Let Z ∼ N(0, 1). Find E(Φ(Z)), where Φ(·) is the cumulative distribution function of

Z. [1/2]

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!

Let Z ∼ N(0, 1). Find E(Φ(Z)), where Φ(·) is the cumulative distribution function of Z.

Learn how to calculate the expected value of the cumulative distribution function (CDF) of a standard normal variable Z ~ N(0, 1). This problem involves key statistical concepts like normal distribution, expected value, and CDF, with a final result of E[Φ(Z)] = 1/2.

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