Calculate $E[\Phi(aX+b)^n]$

Question

I am working on a project, where this expected value shows up again and again, where $X \sim N(0,1)$ and $\Phi$ is the cdf of a standard normal distribution.
$$E[\Phi(aX+b)^n]$$ However, I have not yet been able to solve it.

I am aware of the solution if $n=1$, given for instance in the answer to this post.
I am also aware that if $a=1$ and $b=0$, then $\Phi(X)$ would be uniformly distributed and it would then be easy to find.

Thank you very much for your time and help!

Answer 1

Let us denote $(Y_i)_{i=1,..,n}$ the $n$ i.i.d random variables following the normal distribution $\mathcal{N}(0,1)$ and are independent to $X$. We have $$\begin{align} \mathbb{E}(\Phi^n(aX+b))& = \mathbb{E}\left(\prod_{i=1}^n \mathbb{P}(Y_i\le aX+b|X) \right) \\ &= \mathbb{E}\left(\prod_{i=1}^n \mathbb{E}(\mathbf{1}_{\{ Y_i\le aX+b \}}|X) \right)\tag{1} \end{align}$$

We know that, if $Y, Z$ are independent to each other, then $$ \mathbb{E}(Y)\cdot \mathbb{E}(Z) = \mathbb{E}(YZ) \tag{2}$$

Given $X$, these random variables $\left(\mathbf{1}_{\{ Y_i\le aX+b \}}|X \right)_{i=1,..,n}$ are independent to each other. Then, we apply $(2)$ to $(1)$: $$\begin{align} \mathbb{E}(\Phi^n(aX+b))&= \mathbb{E}\left(\prod_{i=1}^n \mathbb{E}(\mathbf{1}_{\{ Y_i\le aX+b \}}|X) \right) \\ &= \mathbb{E}\left( \mathbb{E}\left(\left.\prod_{i=1}^n\mathbf{1}_{\{ Y_i\le aX+b \}}\right|X\right) \right) \\ &= \mathbb{E}\left( \mathbb{E}\left(\left.\prod_{i=1}^n\mathbf{1}_{\{ Y_i-aX\le b \}}\right|X\right) \right) \\ &= \mathbb{E}\left( \prod_{i=1}^n\mathbf{1}_{\{ Y_i-aX\le b \}} \right) \\ &= \mathbb{P}\left(\bigcap_{i=1}^n\{ Y_i-aX\le b \}\right) \tag{3} \end{align}$$

Let us denote $Z_i = Y_i -aX$ for $i =1,..,n$. Then $(Z_i)_{i=1,...,n}$ follows the multivariate normal distribution $\mathcal{N}_n(\mathbf{0}_n;\Sigma)$ avec $$\Sigma_{ij} = \text{Cov}(Z_i,Z_j) = \left\{ \begin{array}{ll} a^2+1 & \mbox{if } i=j \\ a^2 & \mbox{if } i\ne j \\ \end{array} \right. \tag{4} $$ From $(3)$, we deduce that

$$\color{red}{\mathbb{E}(\Phi^n(aX+b)) = \Phi_n \left(\mathbf{l}, \mathbf{u};\mathbf{0}_n;\Sigma \right)}$$ with

• $\Phi_n() $ the multivariate normal probability
• $\mathbf{l} \in \mathbb{R}^n$ vector with all elements are equal to $-\infty$
• $\mathbf{u} \in \mathbb{R}^n$ vector with all elements are equal to $b$
• $\mathbf{0}_n \in \mathbb{R}^n$ vector of $0$
• $\Sigma$ is defined in $(4)$