To solve this question, we need to understand the components involved:
- The expression given inside the expectation, \(\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{x} e^{-z^2/2} \, \mathrm{dz}\), is the cumulative distribution function (CDF) of the standard normal distribution, which is commonly denoted as \(\Phi(x)\).
- The problem states that \(r \cdot v \, X \sim N(0, 1)\). Here, \(X\) represents a standard normally distributed random variable, meaning it follows \(\mathcal{N}(0, 1)\).
- Since \(X\) is standard normal, \(\Phi(x)\) represents the probability that a normal random variable is less than or equal to \(x\). For a standard normal distribution \(X\), the mean is 0.
- Therefore, the expectation of the CDF at \(x = 0\) is what we are interested in calculating.
Now, let's compute this:
- The CDF of a standard normal distribution at \(x = 0\) is \(\Phi(0)\).
- By definition of the standard normal distribution, \(\Phi(0) = 0.5\), because the standard normal distribution is symmetric around 0, and the area to the left of zero is exactly half.
Hence, the expected value \(E\left( \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{x} e^{-z^2/2} \mathrm{dz} \right)\) when \(x = 0\) is:
Therefore, the correct answer is \(\frac{1}{2}\).
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