Exercise 1.12 (Expectation does not depend on the choice of model)

Question

Let $X$ be a random variable taking values in $[0,+\infty]$, and let $Y$ be a simple unsigned random variable such that $0 \leq Y \leq X$ is surely true. Show that there exists a function $f: [0,+\infty] \rightarrow [0,+\infty]$ taking on finitely many values such that $Y \leq f(X) \leq X$ is surely true. Conclude in particular that the above definition of expectation does not depend on the choice of model $\Omega$.

Elu Thingol · Accepted Answer

Let $\Omega$ be the underlying randomness source of both $X$ and $Y$. By definition, $Y$ has a simple representation $Y=\sum_{i=1}^{n}a_i 1_{E_i}$ for non-negative reals $a_i$ and disjoint measurable sets $E_1,\ldots,E_n$. Define the function
\begin{equation*}
	f:[+\infty] 	o [+\infty] \qquad
	f(x) = \max\left\{a_i \in [0,+\infty] \mid a_i \leq x, 1 \leq i \leq n ight\}.
\end{equation*}
This function
\begin{itemize}
\item is well defined, as $\{a_i \leq x\}$ is never empty since $Y \leq X$ and $Y$ is defined on whole $\Omega$;
\item takes on at most $n$ non-negative values.
\end{itemize}
We now verify that
$$\forall \omega \in \Omega: \quad Y(\omega) \leq f(X(\omega)) \leq X(\omega).$$
But this is obvious since
\begin{align*}
	X(\omega) &\geq \max\left\{a_i \mid a_i \leq X(\omega) ight\}\
	&= f(X(\omega))\
	&\geq \max\left\{a_i \mid a_i \leq Y(\omega) ight\}\
	&= Y(\omega).
\end{align*}

Exercise 1.12 (Expectation does not depend on the choice of model)

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Exercise 1.12 (Expectation does not depend on the choice of model)

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