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Function of a random variable

$\Omega\overset{X}{\rightarrow}\mathbb{R}\overset{g}{\rightarrow}\mathbb{R}$

In the next example of a function of a random variable, $P_Y(y)= P_X(g^{-1}(y))$.

Example: A system has three output states $\Omega_X = \left \{ -1,0,1 \right \}$ with probabilities $P_X(-1)=\frac{1}{3}$, $P_X(0)=\frac{1}{3}$ and $P_X(1)=\frac{1}{3}$.

What's the probability function of $Y=X^2$?

$\Omega_Y = \left \{ 0,1 \right \}$.

$P_Y(0)= P_X(g^{-1}(0))=P_X(0) =\frac{1}{3}$ and $P_Y(1)= P_X(g^{-1}(1))=P_X(\left \{ -1,1 \right \}) =\frac{2}{3}$ (End of the example)

If $P_Y(1)= P_X(g^{-1}(1))$ then the inverse function $g^{-1}(y)$, has for an element of its domain, two elements of its codomain $P_X(\left \{ -1,1 \right \})$. By the definition of a function

"In mathematics, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output",

then $g^{-1}(y)$ is not a function.

One can say that the image of $P_X(g^{-1}(1))$ is the addition ($P_X(-1) + P_X(1)$), but the addition ($P_X(-1) + P_X(1)$) is not the image of $P_X(g^{-1}(1))$. The images of $P_X(g^{-1}(1))$ are two: $P_X(-1)$ and $P_X(1)$, the addition would be a completely different element.

Is it or isn't $g^{-1}(y)$ a function? And if it is not, why do they use $g^{-1}(y)$as a notation of this relation?

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$g^{-1}(A)$ doesn't mean the inverse function in this setting but the preimage of $A$ under $g$ so $g^{-1}(A) = \{ x \in \Bbb R | g(x) \in A \}$.

If $g$ is a measurable function then $g^{-1}(A)$ is measurable so $P_X(A)$ is well defined.

And because you are in a discrete setting ofc it holds that $g^{-1}(A)$ can be written as a disjoint union of elements of $g^{-1}(A)$ so $$g^{-1}(A) = \bigcup_{x \in g^{-1}(A)}\{x\}$$ and because it's countable and disjoint it also holds $$P_X\left(g^{-1}(A)\right) = \sum_{x \in g^{-1}(A)} P_X(\{x\})$$

Setting $A = \{1\}$ you get your result.

Indeed it's a bit "handwaving" to write $P_Y(1)$ instead of $P_Y(\{1\})$ but it's clear in this setting what is meant…

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  • $\begingroup$ Thanks a lot! Can you please expand on "it's a bit "handwaving" to write $P_Y(1)$ instead of $P_Y(\{1\})$, but it's clear in this setting what is meant..." $P_Y(\{1\})$ I think it means the element $1$ from the set $A$. What does $P_Y(1)$ mean? I can't see how this change in the notation makes any difference in the interpretation of $g^{-1}$ .Can you please explain? And finally how do you know that "$g^{-1}(A)$ doesn't mean the inverse function but the preimage of "A" under "g"? $\endgroup$ – match6 Nov 29 '16 at 20:08
  • $\begingroup$ In your first posting you wrote "$P_Y(1)= P_X(g^{-1}(1))$" but $P_Y$ is a probability measure so as an argument only measurable sets are possible. Although 1 is a real number and not a set (and so $P_Y(1)$ is strictly speaking not well defined it's well known that for a single elemented set the set braces can be omnited. So $P_Y(x)$ means $P_Y(\{x\})$ for a real number x. $\endgroup$ – Gono Nov 30 '16 at 8:21
  • $\begingroup$ To your $g^{-1}$ question: You claim, that $g^{-1}$ means the inverse function, so I could also ask you how do you know that? Just as well $g^{-1}$ could mean the reciprocal of g. But it doesn't. I just told you that your claim it's the inverse function is wrong. And here it's the same with the notation stuff above: For real numbers x and the preimage by $g^{-1}(x)$ always $g^{-1}(\{x\})$ is meant if you are technically correct. For a inverse function writing $g^{-1}(x)$ is totally ok because you have there one unique solution so you can write $y = g^{-1}(x)$. But not for measurable functions… $\endgroup$ – Gono Nov 30 '16 at 8:26
  • $\begingroup$ … in general because the preimage is not always a unique element and if you consider the definition of a measurable function you will see, that it's defined by preimages not by the inverse function. $\endgroup$ – Gono Nov 30 '16 at 8:27
  • $\begingroup$ Thanks. It helped me a lot. $\endgroup$ – match6 Nov 30 '16 at 13:22

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