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

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.

• $\phi(x)^2$ is proportional to $\phi(x\sqrt{2})$ which is proportional to the density of $X\sqrt{2}.$
– whuber
Commented Jul 2, 2019 at 18:18
• Write the exact pdf instead of $\phi$ for the final integration. Commented Jul 2, 2019 at 19:24
• Here you'll find the general solution to the last integral with steps. Also, the last integral is a variant of the Gaussian integral. The Wikipedia page has several derivations. Commented Jul 3, 2019 at 6:57

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}

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.

• Okay, thanks. But I don't just want to check the answer, I want some hint how to proceed after the last step I have done. Commented Jul 2, 2019 at 22:08
• Then your question is not about finding the correlation coefficient per se, but really about how to solve a particular integral. If so, that question might be better placed at math.se Commented Jul 3, 2019 at 4:48

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, $$$$\mathbb{E}(\Phi(X)^2)=\mathcal{MVN}\left(x=\{0,0\},\mu=\{0,0\},\Sigma=\begin{bmatrix}2 & 1\\1& 2 \end{bmatrix} \right)$$$$

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.