Math Problem Statement

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

Solution

Let ZN(0,1)Z \sim N(0, 1), meaning ZZ follows the standard normal distribution with mean 00 and variance 11. We are tasked with finding the expected value E(Φ(Z))E(\Phi(Z)), where Φ()\Phi(\cdot) is the cumulative distribution function (CDF) of a standard normal random variable ZZ.

Step-by-Step Solution

  1. Understanding Φ(Z)\Phi(Z):

    • The function Φ(Z)\Phi(Z) represents the probability that a standard normal variable is less than or equal to ZZ, i.e., Φ(Z)=P(ZZ)\Phi(Z) = P(Z' \leq Z) where ZN(0,1)Z' \sim N(0, 1).
    • Since ZN(0,1)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 ZZ, i.e., E(Z)E(Z), is 0. This symmetry plays a key role in simplifying the problem.
  3. The Nature of Φ(Z)\Phi(Z):

    • The function Φ(Z)\Phi(Z) gives values between 00 and 11, because it represents cumulative probabilities.
    • The probability Φ(Z)\Phi(Z) for a normally distributed variable ZZ is uniformly distributed in the interval (0,1)(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)(0, 1) is 0.50.5. Therefore, the expected value of Φ(Z)\Phi(Z) is: E(Φ(Z))=0.5E(\Phi(Z)) = 0.5

Conclusion:

E(Φ(Z))=0.5E(\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 Φ1(x)\Phi^{-1}(x), the inverse CDF of the standard normal distribution?
  4. How would the expected value change if ZN(μ,σ2)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 XX gives the probability that XX will take a value less than or equal to a specific value. It’s especially useful when working with probabilities across a distribution.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Cumulative Distribution Function (CDF)
Standard Normal Distribution
Expected Value

Formulas

E(Φ(Z)) = 0.5

Theorems

Probability Integral Transform Theorem

Suitable Grade Level

Undergraduate - Probability and Statistics