Conditional Variance: Is it a random variable?

Question

Let's X, Y are random variables (i.e. each one maps elements of a sample space to real numbers). In particular, let's X is such that $X:\Omega \rightarrow \mathcal{R}$, where $\Omega=\{\omega_1, ..., \omega_n \}$ and $X(\omega_i)=i$

Then the conditional variance $Var(Y|X(\omega_i))$ is a random variable because it maps an element of a sample space of $\Omega$ to $\mathcal{R}$.

Now consider $Var(Y|X(\omega_i)=1 \bigcup X(\omega_i)=2)$ This expression maps a set $\{\omega_1, \omega_2 \}$ to $\mathcal{R}$. So, it seems no more to satisfy the definition of a random variable. Then what is it? A measure? Perhaps no. Thank you for answering.

Answer 1

By definition, $$ \textrm{Var}(Y\mid X):=\mathbb E\bigl[\bigl(Y-\mathbb E(Y\mid X)\bigr)^2\mid X\Bigr]=\mathbb E(Y^2\mid X)-\mathbb E(Y\mid X)^2. $$ Thus, the conditional variance is a random variable, in the same way that the conditional expectation $\mathbb E(Y\mid X)$ is. Conceptually, the variance is the "same type of object" as the expectation, in this regard.

Now, one may also consider an event $A\subseteq \Omega$ (the sample space) and ask what is $\textrm{Var}(Y\mid A)$. And it follows the exact same behavior as the conditional expectation, namely that we define $$ \textrm{Var}(Y\mid A):=\mathbb E\bigl[\bigl(Y-\mathbb E(Y\mid A)\bigr)^2\mid A\Bigr]=\mathbb E(Y^2\mid A)-\mathbb E(Y\mid A)^2. $$

By definition, $$\mathbb E(Y\mid A):=\frac{\mathbb E(Y\cdot 1_A)}{\mathbb E(1_A)},$$ where $1_A$ denotes the indicator of the set $A$. It is a random variable taking the value $1$ on $A$ and $0$ off $A$. Note also that $\mathbb E(1_A)=\mathbb P(A)$, I just wrote it that way in the denominator of the formula for consistency with the numerator.

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Per the discussion below, there was an even more basic question that I should clarify. A random variable is a function from the sample space $\Omega$ to the real numbers. This means it assigns a real number to each element $\omega\in \Omega$. On the other hand, when we condition on an event we obtain a set function on $\Omega$, or in other words, a function that assigns values to subsets of $\Omega$ and not to individual elements of $\Omega$. In this case, being even more precise, we have a partially defined set function which means that not every subset is assigned a value - it is only those subsets which are measurable and are assigned a positive measure for which the conditional variance is defined.

To compare and contrast the two types of mathematical objects, conditional variance with respect to a random variable is a function from $\Omega$ to $\mathbb R$, whereas conditional variance with respect to an event is a partially defined function from $P(\Omega)$ to $\mathbb R$ (the power set of $\Omega)$.

Answer 2

Actually as $\{ \omega_1, \omega_2 \} \in \Omega \oplus \Omega$ and that the latter is an equally valid sample space, it would no less satisfy the definition of a random variable.