Question

Suppose \(Y\) is \(\mathscr{B}\)-measurable, and \(E[|Y|]<\infty\). Show that \(E[Y Z] \geq 0\) for all bounded nonnegative \(\mathscr{B}\)-measurable random variables \(Z\) implies \(P[\\{\omega: Y(\omega) \geq 0\\}]=1\).

Short answer

To summarize, the statement \(P[\{\omega: Y(\omega) \geq 0\}]=1\) is proven by analyzing the expected value \(E[Y Z]\) for a specially defined measurable function \(Z\) and observing that it contradicts the given condition \(E[Y Z] \geq 0\) when \(Y(\omega) < 0\). Therefore, the measure of the set \(\omega\) for which \(Y(\omega) < 0\) must be zero, leading to the conclusion of the proof.

Step-by-step solution

Define the Set

Define the set \(A = \{\omega: Y(\omega) < 0\}\). This set contains all elements \(\omega\) for which \(Y(\omega)\) is less than zero.

Construct a Bounded Nonnegative Measurable Function

Construct a bounded nonnegative measurable function \(Z\) such that \(Z(\omega) = 1\) if \(\omega \in A\) and \(Z(\omega) = 0\) otherwise, which means that \(Z\) is an indicator function of the set \(A\). Therefore, \(Z\) is a \(\mathscr{B}\)-measurable random variable because \(A\) is a Borel set.

Evaluate the Expected Value

Now evaluate \(E[Y Z]\). As per the problem statement, \(E[Y Z]\) should be greater than or equal to zero. Since \(Z = 1\) for \(\omega \in A\) and \(Z = 0\) otherwise, \(E[Y Z] = E[Y Z \{\omega: Y(\omega) < 0\}]\). This expression is less than or equal to 0 since \(Y(\omega) < 0\) for \(\omega \in A\). Therefore, we have a contradiction to the statement \(E[Y Z] \geq 0\). This contradiction implies that the measure of the set \(A\) must be zero.

Conclude the Proof

Since the measure of set \(A\) is zero, the probability \(P[\{\omega: Y(\omega) < 0\}]=0\). This implies the statement \(P[\{\omega: Y(\omega) \geq 0\}]=1\), completing the proof.