Suppose $X$ is a normal random variable with mean 0, and variance $\sigma^2$. $F(x)$ is the CDF(cumulative distribution function) of a standard normal random variable(mean 0 and variable 1), how to calculate the expectation of $ F(X+a)$, where $ a>0 $.
This was a quant interview question. I know how to calculate the expectation of F(X), i.e when $a=0$, but I have no idea when $a \neq 0$.
My solution for $a=0$:
Method 1: Since F(x) is the CDF of a normal random variable with mean 0, and variance $\sigma^2$. We will have F(x)=1-F(-x). Suppose that f(x) is the corresponding pdf, and f(x)=f(-x). Then \begin{align} \mathbb{E}[F(X)]&=\int_{-\inf}^{+\inf} F(s)f(s)ds\\ &=\int_{-\inf}^{+\inf} (1-F(-s))f(s)ds\\ &=1-\int_{-\inf}^{+\inf} F(-s)f(s)ds\\ &=1-\int_{-\inf}^{+\inf} F(m)f(m)dm \end{align} Hence we have $$ \int_{-\inf}^{+\inf} F(s)f(s)ds=\frac{1}{2}$$
Method 2(This method seems it didn't require that X is normal random variable): Let's first compute the distribution for $F(X)$: \begin{align} \mathbb{P}\{F(X) \leq y\}&=\mathbb{P}\{X \leq F^{-1}(y)\}\\ &=F\cdot F^{-1}(y)=y \end{align} So $F(X)$ is uniformly distributed, hence the mean is $\frac{1}{2}$.