# Expectation of squared normal cdf

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!

• This is probably true. Expressions like that are ubiquitous in the literature using the Gaussian copula. Just google. Commented Dec 8, 2023 at 10:41
• I tried, but was unable to find something that allows me to fully prove this relationship analytically.
– BMBE
Commented Dec 8, 2023 at 10:49
• Looking more carefully. Are you sure you did not prove numerically $$E[p(Y)] = \Phi_2\Big(\Phi^{-1}(p),\Phi^{-1}(p),\rho\Big)\,?$$ (No square around $p(Y)\,.$) It might also be a good idea not to use the letter $p$ for a fixed parameter and for a function. Plus the similarity to the correlation $\rho$ is not what I would have chosen. Commented Dec 8, 2023 at 10:57
• No, I have checked again the code, and it is precisely (I changed the notation following your suggestion): $$E[f(Y)^2] = \Phi_2\Big(\Phi^{-1}(a),\Phi^{-1}(a),b\Big)$$
– BMBE
Commented Dec 8, 2023 at 11:17

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 $$\mathbb{E}(f^2(Y)) =\Phi\left(\pmatrix{\Phi^{-1}(a)\\\Phi^{-1}(a)}; \pmatrix{0\\0},\Sigma \right) =\Phi_2\Big(\Phi^{-1}(a),\Phi^{-1}(a),b\Big)$$

• You mean: 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?
– BMBE
Commented Dec 8, 2023 at 13:44
• @BMBE the typo (sum <-> product) was corrected. Yes, and this statement holds true also for expectation (see theorem 5, page 4).
– NN2
Commented Dec 8, 2023 at 14:42
• NN2: Yes, then I agree, as the "sum property" derives from the linearity of the expectation operator, nothing to do with independence, it is the product that can be decomposed that way thanks to the independence property. Thank you so much for your help !
– BMBE
Commented Dec 8, 2023 at 15:19
• @BMBE You're welcome!
– NN2
Commented Dec 8, 2023 at 15:30
• May I ask you another question? to prove that the pair $(V_1,V_2)$ is standard bivariate normal you simply used the fact that any vector $(V_1,V_2)$ of standard normal random variables, follows a standard bivariate normal distribution if and only if the pair $V_1$ and $V_2$ can be written as $\mu + AW$, where $\mu$ is the vector of means, $A$ is a matrix of constants such that $AA^T = \Sigma$, and $W$ is the following vector of standard normal iid random variables $W := Z_1,Z_2,Y$ ?
– BMBE
Commented Dec 8, 2023 at 19:03