In the context of Proposition 2.13, prove that the mean-variance function (V: mathbb{R} times mathbb{R}_{+} ightarrow

In the context of Proposition 2.13, prove that the mean-variance function \(V: \mathbb{R} \times \mathbb{R}_{+} \rightarrow \mathbb{R}\) is increasing with respect to the first argument and decreasing with respect to the second one.

Data From Proposition 2.13

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: Proposition 2.13 Let u RR be an increasing and concave utility function and let x be a normally distributed random variable with mean and variance o. Then there exists a function V: Rx R+R, increasing with respect to the first argument and decreasing with respect to the second, such that E[u(x)] = V(, o). Proof Let denote a normally distributed random variable with zero mean and unitary variance and define the function V: Rx R+ R by V(x, y) = E[u(x + yz)], for xe R and y = R+. The representation E[u(x)] = V(u,o) follows by noting that + oz = x. The sign of the two partial derivatives of V is established in Exercise 2.14. As a particular case, let us consider the case of an exponential utility function u(x) = exp(-ax), for a > 0, and a normally distributed random variable x with mean j and variance o. Recall that the random variable exp(ax) is distributed according to a log-normal distribution with expected value exp(au + ao/2). This implies that the expected utility admits the following representation: (- E [exp(-a)] = = exp ( a + 0) =: V(o). U(x) == = a