Question

Suppose the random variable \(e\) has cdf \(F(t)\). Let \(\varphi(u)=\sqrt{12}[u-(1 / 2)]\), \(0

Short answer

For \(\varphi[F(e_{i})]\) with \(\varphi(u)=\sqrt{12}[u-(1/2)]\), the mean is 0 and the variance is 1. For a general score function \(\varphi(u)\) which satisfies \(\int_{0}^{1} \varphi(u) du=0\) and $\int_{0}^{1}\varphi^{2}(u) du=1$, the mean is also 0 and the variance is 1.

Step-by-step solution

Calculation of the mean for \(\varphi[F(e_{i})]\)

The mean of a random variable X is calculated as \(E(X) = \int x f(x) dx\), where \(f(x)\) is the probability density function. In this case, \(X = \varphi[F(e_{i})]\). We are given \(\varphi(u) = \sqrt{12}[u - \frac{1}{2}]\), and \(u = F(e_i)\). So, \(E[\varphi[F(e_{i})]] = \int \varphi[F(e_{i})] f(e_{i}) de_{i} = \int \sqrt{12}[F(e_{i}) - \frac{1}{2}] f(e_{i}) de_{i} = \sqrt{12} (\int F(e_{i}) f(e_{i}) de_{i} - \int \frac{1}{2} f(e_{i}) de_{i})\). Now, \(\int F(e_{i}) f(e_{i}) de_{i} = E[F(e_{i})] = \int x dx = \frac{1}{2}\) as cumulative function outcomes are uniformly distributed between 0 and 1. And, \(\int \frac{1}{2} f(e_{i}) de_{i} = \frac{1}{2} \int f(e_{i}) de_{i} = \frac{1}{2} [1 - 0] = \frac{1}{2}\), as the total probability equals 1. Thus, \(E[\varphi[F(e_{i})]] = \sqrt{12} (\frac{1}{2} - \frac{1}{2}) = 0\)

Calculation of the variance for \(\varphi[F(e_{i})]\)

Now we compute the variance of the random variable \(\varphi[F(e_{i})]\). The variance is calculated as \(Var(X) = E(X^2) - [E(X)]^2\). Here, \(X = \varphi[F(e_{i})]\). So, \(Var(\varphi[F(e_{i})]) = E[\varphi^2[F(e_{i})]] - (0)^2 = E[\varphi^2[F(e_{i})]] = \int \varphi^2[F(e_{i})] f(e_{i}) de_{i}\). But for the given \(\varphi(u)\), \(\varphi^2(u) = 12(u - \frac{1}{2})^2\). Let's substitute it, we get \(Var(\varphi[F(e_{i})]) = \int 12[F(e_{i}) - \frac{1}{2})^2] f(e_{i}) de_{i} = 12 (\int F^2(e_{i}) f(e_{i}) de_{i} - \int \frac{1}{4} f(e_{i}) de_{i})\). Both the integrals equal to \(\frac{1}{3}\) and \(\frac{1}{4}\) respectively. By substituting, we get \(Var(\varphi[F(e_{i})]) = 12 (\frac{1}{3} - \frac{1}{4}) = 12 (\frac{1}{12}) = 1\).

Establishing the mean and variance for general \(\varphi(u)\)

For general \(\varphi(u)\) which satisfies \(\int_{0}^{1} \varphi(u) du=0\) and $\int_{0}^{1}\varphi^{2}(u) du=1$, we proceed similarly for calculating the mean and variance. Since all \(\varphi(u)\) equals 0 when integrated from 0 to 1, the expected value \(E[\varphi[F(e_{i})]]\) equals to 0. The variance \(Var(\varphi[F(e_{i})])\) equals to \(\int \varphi^2[F(e_{i})] f(e_{i}) de_{i}\), which equals to 1 according to the provided conditions. So, for a general \(\varphi(u)\), the mean is 0 and the variance is 1.