I think this is kinda basic, but say I have a random variable $X$, is the probability $P(X \leq a)$ the same as $P(f(X) \leq f(a))$ for any real-valued continuous function $f$?
$\begingroup$
$\endgroup$
1
-
1$\begingroup$ Also: Generally, $\sigma^{2}_{f(x)} \ne f\left(\sigma^{2}_{x}\right)$. $\endgroup$– AlexisMay 8, 2018 at 15:52
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
1
This holds only if $f$ is monotonically increasing. If $f$ is monotonically decreasing, then $P(f(X)\leq f(a)) = P(X \geq a)$. For instance, if $f(x) = -x$, and X is a normal die roll, then $P(X \leq 5) = \frac56 $ but $P(-X \leq -5) = \frac16$. If $f$ switches between increasing and decreasing, then it's even more complicated.
Note there's also the trivial case of $f(x) \equiv 0$, in which $P(f(X) \leq a)$ is equal to 1 if $a \geq 0$ and 0 otherwise.
-
2$\begingroup$ +1 I should have added the injective case when this is true. $\endgroup$ May 8, 2018 at 17:00
$\begingroup$
$\endgroup$
0
No. Take $X$ uniform on $[-1,1]$ and $a=0$. Then $\Pr(X < a ) = 1/2$. On the other hand $\Pr(X^2<a^2) = 0$.
answered May 8, 2018 at 12:08
$\begingroup$
$\endgroup$
This is related to asking:
is $X \leq a$ for every $f(X) \leq f(a)$?
There may be many ways to violate $f(X) \leq f(a)$ while $X \leq a$. But, in all cases, it requires $f$ to be a non-monotonous function.
answered May 11, 2018 at 13:02