4
$\begingroup$

Let $W$ be a random variable that only takes on the values $1$ or $0$. Let $X$ and $Y$ be two other random variables. I came across the following: $$\mathbb{E}(Y|W=1, X)$$ How is this 'conditional expectation' even defined? It doesn't seem to fit the usual definition of a conditional expectation because $W=1$ is an event while $X$ is a random variable.

I was hoping to find a definition similar to that of $\mathbb{E}(Y|X)$, which has the property that $$\int_{B} \mathbb{E}(Y|X) dP=\int_{B} Y dP$$ for all $B \in\sigma(X)$.

Additional context:

Suppose we have observations $Y_i=W_i Y_i(1)+(1-W_i)Y_i(0)$ where the random variable $W_i \in\{0,1\}$ and the two random variables $Y_i(0), Y_i(1)$ can each be described by a regression model, that is $$Y_i(w)=\beta_{w,0}+\beta_{w,1}X_i+\epsilon_{w,i}, \hspace{5mm} w\in\{0,1\}$$ When the author writes $E(Y_i|W_i=1, X_i)$, I am pretty sure he means that $$E(Y_i|W_i=1, X_i)=\beta_{1,0}+\beta_{1,1}X_i$$ but I am wondering if this conditional expectation is well defined or made up notation.

$\endgroup$
5
  • $\begingroup$ $E(Y|W=1,X)$ may be shorthand notation for $g(X)$ where $g(x)=E(Y|W=1, X=x)$. So it's a function of the random variable $X$ and thus itself a random variable. $\endgroup$ Jun 17, 2018 at 12:55
  • $\begingroup$ @JarleTufto does the tower law hold in that case, that is $\mathbb{E}(\mathbb{E}(Y|W=1, X))=\mathbb{E}(Y|W=1)?$ $\endgroup$
    – Joogs
    Jun 17, 2018 at 16:02
  • $\begingroup$ Yes, all laws of probability also hold conditionally. $\endgroup$ Jun 17, 2018 at 16:40
  • 1
    $\begingroup$ @JarleTufto , it appears I made a mistake in the above, it should be $$\mathbb{E}(\mathbb{E}(Y|W=1, X)|W=1)=\mathbb{E}(Y|W=1)$$ which looks very similar to the tower law $$\mathbb{E}(\mathbb{E}(Y|W, X)|W)=\mathbb{E}(Y|W)$$, but how do we know that the first expression is true as well? It seems that your general statement "all laws of probability also hold conditionally" plays a role here: namely, the tower law also holds conditionally on the event $W=1$. Do you have any reference explaining why this is the case? $\endgroup$
    – Joogs
    Jun 17, 2018 at 18:09
  • 1
    $\begingroup$ @Joogs: Most introductory books in probability theory will establish that a conditional probability with a given condition is a probability measure (i.e., obeys the Kolmogorov axioms). Once that is established, all theorems applying to probability measures hold. $\endgroup$
    – Ben
    Jun 28, 2018 at 4:56

2 Answers 2

2
$\begingroup$

This is a fairly standard notation, $\mathbb{E}[Y|W,X]$ being the random variable derived from the pair $(W,X)$. When the realisation of $W$ is one (1) and the realisation of $X$ is $x$, this random variable has as realisation $\mathbb{E}[Y|1,x]$, or in a clearer representation $\mathbb{E}[Y|W=1,X=x]$. Since the only possible realisations of $W$ are $0$ and $1$, the random variable $\mathbb{E}[Y|W,X]$ can only take as values $\mathbb{E}[Y|W=0,X]$ and $\mathbb{E}[Y|W=1,X]$.

$\endgroup$
3
  • $\begingroup$ So you are saying that $\mathbb{E}(Y|W=1,X)$ should be intepreted as $f(W=1, X)$ where $f$ is the function in $\mathbb{E}(Y|W, X)=f(W,X)$, which makes sense. Can you give a measure theoretic definition of this expression (similar to the usual conditional expectation $\mathbb{E}(Y|X)$ I gave above) ? Also, does it hold that $\mathbb{E}(\mathbb{E}(Y|W=1, X))=\mathbb{E}(Y|W=1)?$ $\endgroup$
    – Joogs
    Jun 17, 2018 at 15:57
  • $\begingroup$ It should be $$\mathbb{E}(\mathbb{E}(Y|W=1, X)|W=1)=\mathbb{E}(Y|W=1)?$$ $\endgroup$
    – Xi'an
    Jun 17, 2018 at 16:39
  • $\begingroup$ This looks very similar to the the tower property which says that $$\mathbb{E}(\mathbb{E}(Y|W,X)|W)=\mathbb{E}(Y|W)$$, but it is definitely not the same since the RHS in the usual tower property, $\mathbb{E}(Y|W)$, is a random variable while $\mathbb{E}(Y|W=1)$ is just a real number. How do we know $$\mathbb{E}(\mathbb{E}(Y|W=1, X)|W=1)=\mathbb{E}(Y|W=1)$$ is true? $\endgroup$
    – Joogs
    Jun 17, 2018 at 18:04
0
$\begingroup$

I will post this as an answer to my own question, but I am not entirely sure if it is correct.

$\mathbb{E}(Y|W=1, X)$ is the random variable defined by having the property:

$$\int_B \mathbb{E}(Y|W=1,X) dP=\int_B Y dP$$ for all $B$ in $\sigma(\{W=1\}\cap C: C \in\sigma(X))$.

From this property it immediately follows that the 'conditional' tower law holds since by definition of the property above $$\mathbb{E}(I_{\{W=1\}}E(Y|W=1,X))=E(I_{\{W=1\}}Y)$$ which implies that $$\mathbb{E}(\mathbb{E}(Y|W=1,X)|W=1)=\frac{\mathbb{E}(I_{\{W=1\}}E(Y|W=1,X))}{P(W=1)}\\=\frac{\mathbb{E}(I_{\{W=1\}}Y)}{P(W=1)}=E(Y|W=1)$$

$\endgroup$

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.