Upper bound on the expectation of a random variable indexed by a random variale

Question

Imagine that we are given $X(u)$, a random variable indexed by some parameter $u \in I \subset \mathbb{R}$. Furthermore, assume that we know that: \begin{equation*} \mathbb{E}(|X(u)|)\leq 1 \quad \text{$\forall$ $u\in I$.} \end{equation*} Now, let $\theta$ be a random variable such that $\mathbb{P}(\theta \in I) = 1$. We do not suppose that $\theta$ and $X(u)$ are independent. Does it hold that: \begin{equation*} \mathbb{E}(|X(\theta)|)\leq 1 \quad \text{?} \end{equation*}

Answer 1

At least one declared downvote for extramathematical reasons, by user @stuartstevenson. This sucks.

The result does not hold without independence, as the simple example below shows.

Consider the parameter space $I=[0,1)$ and the probability space $(\Omega,\mathcal F,P)$ defined by $\Omega=[0,1)$, $\mathcal F=\mathcal B(\Omega)$, and $P$ the Lebesgue measure restricted to $\Omega$.

For every $u$ in $I$ and $\omega$ in $\Omega$, let $X(u)(\omega)=2\cos^2(2\pi(\omega+u))$. For every $\omega$ in $\Omega$, let $\theta(\omega)=1-\omega$.

Then, $E(X(u))=1$ for every $u$ in $I$ but the random variable $Y=X(\theta)$ is, by definition, such that $Y(\omega)=X(1-\omega)(\omega)=2$ for every $\omega$ in $\Omega$ hence $E(Y)=2$.

Answer 2

Yes, it holds.

If $\theta \in I$ with probability $1$ then $\theta \in I$.

The rest follows from definition.

The only issue I might see is if this is in some kind of limiting sense in which case you might use the suffix, "in the limit" or "almost surely" depending on the context.

Answer 3

As Did says, this is false.

Suppose that $I=\{0,1\}$ and $X(u)$ and $\theta$ are determined by a fair coin toss:

$$ \begin{align} \text{heads}&\implies X(0)=2, X(1)=0, \theta=0\\ \text{tails}&\implies X(0)=0, X(1)=2, \theta=1 \end{align} $$

Then $E(X(0))=E(X(1))=1$ and $E(X(\theta))=2$.