Expectation of normal c.d.f. of a normal random variable

Question

Let $a$ be a constant, X is a standard normal and $\Phi$ is the c.d.f. of standard normal.

1. what is the $$E[\Phi(aX)]$$
2. what is the $$E[\Phi(X + a])]$$

my approach would be to derive the density of $g(X) = Y = aX$ and of $g(X)=Y=X+a$ (which should result in $1/a$ and $1$, respectively. Please let me know if this sounds correct.)

Is there a faster way? I found the below answer but do not take into account the role of the constant. Once we have a constant we can't cancel out the two c.d.f.

In my approach i get to $$\Phi\large(g^{-1}(x)\large)$$ and then using chain rule to arrive to the density. Results that i get doing this are indeed $1/a$ and $1$.

https://math.stackexchange.com/questions/1331093/expectation-of-cumulative-distribution-function-of-a-standard-normal-distributed#:~:text=Let%20X%20be%20a%20normally,√2πdt.

Answer 1

Here is another way:

By writing down the integral you can notice that $$\mathbb{E}[\Phi(X+a)] = P(Y \leq Z + a)$$ where $Y, Z \sim \mathcal{N}(0,1)$ are independent. Then $Y-Z\sim \mathcal{N}(0, 2)$ so $$\mathbb{E}[\Phi(X+a)] = P(Y - Z \leq a) = \Phi\left(\frac{a}{\sqrt{2}}\right)$$

The case of $\mathbb{E}[aX]$ is even easier! As before, notice that $$\mathbb{E}[\Phi(aX)] = P\left(Y \leq aZ\right)$$ with $Y,Z \sim \mathcal{N}(0,1)$ independent. But $Y-aZ \sim \mathcal{N}(0,1+a^2)$ and from there you get $$\mathbb{E}[\Phi(aX)] = P(Y-aZ \leq 0) = \Phi(0) = \frac{1}{2}$$

EDIT: Here is a more detailed calculation:

Let $\phi(x)$ be the CDF of a standard normal RV. Then LOTUS tells us that $$\mathbb{E}[\Phi(X+a)]=\int_{-\infty}^\infty \Phi(x+a)\phi(x)dx$$ By definition, $$\Phi(x) = \int_{-\infty}^x \phi(t)dt$$ so $$\mathbb{E}[\Phi(X+a)]=\int_{-\infty}^\infty \int_{-\infty}^{x+a}\phi(t)\phi(x)\,dtdx$$ which is exactly what you would get if you were to calculate the probability $P(Y\leq Z+a)$ (up to renaming of the dummy integration variables).

Answer 2

Let's consider the second case, namely $Y = g(X) = \Phi(X+a)$. You may indeed determine the new density function from the usual change of variable, so that $$ f_Y(y) = f_X(g^{-1}(y))\left|\frac{\partial g^{-1}(y)}{\partial y}\right|, $$ where the last factor is nothing else than the Jacobian resulting from the chain rule. In the present case, computations promise to be tedious because of the form of $g$.

Instead, if I were you, I would "cheat" by trying to produce a differential equation with respect to the parameter $a$, as done below : $$ \begin{align} \frac{\partial}{\partial a} \mathbb{E}[\Phi(X+a)] &= \mathbb{E}\left[\frac{\partial}{\partial a}\Phi(X+a)\right] \\ &= \mathbb{E}[\phi(X+a)] \\ &= \int_\mathbb{R} \phi(x)\phi(x+a) \,\mathrm{d}x \\ &= \int_\mathbb{R} \frac{\mathrm{d}x}{2\pi} e^{-\frac{1}{2}x^2}e^{-\frac{1}{2}(x+a)^2} \\ &= \int_\mathbb{R} \frac{\mathrm{d}x}{2\pi} e^{-(x^2+ax+\frac{a^2}{2})} \\ &= \frac{e^{-\frac{1}{4}a^2}}{2\sqrt{\pi}} \end{align} $$ hence, after integration (by separation of the variables), $$ \mathbb{E}[\Phi(X+a)] = \int_{-\infty}^a \frac{\mathrm{d}\alpha}{2\sqrt{\pi}} e^{-\frac{1}{4}\alpha^2} = \int_{-\infty}^{a/\sqrt{2}} \frac{\mathrm{d}t}{\sqrt{2\pi}} e^{-\frac{1}{2}t^2} = \Phi\left(\frac{a}{\sqrt{2}}\right), $$ where we used the boundary condition $\mathbb{E}[\Phi(X+a)]|_{a=-\infty} = 0$, since $\Phi(-\infty) = 0$.

Of course, you may proceed in the same way with the first case, i.e. $Y = \Phi(aX)$. You will end up with the following expression : $$ \frac{\partial}{\partial a} \mathbb{E}[\Phi(aX)] = \mathbb{E}[X\phi(aX)] = \int_\mathbb{R} \frac{\mathrm{d}{x}}{\sqrt{2\pi}} x\phi(x)\phi(ax) \,\mathrm{d}x = \int_\mathbb{R} \frac{\mathrm{d}{x}}{2\pi} xe^{-\frac{1}{2}(1+a^2)x^2} = 0, $$ because the integrand is odd. In consequence, $\mathbb{E}[\Phi(aX)]$ is constant with respect to the parameter $a$, hence $$ \mathbb{E}[\Phi(aX)] = \mathbb{E}[\Phi(aX)]|_{a=0} = \mathbb{E}[\Phi(0)] = \frac{1}{2}. $$