Function of a random variable


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?


$g^{-1}(A)$ 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…

  • $\begingroup$ 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

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.