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 $\mathbb{P}(g(x,Z))$?

  • 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


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 $$ \begin{align} P(g(X,Z)\in A\mid X=x)&={\rm E}[U\mid X=x]=\int_\mathbb{R} \psi(x,z)\,P_Z(\mathrm dz\mid X=x)\\ &=P(g(x,Z)\in A\mid X=x). \end{align} $$

  • $\begingroup$ 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

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