Intuition of expectation on a set vs. conditional expectation Suppose $X$ on a probability space $(\Omega,P_X)$. Assume that $X$ is integrable with respect to the measure $P_X$. I am interested in the following two objects, for $A$ a subset of the probability space:

*

*$\mathbb{E}(X\mathbb{I}_A)=\int_AXdP_X$,


*$\mathbb{E}(X\mid A)=\int_A X d P_{X\mid A}$.
Although I mathematically understand the difference between them, I am struggling with the intuition as of what each of them represents. I understand that 1) is the integral of $X$ limited to set $A$ with respect to $P$, whereas 2) is the integral of $X$ limited to set $A$ with respect to $P$ normalized as a probability measure on such a set.
Question: Any help clarifying the interpretation/intuition that differentiates these two objects, especially related to 1) is much appreciated. Should I just re-interpret $X\mathbb{I}_A$ as a truncated version of $X$ (and hence, a different random variable)? The fact that $\frac{\mathbb{E}(X\mathbb{I}_A)}{P(A)}=\mathbb{E}(X\mid A)$ only adds to my confusion.
 A: First, I believe the two following facts should be clear if you want to understand the difference between the two expected values you mention:

*

*The expected value of a random variable only depends on the distribution of that random variable.


*The distribution of a random variable depends on the underlying probability measure $\mathbb P$ which endows the universe $\Omega$.
I think $\mathbb E[X1_A]$ is easy to understand: it is the expected value of the random variable that coincides with $X$ on $A$ and is equal to $0$ elsewhere:
$$
\mathbb E[X1_A]=\int_AX(\omega)\,\mathbb P(d\omega).
$$
On the other hand, if $A$ has non-zero probability, that is $\mathbb P(A)>0$, then $\mathbb E[X\vert A]$ is by definition the expected value of $X$ under the conditionnal probability $\mathbb P_A$, which is defined for all measurable subset $B\subset\Omega$ by
$$
\mathbb P_A(B)=\frac{\mathbb P(A\cap B)}{\mathbb P(A)},
$$
often denoted $\mathbb P(B\vert A)$. In other words,
$$
\mathbb E[X\vert A]=\int_\Omega X(\omega)\,\mathbb P_A(d\omega).
$$
For anymeasurable subset $B\subset\Omega$ you have
$$
\int_\Omega 1_B(\omega)\,\mathbb P_A(d\omega)=\mathbb P_A(B)=\frac{\mathbb P(A\cap B)}{\mathbb P(A)},
$$
hence
$$
\mathbb E[1_B\vert A]=\frac{\mathbb E[1_B1_A]}{\mathbb P(A)}\cdot
$$
Now remember that any nonnegative random variable $X=\Omega\to\mathbb R$ can be written $X=\sum_{n=0}^{+\infty}a_n1_{B_n}$ where $(a_n)_{n\in\mathbb N}$ is a sequence of nonnegative real numbers and  $(B_n)_{n\in\mathbb N}$ a sequence of measurable subsets of $\Omega$. Then with standard rules on the expected values you find
$$
\mathbb E[X\vert A]=\sum_{n=0}^{+\infty}a_n\mathbb E[1_{B_n}\vert A]=\sum_{n=0}^{+\infty}a_n\frac{\mathbb E[1_{B_n}1_A]}{\mathbb P(A)}=\frac{\mathbb E[X1_A]}{\mathbb P(A)}\cdot
$$
A: For intuition, we can turn to concrete examples.
Let's say you're a baker who bakes cakes for a living. For simplicity, we'll assume you bake exactly one (very very fancy) cake per day for the first person who orders a cake from you. Let

*

*$\mathbf X$ be the amount of money you earn tomorrow;

*$A$ be the event that Alice (one particular person) is the person who orders cake.

Then $\mathbb E[\mathbf X \mathbb I_A]$ is the expected amount of money Alice pays you tomorrow: if Alice orders cake, $\mathbf X \mathbb I_A$ is the amount she pays you for it, and if Alice doesn't order cake, $\mathbf X \mathbb I_A = 0$.
Meanwhile, $\mathbb E[\mathbf X \mid A]$ is the expected amount of money Alice pays you if she is the one to order cake from you.
Suppose you have $100$ potential clients who are equally likely to order cake from you each day, and Alice is a very high tipper who pays you an average of $\$500$ for cake. Then $\mathbb E[\mathbf X \mid A] = 500$: if Alice orders cake from you tomorrow, you expect to make $\$500$ on average. However, $\mathbb E[\mathbf X \mathbb I_A] = 5$: though Alice pays you $\$500$ per cake when she orders from you, she only comes by once around every $100$ days, so you make an average of $\$5$/day from Alice.
