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

Let \( Z \sim N(0, 1) \), meaning \( Z \) follows the standard normal distribution with mean \( 0 \) and variance \( 1 \). We are tasked with finding the expected value \( E(\Phi(Z)) \), where \( \Phi(\cdot) \) is the cumulative distribution function (CDF) of a standard normal random variable \( Z \).

### Step-by-Step Solution

1. **Understanding \( \Phi(Z) \):**
   - The function \( \Phi(Z) \) represents the probability that a standard normal variable is less than or equal to \( Z \), i.e., \( \Phi(Z) = P(Z' \leq Z) \) where \( Z' \sim N(0, 1) \).
   - Since \( Z \sim N(0, 1) \), the problem asks for the expected value of the CDF of a standard normal random variable evaluated at another standard normal random variable.

2. **Symmetry of the Standard Normal Distribution:**
   - The standard normal distribution is symmetric around 0. The expected value of \( Z \), i.e., \( E(Z) \), is 0. This symmetry plays a key role in simplifying the problem.

3. **The Nature of \( \Phi(Z) \):**
   - The function \( \Phi(Z) \) gives values between \( 0 \) and \( 1 \), because it represents cumulative probabilities.
   - The probability \( \Phi(Z) \) for a normally distributed variable \( Z \) is uniformly distributed in the interval \( (0, 1) \). This result comes from the probability integral transform theorem.

4. **Expected Value of a Uniform Distribution:**
   - The expected value of a uniform random variable on \( (0, 1) \) is \( 0.5 \). Therefore, the expected value of \( \Phi(Z) \) is:
     \[
     E(\Phi(Z)) = 0.5
     \]

### Conclusion:
\[
E(\Phi(Z)) = 0.5
\]

Would you like further clarification or additional details on this? Here are 5 related questions for deeper exploration:

1. What is the intuition behind the probability integral transform theorem?
2. How does the symmetry of the normal distribution impact various calculations?
3. What is the interpretation of \( \Phi^{-1}(x) \), the inverse CDF of the standard normal distribution?
4. How would the expected value change if \( Z \sim N(\mu, \sigma^2) \) instead?
5. Can you derive the CDF of a normal distribution from first principles?

**Tip:** The cumulative distribution function (CDF) of any random variable \( X \) gives the probability that \( X \) will take a value less than or equal to a specific value. It’s especially useful when working with probabilities across a distribution.

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

Learn how to find the expected value of the cumulative distribution function (CDF) for a standard normal variable Z, where Z ~ N(0, 1). This solution uses key concepts from probability theory, including the probability integral transform theorem, to show that E(Φ(Z)) = 0.5.

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