Calculate the expectation of a shift CDF

Question

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 $\frac{1}{2}$.

Answer 1

This leads to the same result as Alexeys answer. However, my reasoning is different. $$ E[F_X(X+a)]=\int_{-\infty}^{\infty} F_X(x+a) f_X(x)dx=\int_{-\infty}^{\infty} \int_{-\infty}^{x+a}f_X(y)dy f_X(x)dx=\\ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} 1_{(-\infty,x+a]}(y) f_X(y) f_X(x)dydx= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} 1_{_{\{y-x\le a\}}}(y) f_X(y) f_X(x)dydx. $$

The product of the two densities in the integral is the density of a bivariate Gaussian vector (X,Y), whose components are independent and follow a normal distribution $N(\mu,\sigma^2)$. Hence this integral is the same as

$$ E[1_{\{Y-X\le a\}}]=P[\{Y-X\le a\}], $$

where $Y,X$ are iid $N(\mu,\sigma^2)$. Thus $Y-X$ has a $N(0,2\sigma^2)$ distribution. We get

$$ E[F_X(X+a)]=F_{Y-X}(a)=\Phi\left(\frac{a}{\sqrt{2}\sigma}\right). $$

Answer 2

As you already mentioned you can get the CDF of this distribution using distribution transform:

$$P(F(X+a)\le y)=P(F^{-1}(F(X+a))\le F^{-1}(y))=P(X+a\le F^{-1}(y))=P(X\le F^{-1}(y)-a)=F(F^{-1}(y)-a)$$

Then you could write down expectation in integral form, but on the first glance it seems not so trivial how to get an explicit expression. Probably they didn't expect you to do this.

You can also try to run quick Monte-Carlo in R to see what your distribution looks like:

  a <- 1
  sigma <- 2
  m <- pnorm(rnorm(10000,0,sigma)+a,0,sigma)
  hist(m)  
  plot(ecdf(m))
  mean(m)

You will definitely see that it's not uniform for the values of $a$ that are significantly bigger than 0.

Answer 3

Hopefully this is correct...

So firstly, what is $F(X)$ when $X$ is a random variabel?

This is where I might be completely wrong, but at least it gives the correct answer in the case of a=0, so let's try that one first.

We have two random variables with the same normal distribution, mean 0 and sd $\sigma^2$. $X_1, X_2$

So we are supposed to calculate $F_{X_1}(X_2) = P(X_1 \leq X_2) = P(X_1 - X_2 \leq 0)$

Since both $X_1$ and $X_2$ are both normal we have that $X =X_1 - X_2$ is also normal with mean $0$ and s.d. $2\sigma^2$

So we have $P(X \leq 0)$ which is 1/2.

For the case of $a \neq 0$ we would get $P(X \leq a)$ which is $F_{X}(a)$. However it seems like it would always end up as a constant, so the expected value seems a bit 'redundant', which makes me think that this might not be the correct solution..

edit: Also, in $P(X \leq a)$ you could divide to get, since mean is 0, divide to get standard normal and you would get $\Phi(\frac{a}{\sqrt2sigma})$ or smth like that.

edit2:

sigma = 2;
n = 5000000;
a=5;
mean(normcdf(normrnd(0,sigma,1,n)+a,0,sigma))
normcdf(a/(sqrt(2)*sigma),0,1)