Let $Z \sim N(0,1)$ be a standard normal random variable, and let $\Phi(\cdot)$ represent the cumulative distribution function (CDF) of the standard normal distribution. We are tasked with finding $E[\Phi(Z)]$, which is the expected value of $\Phi(Z)$.

Step-by-Step Solution:

1. Interpretation of the Expectation: We need to compute: $E[\Phi(Z)] = \int_{-\infty}^{\infty} \Phi(z) f_Z(z) dz$ where $f_Z(z) = \frac{1}{\sqrt{2\pi}} e^{-z^2 / 2}$ is the probability density function (PDF) of the standard normal distribution.

2. Properties of $\Phi(Z)$: By the definition of $\Phi(Z)$, for any real value $z$: $\Phi(z) = P(Z \leq z) = \int_{-\infty}^{z} \frac{1}{\sqrt{2\pi}} e^{-t^2 / 2} dt$ Since $Z$ is symmetrically distributed around 0, its cumulative distribution at $Z$ is centered around 0. Hence, the expectation of $\Phi(Z)$ can be thought of as the average probability that a standard normal random variable is less than some value $Z$.

3. Symmetry and Mean: The random variable $Z$ is symmetric, and the expected value of the CDF of $Z$ over the entire distribution of $Z$ should be $\frac{1}{2}$. This can be intuitively understood since $Z$ has an equal probability of being less than or greater than 0, and the value of $\Phi(0) = \frac{1}{2}$.

Thus, the result is:

$E[\Phi(Z)] = \frac{1}{2}$

This is the final answer.

Would you like any additional details or clarifications?

Here are 5 related questions to expand your understanding:

1. How do you derive the expected value of a function applied to a random variable?
2. What is the significance of the cumulative distribution function in probability theory?
3. How does the symmetry of a normal distribution affect expected values?
4. Can this result generalize to non-standard normal distributions?
5. What role does the mean and variance play in the shape of the normal distribution?

Tip: The symmetry of the normal distribution often simplifies complex expectations. Always check if symmetry can help reduce computational effort!