Let's assume a non-linear, strictly decreasing function of a random variable $Y\sim N(0,1)$ of the form :
$$f(Y) = \Phi\left(\frac{\Phi^{-1}(a)-\sqrt{b} Y}{\sqrt{1-b}}\right)$$
where $\Phi(\cdot)$ denotes the standard normal cumulative distribution function, $a$ is a probability and $b$ is a real number between 0 and 1.
I was able to prove numerically, in R, that the expectation of its square converges to a bivariate standard normal distribution with the following inputs:
$$E[f(Y)^2] = \Phi_2\Big(\Phi^{-1}(a),\Phi^{-1}(a),b\Big)$$
Hence, it is like if the expectation of a squared standard normal cdf converges to a bivariate standard normal cdf. I have a "flair" about why this should be the case. However, I would like to prove this relationship analytically, the first step should be something like:
$$E[f(Y)^2] = E\left[\Phi\left(\frac{\Phi^{-1}(a)-\sqrt{b} Y}{\sqrt{1-b}}\right) \times \Phi\left(\frac{\Phi^{-1}(a)-\sqrt{b} Y}{\sqrt{1-b}}\right)\right]$$
Any help would be much appreciated!
Best Answer
We will use the conditional expectation to prove $$\mathbb{E}(f^2(Y)) = \Phi_2\Big(\Phi^{-1}(a),\Phi^{-1}(a),b\Big)$$
We have:
$$\begin{align} \mathbb{E}(f^2(Y))&= \mathbb{E}\left(\Phi\left(\frac{\Phi^{-1}(a)-\sqrt{b} Y}{\sqrt{1-b}}\right)\cdot\Phi\left(\frac{\Phi^{-1}(a)-\sqrt{b} Y}{\sqrt{1-b}}\right)\right) \\ &=\mathbb{E}\left(\mathbb{P}\left(\left. Z_1\le\frac{\Phi^{-1}(a)-\sqrt{b} Y}{\sqrt{1-b}}\right|Y\right)\cdot\mathbb{P}\left(\left.Z_2\le\frac{\Phi^{-1}(a)-\sqrt{b} Y}{\sqrt{1-b}}\right|Y\right)\right) \\ &=\mathbb{E}\left(\mathbb{E}\left(\left.\mathbf{1}_{\left\{Z_1\le\frac{\Phi^{-1}(a)-\sqrt{b} Y}{\sqrt{1-b}}\right\}}\right|Y\right)\cdot\mathbb{E}\left(\left.\mathbf{1}_{\left\{Z_2\le\frac{\Phi^{-1}(a)-\sqrt{b} Y}{\sqrt{1-b}}\right\}}\right|Y\right)\right) \end{align}$$
with $Z_1$ and $Z_2$ following the standard normal distribution $\mathcal{N}(0,1)$ and being independent with each other and independent to $Y$.
Conditional on $Y$, the two indicator functions are independent, so the product of their conditional expectations is equal to the conditional expectation of their product.
$$\begin{align} \mathbb{E}(f^2(Y)) &=\mathbb{E}\left(\mathbb{E}\left(\left.\mathbf{1}_{\left\{Z_1\le\frac{\Phi^{-1}(a)-\sqrt{b} Y}{\sqrt{1-b}}\right\}}\cdot \mathbf{1}_{\left\{Z_2\le\frac{\Phi^{-1}(a)-\sqrt{b} Y}{\sqrt{1-b}}\right\}}\right|Y\right)\right) \\ &=\mathbb{E}\left(\mathbf{1}_{\left\{Z_1\le\frac{\Phi^{-1}(a)-\sqrt{b} Y}{\sqrt{1-b}}\right\}}\cdot \mathbf{1}_{\left\{Z_2\le\frac{\Phi^{-1}(a)-\sqrt{b} Y}{\sqrt{1-b}}\right\}}\right) \\ &=\mathbb{E}\left(\mathbf{1}_{\left\{\sqrt{1-b} Z_1 +\sqrt{b} Y \le \Phi^{-1}(a)\right\}}\cdot \mathbf{1}_{\left\{\sqrt{1-b} Z_2 +\sqrt{b} Y \le \Phi^{-1}(a)\right\}}\right) \\ &=\mathbb{P}\left({\left\{\underbrace{\sqrt{1-b} Z_1 +\sqrt{b} Y}_{=: V_1} \le \Phi^{-1}(a)\right\}}\cap {\left\{\underbrace{\sqrt{1-b} Z_2 +\sqrt{b} Y}_{=: V_2} \le \Phi^{-1}(a)\right\}}\right) \\ \end{align}$$
It's easy to verify that $(V_1,V_2)$ follows a bivariate normal distribution with zero mean and covariance matrix $\Sigma= \pmatrix{1&b\\b&1}$, by consequence $$\mathbb{E}(f^2(Y)) =\Phi\left(\pmatrix{\Phi^{-1}(a)\\\Phi^{-1}(a)}; \pmatrix{0\\0},\Sigma \right) =\Phi_2\Big(\Phi^{-1}(a),\Phi^{-1}(a),b\Big)$$