How to define $\mathbb{E}[X|X<y]$ formally

Question

In my introduction to probability theory lecture we defined $\mathbb{E}[X|Y]$ to be the random Variable Z with the properties:

1. $Z=g(Y)$ where g is measureable
2. $\int_{\{Y\in B\}}ZdP=\int_{\{Y\in B\}}XdP$

Which ends up being a.s. unique and in the special case where X and Y are absolute continuous distributed you get $\mathbb{E}[X|Y]=\int x\phi(x|Y)dx$ where $\phi(x|y)=\frac{\phi(x,y)}{\phi_Y(y)}$ with $\phi(x,y)$ the common distribution and $\phi_Y(y)$ the distributino of Y.

You can also define $P(A|Y)=\mathbb{E}[1_A|Y]$. So I could write $P(X<y\mid Y)$.

But the question is kind of how to do the opposite: $\mathbb{E}[X|X<y]$ or more generally $\mathbb{E}[X|A]$ which was used in a game theory lecture. Where the lecturer basically just defined $X|X<y$ to have the density $\frac{\phi_X(x)1_{x<y}}{P(X<y)}$.

And this kind of makes intuitive sense but I wonder how to unify this with the abstract definition of an expected value I learned in my probability theory lectures.

Answer 1

It depends on what you want I think.

Your lecturer aims at the conditional random variable $X$ which takes on only values below $y$. In that sense, the shape of the density of $X$ is the same, only its support is now on the interval $(-\infty,y]$ which gives the scaling factor of $1/P(X<y)$.

On the other hand, you might view $X<y$ as a random variable which takes on the value $1$ if $X<y$ and $0$ if $X\geq y$. In that case, which is essentially the more formal one. You get: $$ E(X|X<y) = \frac{1}{P(X<y)} \int_{-\infty}^{y} x f_X(x)\,\mathrm{d} x\cdot \mathbb{1}_{X<y} + \frac{1}{P(X\geq y)} \int_{y}^{\infty} x f_X(x)\,\mathrm{d} x\cdot \mathbb{1}_{X\geq y}. $$

Edit: Even more formally, you actually have to find the sigma algebra that corresponds to $X<y$. In this case, it is $$\mathcal{A}=\{\varnothing,\{\omega:\ X(\omega)<y\},\{\omega:\ X(\omega)\geq y\},\Omega\}. $$ The conditional expectation is now equal to $$ E(X|X<y) = E(X|\mathcal{A}) $$ You now have to check that for all $A\in\mathcal{A}$, we have $$ \int_A \left(\frac{1}{P(X<y)} \int_{-\infty}^{y} x f_X(x)\,\mathrm{d} x\cdot \mathbb{1}_{X<y}(\omega) + \frac{1}{P(X\geq y)} \int_{y}^{\infty} x f_X(x)\,\mathrm{d} x\cdot \mathbb{1}_{X\geq y}(\omega)\right) \,\mathrm{d} \omega = \int_A X(\omega) \,\mathrm{d}\omega $$ Let's take for example $A=\{\omega:\ X(\omega)<y\}$ (the empty set and the complete set are easy, $A^\complement$ is similar), then we get here that $\mathbb{1}_{X\geq y}(\omega) = 0$ on $A$, $\mathbb{1}_{X<y}(\omega) = 1$ on $A$. Note that $P(X < y) = P(A)$ and we get on the left-hand side $$ \frac{1}{P(A)} \int_A \int_{-\infty}^{y} x f_X(x)\,\mathrm{d} x \cdot \mathbb{1}_{X < y}(\omega)\,\mathrm{d} \omega = \int_{-\infty}^{y} x f_X(x)\,\mathrm{d} x \cdot \frac{1}{P(A)} \int_A \mathbb{1}_{X < y}(\omega)\,\mathrm{d} \omega = \int_{-\infty}^{y} x f_X(x)\,\mathrm{d} x = E(X\mathbb{1}_{X<y}). $$ Moreover, the right-hand side can be seen as follows (note that $\omega \in A$ is equivalent to $X(\omega) < y$) $$ \int_A X(\omega)\,\mathrm{d}\omega = \int_\Omega X(\omega)\mathbb{1}_{\omega \in A}\,\mathrm{d}\omega = \int_\Omega X(\omega)\mathbb{1}_{X(\omega)\leq y}\,\mathrm{d}\omega = E(X\mathbb{1}_{X\leq y}). $$

Edit2: You also have to note that the expression for the conditional expectation is measurable. This is of course quite trivial for me since I took a few courses in these topics. However, to give a complete answer, I will give the reasoning for this argument. Since $X$ is a random variable, the indicator 'random variables' $\mathbb{1}_{X<y}$ and $\mathbb{1}_{X\geq y}$ are measurable. Moreover, linear combinations of measurable functions are measurable. Hence, it is indeed measureable.

Answer 2

In general, if you want to represent everything in terms of the $\sigma$-algebra notion of conditional expectation, then $E[X \mid A]$ where $A$ is a shorthand for $E[X \mid \sigma(1_A)](\omega)$ where $\omega$ is any element of $A$. The fact that $E[X \mid \sigma(1_A)]$ is $\sigma(1_A)$ measurable forces it to be constant on $A$ (indeed this is the generalization of your first property of $E[X \mid Y]$).

When $A$ is null you have to be a bit careful, however, because in this case your second property doesn't care about the values of $E[X \mid 1_A]$ on $A$ since they wash out upon integration. In this situation you generally need to use some kind of limiting setup: for example, to condition on $Y=y$ you might instead condition on $Y \in (y-\varepsilon,y+\varepsilon)$ and send $\varepsilon \to 0$.