Question

Question: Suppose that X and Y are random variables and that X and Y are non-negative for all points in a sample space S. Let Z be the random variable defined by Z(s) = max(X(s), Y(s)) for all elements\(s \in S\). Show that\(E(Z) \le E(X) + E(Y)\)

Short answer

Answer:

The inequality \(E(Z) \le E(X) + E(Y)\) has been proved.

Step-by-step solution

Note the given data

Let Z be the random variable defined by Z(s) = max(X(s), Y(s))for all elements\(s \in S\).

Theorem  

Theorem : If X is a random variable and p(X =r) is the probability that X =r then.\(E\left( X \right) = \sum\limits_{r \in X(s)} {p(X = r)r} \)

Calculation

By the definition.

For a random variable U,\(E\left( U \right) = \sum\limits_{} {rp(U = r)} \)for all\(r \in \).sample space.

Since X, Y are non-negative, E(X) and E(Y) are nonnegative since the expectation is the product of the random variable and probabilities both of which are non-negative.

Consider the set A, of all s such that\(X\left( s \right) \ge Y\left( s \right)\)

For \(s \in A\), \(Z\left( s \right) = X\left( s \right)\)

Which means\(E\left( Z \right) = E\left( X \right)\)

Hence\(E(Z) \le E(X) + E(Y)\)as\(E(Y)\)is nonnegative.

Similarly, Consider the set B, of all s such that\(X\left( s \right) < Y\left( s \right)\)

For \(s \in B\), \(Z\left( s \right) = Y\left( s \right)\)

Which means\(E\left( Z \right) = E\left( Y \right)\)

Hence\(E(Z) \le E(X) + E(Y)\)as\(E(X)\)is nonnegative

Since X and Y are defined for all points s in S, the union of all points s such that \(X\left( s \right) \ge Y\left( s \right)\) and all points such that \(X\left( s \right) < Y\left( s \right)\) gives the sample space S.

Conclusion  

Thus, we have shown that the inequality is true for all s in the sample space.