I have three random variables. Let $X$ be independent from $(Y,Z)$ ($:=\sigma(Y,Z))$, $Y\neq0$ a.s.

Q: How can I proof (if it holds), that $E\left(\frac{X}{Y} \mid Z\right)=E(X) E\left(\frac{1}{Y}\mid Z\right)$ ?

From answers to an older question that this follows under the assumption X independent from $(\frac{1}{Y}, Z)$, however I would really like to avoid to have to assume this directly.

I know that there is a propositon that for $X_i\colon \Omega \to E_i$ independent, $\varphi_i\colon E_i \to E'_i$ measurable, $\varphi_i \circ X_i$ are independent as well, this used for $(X,Y,Z)$, (id, $x \mapsto \frac{1}{x}$, id$)$, I have $X,\frac{1}{Y}, Z$ independent, but this means that $\sigma(X), \sigma(\frac{1}{Y}), \sigma(Z)$ are independent, and not $\sigma(X), \sigma(\frac{1}{Y},Z)$. But maybe it can still be used?

  • $\begingroup$ That older question says $X$ is independent of $Y$ and $X$ is independent of $Z$, but stops short of saying $X$ is independent of $(Y,Z)$, which is a stronger condition. But I suspect that that stronger condition may have been intended. However, if $X$ is independent of $(Y,Z)$, then $X$ is independent of $(1/Y,Z)$, so you get the result you want. $\endgroup$ Commented Aug 18, 2011 at 15:16
  • $\begingroup$ So from @Didier's answer, it follows that $X$ and $(1/Y,Z)$ are independent if $X$ and $(Y,Z)$ are. $\endgroup$
    – Johannes L
    Commented Aug 31, 2011 at 13:46

1 Answer 1


The property holds and indeed, it is a consequence of the fact that $X$ is independent on $(Y,Z)$ (which, by the way, is a synonym of the fact that $\sigma(X)$ and $\sigma(Y,Z)$ are independent sigma-algebras). To see this, let us first state a slightly more general result.

Lemma Let $T$ denote any random variable and $u$ any measurable function. Then, $\sigma(u(T))\subseteq\sigma(T)$.

How does this apply to your setting? Take $T=(Y,Z)$ and $u:(y,z)\mapsto(1/y,z)$ and let $U=1/Y$. Then $u(T)=(U,Z)$. Since $u$ is one-to-one on $\mathbb R^*\times\mathbb R$, $u\circ u$ is one-to-one as well and the lemma applies to $u$ and to $u\circ u$. But $u\circ u$ is in fact the identity, hence $\sigma(Y,Z)\subseteq\sigma(U,Z)\subseteq\sigma(Y,Z)$, which proves that $\sigma(U,Z)=\sigma(Y,Z)$.

Finally, $X$ is independent on $\sigma(Y,Z)$ if and only if $X$ is independent on $\sigma(U,Z)$ and this fact is a tautology.

Now to the proof of the lemma. Assume the event $A$ is in $\sigma(u(T))$. As you know, this means there exists a measurable $B$ such that $A=[u(T)\in B]$. Recall that $u^{-1}(B)=\{t\mid u(t)\in B\}$ and let $C=u^{-1}(B)$. Then $C$ is measurable and $A=[T\in C]$, hence $A$ belongs to $\sigma(T)$ and the lemma holds.

Edit This is to answer the OP's comment below, without the length limitations associated to comments since it appears necessary to come back in details to some definitions.

Consider a set $\Omega$ and a function $\xi:\Omega\to\mathbb R$. To say that $\xi$ is a random variable is simply to add to this fact three conditions:

  1. The source space $\Omega$ is endowed with a sigma-algebra, say $\mathcal F$.
  2. The target space $\mathbb R$ is endowed with a sigma-algebra, usually the Borel sigma-algebra $\mathcal B$.
  3. And $[\xi\in B]$ is an event for every $B$ in $\mathcal B$.

Recall that $[\xi\in B]$ is a notation for $\xi^{-1}(B)$ and that to be an event is probabilistic jargon for to belong to $\mathcal F$ (once the sigma-algebra $\mathcal F$ has been fixed).

Likewise, for any function $\Xi:\Omega\to\mathbb R^n$ with $n\geqslant1$, to ask that $\Xi$ is a (vector-valued) random variable would be to ask that $[\Xi\in B]$ is in $\mathcal F$, for every $B$ in the Borel sigma-algebra $\mathcal B_n$ on $\mathbb R^n$.

Right. It happens that every function $\xi:\Omega\to\mathbb R$ (random variable or not) defines a sigma-algebra $\sigma(\xi)$ on the source space $\Omega$, which is the collection of the subsets $[\xi\in B]$ for $B$ in $\mathcal B$. Yes, this is a sigma-algebra, always. Hence, to ask that $\xi$ is a random variable is to ask that $$ \sigma(\xi)\subseteq\mathcal F. $$ Likewise, for $\Xi:\Omega\to\mathbb R^n$ with $n\geqslant1$, to ask that $\Xi$ is a (vector-valued) random variable is to ask that $\sigma(\Xi)\subseteq\mathcal F$, where $\sigma(\Xi)$ is the collection of the subsets $[\Xi\in B]$ for $B$ in $\mathcal B_n$.

Rrrright. Now to your comment.

You consider two random variables $Y$ and $Z$ and the function $T=(Y,Z)$ defined, for every $\omega$ in $\Omega$, by $T(\omega)=(Y(\omega),Z(\omega))$. First note that, since you assumed that $Y$ and $Z$ are random variables, this is enough to guarantee that $T$ is a (vector-valued) random variable. Why? Well, to check directly that $[T\in B]$ is in $\mathcal F$ for every $B$ in $\mathcal B_2$ might not be easy but there is a two-steps way to do so, that is both classical and easy:

  1. Check this for products $B=B_1\times B_2$ with $B_1$ and $B_2$ in $\mathcal B$. To do so, simply write $[T\in B_1\times B_2]$ as $[Y\in B_1]\cap[Z\in B_2]$ and use the definitions.
  2. Consider the collection of events $\mathcal C=\{B\in\mathcal B_2\mid [T\in B]\in\mathcal F\}$. You just proved that $\mathcal B\times\mathcal B\subseteq\mathcal C$ and you know by definition that $\mathcal C\subseteq\mathcal B_2$. Since the sigma-algebra $\mathcal B_2$ is generated by $\mathcal B\times\mathcal B$, you know that $\sigma(\mathcal C)=\mathcal B_2$. But since $\mathcal C$ is a sigma-algebra (yes, always), $\mathcal C=\sigma(\mathcal C)$. Hence $\mathcal C=\mathcal B_2$, which is a rewriting of the result you wanted to prove.

Thus, $Y$ and $Z$ are random variables and $T=(Y,Z)$ is a (vector-valued) random variable and you know the definitions of $\sigma(Y)$, $\sigma(Z)$ and $\sigma(T)$, which are all sub-sigma-algebras of $\mathcal F$.

Well, it seems you are ready for the Grand final:

Let $\mathcal G=\sigma(\sigma(Y)\cup\sigma(Z))$. Then $\mathcal G=\sigma(T)$ $(\ast)$.

In other words, the two meanings of $\sigma(Y,Z)$ one can think about are in fact equivalent! We happy.

Recall one last time that $\mathcal G=\sigma(Y^{-1}(\mathcal B)\cup Z^{-1}(\mathcal B))$ and $\sigma(T)=\sigma(T^{-1}(\mathcal B_2))$ and that these are two sigma-algebras on $\Omega$.

To prove $(\ast)$, note that $[Y\in B]=[T\in B\times\mathbb R]$ and likewise for $Z$, hence $\sigma(Y)\cup\sigma(Z)\subseteq\sigma(T)$ hence $\mathcal G\subseteq\sigma(T)$. For the other inclusion, use once again the idea explained above and consider $$ \mathcal D=\{B\in\mathcal B_2\mid [T\in B]\in\mathcal G\}. $$ By definition, $\mathcal D\subseteq\mathcal B_2$ and almost by definition, $\mathcal B\times\mathcal B\subseteq\mathcal D$, hence you can probably finish this yourself.

  • $\begingroup$ What is confusing me: On the one hand, $(Y,Z):=\sigma(Y,Z)$($=\sigma(X^{-1}(\mathcal{F}) \cup Y^{-1}(\mathcal{F}))$) ($(\Omega, \mathcal{P}, \mathbb{P})$ is our probability space) is then generated $\sigma$-algebra, on the other hand with random variable $T:=(Y,Z)$ we use $(Y,Z)$ for the vector of the two random variables $Y$ and $Z$ $\endgroup$
    – Johannes L
    Commented Aug 29, 2011 at 13:28
  • $\begingroup$ @Johannes, the Edit addresses your question. Please tell me if anything is unclear. $\endgroup$
    – Did
    Commented Aug 30, 2011 at 16:57
  • $\begingroup$ Very nice answer. I could follow your argument, except that I still have to think a little to make it clear to myself that $\mathcal{D}$ (resp. $\mathcal{C}$) is indeed a $\sigma$-algebra. Maybe it's worth to put it as a single question/answer pair so I can award you an "accepted answer" for the question why $\sigma(X,Y)$ ( $(X,Y)$ random vector) $=\sigma(X,Y)$ ($(X,Y)$ as the generated \sigma-algebra) $\endgroup$
    – Johannes L
    Commented Aug 31, 2011 at 9:47
  • $\begingroup$ Re D and C being sigma-algebras: yes, this is the heart of the proofs. Just take the axioms that make a sigma-algebra one at a time and there should be no problem checking them. $\endgroup$
    – Did
    Commented Aug 31, 2011 at 18:38

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