I am not sure if this is correct but I vaguely recall the professor mentioned something like this in class: Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space, and $X, Y_{1}, ..., Y_{n}$ random variables on it. Let $\sigma(Y_{1}, ..., Y_{n})$ be the sigma algebra generated by $Y_{1}, ..., Y_{n}$. Then $X$ is $\sigma(Y_{1}, ..., Y_{n})$-measurable iff $X = H(Y_{1}, ..., Y_{n})$ for some $H$ measurable.

Could someone explain the hard direction?


1 Answer 1


Here is an answer for the case where $X$ is a real random variable, and $Y$ a random variable taking values in any measurable space. In your situation, set $Y = (Y_1,\ldots,Y_n)$.

If $X = 1_A$ for some $A \in \sigma(Y)$ then the result follows from the definition of $\sigma(Y)$ : there exists some measurable subset $B$ such that $A = [Y \in B]$, so $1_A=1_B(Y)$.

By countable linear combinations, the result holds for every $\sigma(Y)$-measurable discrete real random variable.

Last, each $\sigma(Y)$-measurable real random variable $X$ can be written as a limit of $\sigma(Y)$-measurable discrete real random variables. For example, consider $X_n = 2^{-n}\lfloor 2^X \rfloor$. Then $X_n = h_n(Y)$ for some measurable function $h_n$. Then the function $h = \limsup_{n \to +\infty} h_n$ works.

  • $\begingroup$ I have a question about $A$. For a single variable $Y_{1}$, we can say any element in $\sigma(Y_{1})$ has the form $Y_{1}^{-1}(B)$, where $B$ is a measurable set in the space where $Y_{1}$ takes values. But the definition of $\sigma(Y_{1}, ..., Y_{n})$ is the smallest sigma algebra that contains $\{Y_{i}^{-1}(B): i = 1, ..., n, B \text{ measurable}\}$. So a set $A$ from this sigma algebra may not have a form $Y_{1}^{-1}(B_{1}) \cap ... \cap Y_{n}^{-1}(B_{n})$ which in your case is $Y^{-1}(B)$. It may be a "generated" set so has a complicated explicit expression? $\endgroup$
    – Tom
    Dec 14, 2022 at 15:21
  • $\begingroup$ @Tom One can also introduce the random variable $Y := (Y_1,\ldots,Y_n)$ and check that $\sigma(Y) = \sigma(Y_1,\ldots,Y_n)$. $\endgroup$ Dec 14, 2022 at 15:31
  • $\begingroup$ I get it. The details involve some product sigma algebra-related arguments about the space where $Y$ takes values. Thank you! $\endgroup$
    – Tom
    Dec 14, 2022 at 16:05

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