12
$\begingroup$

I have a doubt: consider the real valued random variables $X$ and $Z$ both defined on the probability space $(\Omega, \mathcal{F},\mathbb{P})$.

Let $Y:= g(X,Z)$, where $g(\cdot)$ is a real-valued function. Since $Y$ is a function of random variables it is a random variable.

Let $x:=X(\omega)$ i.e. a realisation of $X$.

Is $\mathbb{P}(Y|X=x)=\mathbb{P}(g(X,Z)|X=x)$ equal to $\mathbb{P}(g(x,Z))$?

$\endgroup$
3
  • 2
    $\begingroup$ Because your notation is rather abbreviated, it might be worth pointing out that it implicitly refers to some Borel set $A$, subject to a universal quantifier, and that a fuller rendering of your question therefore would be whether it is the case that $$\forall_A\ \mathbb{P}(Y\in A|X=x)=\mathbb{P}(g(X,Z)\in A|X=x)=\mathbb{P}(g(x,Z)\in A).$$ $\endgroup$
    – whuber
    Commented May 12, 2014 at 21:42
  • $\begingroup$ @whuber: your last equality is valid only if $X$ and $Z$ are independent. $\endgroup$
    – Zen
    Commented Jun 24, 2014 at 15:10
  • 1
    $\begingroup$ OK, you're just considering "whether it is the case that...". $\endgroup$
    – Zen
    Commented Jun 24, 2014 at 15:13

1 Answer 1

8
$\begingroup$

If $g$ is measurable, then $$ P(g(X,Z)\in A\mid X=x)=P(g(x,Z)\in A\mid X=x),\quad A\in\mathcal{B}(\mathbb{R}) $$ holds for $P_X$-a.a. $x$. In particular, if $Z$ is independent of $X$, then $$ P(g(X,Z)\in A\mid X=x)=P(g(x,Z)\in A),\quad A\in\mathcal{B}(\mathbb{R}) $$ holds for $P_X$-a.a. $x$.

This relies on the following general result:

If $U,T$ and $S$ are random variables and $P_S(\cdot \mid T=t)$ denotes a regular conditional probability of $S$ given $T=t$, i.e. $P_S(A \mid T=t)=P(S\in A\mid T=t)$, then $$ {\rm E}[U\mid T=t]=\int_\mathbb{R} {\rm E}[U\mid T=t,S=s]\,P_S(\mathrm ds\mid T=t).\tag{*} $$

Proof: The definition of a regular conditional probability ensures that $$ {\rm E}[\psi(S,T)]=\int_\mathbb{R}\int_\mathbb{R} \psi(s,t)\,P_S(\mathrm ds\mid T=t)P_T(\mathrm dt) $$ for measurable and integrable $\psi$. Now let $\psi(s,t)=\mathbf{1}_B(t){\rm E}[U\mid S=s,T=t]$ for some set Borel set $B$. Then $$ \begin{align} \int_{T^{-1}(B)} U\,\mathrm dP&={\rm E}[\mathbf{1}_B(T)U]={\rm E}[\mathbf{1}_B(T){\rm E}[U\mid S,T]]={\rm E}[\psi(S,T)]\\ &=\int_{\mathbb{R}}\int_{\mathbb{R}}\psi(s,t)\, P_S(\mathrm ds\mid T=t)P_T(\mathrm dt)\\ &=\int_B\varphi(t)P_T(\mathrm dt) \end{align} $$ with $$ \varphi(t)=\int_\mathbb{R}{\rm E}[U\mid T=t,S=s]\,P_S(\mathrm ds\mid T=t). $$ Since $B$ was arbitrary we conclude that $\varphi(t)={\rm E}[U\mid T=t]$.

Now, let $A\in\mathcal{B}(\mathbb{R})$ and use $(*)$ with $U=\psi(X,Z)$, where $\psi(x,z)=\mathbf{1}_{g^{-1}(A)}(x,z)$ and $S=Z$, $T=X$. Then we note that $$ {\rm E}[U\mid X=x,Z=z]={\rm E}[\psi(X,Y)\mid X=x,Z=z]=\psi(x,z) $$ by definition of conditional expectation and hence by $(*)$ we have $$ \begin{align} P(g(X,Z)\in A\mid X=x)&={\rm E}[U\mid X=x]=\int_\mathbb{R} \psi(x,z)\,P_Z(\mathrm dz\mid X=x)\\ &=P(g(x,Z)\in A\mid X=x). \end{align} $$

$\endgroup$
2
  • $\begingroup$ Hello @StefanHansen. I'd be very grateful if you could clarify two or three things. The first thing is, is there a formula that relates the joint law $P_{S,T}$ to $P_S(\cdot\mid T=t)$ and $P_T$? The reason I'm asking is because I don't really understand your first equation in the proof. The second thing is; the very last equality in your answer, i.e. $$\int_\mathbb{R} \psi(x,z)\,P_Z(\mathrm dz\mid X=x)=P(g(x,Z)\in A\mid X=x),$$ how do you motivate this? Is it a definition of something? The third thing is, can you use what you proved to show the same identity holds for expectation? How? $\endgroup$
    – psie
    Commented 15 hours ago
  • $\begingroup$ If you could also explain why ${\rm E}[\mathbf{1}_B(T)U]={\rm E}[\mathbf{1}_B(T){\rm E}[U\mid S,T]]$, I'd be immensely grateful. $\endgroup$
    – psie
    Commented 15 hours ago

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.