# 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 Jul 2 '19 at 18:18
• Write the exact pdf instead of $\phi$ for the final integration. – StubbornAtom Jul 2 '19 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. – COOLSerdash Jul 3 '19 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. – user587389 Jul 2 '19 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 – wolfies Jul 3 '19 at 4:48