sorry about the quality.

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  • $\begingroup$ I understand why I was perplexed. The second depiction is not right $\endgroup$
    – Naz
    Jan 2 '16 at 5:52
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$P(Y\leq y)=P(g(X)\leq g(x))$

When $g$ monotonic increasing, $g(X)\leq g(x)$ if and only if $X\leq x$. We have:

$P(Y\leq y)=P(g(X)\leq g(x))=P(X\leq x)$

When $g$ monotonic decreasing, $g(X)\leq g(x)$ if and only if $X\geq x$. We have:

$P(Y\leq y)=P(g(X)\leq g(x))=P(X\geq x)$

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The probability measure $P$ applies to events, i.e., measurable subsets of the underlying universal set $\Omega.$

In the two cases you are considering the subsets defined by the inequality on both sides of the identity contain the same $\omega\in\Omega,$ so it is not only the probabilities that are equal but the events themselves, as measurable sets.

Written out in full the first identity is actually

$$P(\{\omega\in\Omega;g(X(\omega))\leq g(x)\})=P(\{\omega\in\Omega;X(\omega)\leq x\}).$$

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    $\begingroup$ It is not very obvious to me how the inequalities contain the same $\omega \in \Omega$ $\endgroup$
    – Naz
    Jan 1 '16 at 14:47
  • $\begingroup$ I tried making the first case more explicit. $\endgroup$ Jan 1 '16 at 14:50
  • $\begingroup$ ughhh.... so some (increasing monotonic) function when applied on $X(\omega)$ will have the codomain of $X(\omega)$ for arbitrary $g(x)$? How does this change in the second case? $\endgroup$
    – Naz
    Jan 1 '16 at 14:57
  • $\begingroup$ Not arbitrary, strictly increasing. A strictly increasing function is by definition one-to-one and both it and its inverse have the property that $a\leq b$ iff $f(a)\leq f(b)$. The second case is about a strictly decreasing function. Same situation except that it and its inverse reverse the inequality. $\endgroup$ Jan 1 '16 at 15:53

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