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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.

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    $\begingroup$ $\phi(x)^2$ is proportional to $\phi(x\sqrt{2})$ which is proportional to the density of $X\sqrt{2}.$ $\endgroup$ – whuber Jul 2 '19 at 18:18
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    $\begingroup$ Write the exact pdf instead of $\phi$ for the final integration. $\endgroup$ – StubbornAtom Jul 2 '19 at 19:24
  • $\begingroup$ 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. $\endgroup$ – COOLSerdash Jul 3 '19 at 6:57
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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}

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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 $f(x)$:

enter image description here

The cdf is:

enter image description here

where Erf denotes the error function.

Then the desired correlation can be found immediately with:

enter image description here

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

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    $\begingroup$ 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. $\endgroup$ – user587389 Jul 2 '19 at 22:08
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    $\begingroup$ 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 $\endgroup$ – wolfies Jul 3 '19 at 4:48

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