If $X$ follows standard normal distribution, find the correlation coefficient between $X$ and $\Phi(X)$

Question

If $X$ follows standard normal distribution, find the correlation coefficient between $X$ and $\Phi(X)$, where $\Phi(X)$ is the cdf of $X$.

My attempt is:

First we have to calculate $Cov(X, \Phi(X))$. Since $X$ follows standard normal distribution, $E(X)=0$. Hence, $Cov(X, \Phi(X)) = E(X\Phi(X))$.

Now, $E(X\Phi(X))$

$= \int_{-\infty}^{\infty}x\Phi(x)\phi(x)dx$ (where $\phi(x)dx$ is the pdf of $X$).

$=[\Phi(x)\left\{-\phi(x)\right\}]_{-\infty}^{\infty} -\int_{-\infty}^{\infty}\phi(x)\left\{-\phi(x)\right\}]$ (by using integration by parts and using the fact that $\int x\phi(x)dx = - \phi(x)$.

$=0+\int_{-\infty}^{\infty}(\phi(x))^2dx$

I am getting stuck here. Please anyone help me solve it. Thanks in advance.

Answer 1

You're almost there. As pointed out by @whuber, the trick is to recognise that $\phi(x)^2$ is another Gaussian that integrates to one after normalisation: \begin{align} \int \phi(x)^2 dx &= \int_{-\infty}^\infty \left(\frac1{\sqrt{2\pi}} e^{-\frac12x^2}\right)^2dx \\&= \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty \frac1{\sqrt{2\pi}} e^{-x^2}dx \\&= \frac1{\sqrt{4\pi}}\int_{-\infty}^\infty \frac1{\sqrt{2\pi}/\sqrt{2}} e^{ -\frac12(\frac x{1/\sqrt{2}})^2}dx \\&= \frac1{2\sqrt{\pi}}. \end{align}

Answer 2

If you want to check your work, it only takes a few seconds with a computer algebra system. In your case, $X \sim N(0,1)$ with pdf $f(x)$:

The cdf is:

where Erf denotes the error function.

Then the desired correlation can be found immediately with:

... where I am using the Corr function from the mathStatica package for Mathematica.

Answer 3

For a general case with $a,b$ we can derive the result of following integral, \begin{eqnarray} \int_{-\infty}^{\infty}\Phi(a+bx)^{2}\phi(x)dx\\&=&P\left(z_{1}\leq a+bx,z_{2}\leq a+bx\right)\\&=&P\left(z_{1}-bx\leq -a,z_{2}-bx\leq -a\right)\\&=& \mathcal{MVN}\left(x=\{-a,-a\},\mu=\{0,0\},\Sigma=\begin{bmatrix}b^{2}+1 & 1\\1& b^{2}+1 \end{bmatrix} \right) \end{eqnarray} In this case with $a=0, b=1$ we have, \begin{equation} \mathbb{E}(\Phi(X)^2)=\mathcal{MVN}\left(x=\{0,0\},\mu=\{0,0\},\Sigma=\begin{bmatrix}2 & 1\\1& 2 \end{bmatrix} \right) \end{equation}

After we take Jarle Tufto's result as given and having the fact that $\mathbb{V}(X)=1$, $\mathbb{V}(\Phi(X))=\mathbb{E}(\Phi(X)^2)-\mathbb{E}(\Phi(X))^2$ and $\mathbb{E}(\Phi(X))^2=\frac{1}{4}$, we then obtain final correlation formula \begin{eqnarray} \rho=\frac{\frac{1}{2\pi}}{\sqrt{1}\sqrt{\mathcal{MVN}\left(x=\{0,0\},\mu=\{0,0\},\Sigma=\begin{bmatrix}2 & 1\\1& 2 \end{bmatrix} \right)-\frac{1}{4}}} \end{eqnarray} A quick R-implementation shows,

    sqrt(3/pi)
    [1] 0.977205 

       (1/(2*sqrt(pi)))/sqrt(pmnorm(x = c(0,0), mean = rep(0.,2), 
            varcov = matrix(c(2,1,1,2), ncol=2, byrow=T))-0.25)
    [1] 0.977205

which coincides with wolfies's result.