Question

If $r \cdot v X \sim N(0, 1)$ then $E\left(\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{X} e^{-z^2/2} dz\right)$ equals to

• A. $\infty$
• B. $0$
• C. $1/2$
• D. $1$

Hint:

For any continuous random variable, the expected value of its own CDF is always $1/2$. This is because the CDF transforms the variable into a $U(0,1)$ distribution, which has a mean of $0.5$.

Answer 1

The Correct Option is C

This problem asks for the expected value of the Cumulative Distribution Function (CDF) of a standard normal variable, where the argument of the CDF is the variable itself.

Step 1: \color{redIdentifying the Function
The expression $\Phi(X) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{X} e^{-z^2/2} dz$ is the CDF of the standard normal distribution evaluated at $X$.
We are asked to find $E[\Phi(X)]$.

Step 2: \color{redApplying the Probability Integral Transform
The Probability Integral Transform theorem states that if $X$ is a continuous random variable with CDF $F_X$, then the random variable $U = F_X(X)$ follows a Uniform distribution on the interval $(0, 1)$.
In this case, $X \sim N(0, 1)$ and $\Phi$ is its CDF. Therefore:
$U = \Phi(X) \sim U(0, 1)$

Step 3: \color{redCalculating the Expectation of U
We need to find $E(U)$. For a uniform distribution $U(a, b)$, the mean is $\frac{a+b}{2}$.
For $U(0, 1)$:
$E(U) = \frac{0 + 1}{2} = \frac{1}{2}$
Thus, $E[\Phi(X)] = 1/2$.