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I need to show that $E\{ZΦ(Z)\} = 1 / \left( 2\sqrt{\pi} \right)$. Let $Z$ be a standard normal random variable with density $ϕ$ and distribution function $Φ$

I don't know how to start.

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    $\begingroup$ Welcome to CrossValidated! Is this a homework question? If so, please add the self-study tag to your post and read over its wiki page. We welcome homework questions here but treat them somewhat differently. $\endgroup$ Dec 2 '16 at 15:11
  • $\begingroup$ Write out the expectation explicitly, then think about using integration by parts to evaluate it. $\endgroup$ Dec 2 '16 at 15:33
  • $\begingroup$ Well, this is the problem. I know that E(x) = Integral(x*f(x))dx. In my case x = ZΦ(Z),, what about f(ZΦ(Z)0 ? $\endgroup$
    – May
    Dec 2 '16 at 15:38
  • $\begingroup$ In your equation, $f$ is the density of $x$. As you point out, it's not obvious what the density of $Z\Phi (Z)$ is, so I'd suggest using the alternative equation $E(g(Z))=\int g(z) f(z) dz$ where $g(z)=z\Phi (z)$ $\endgroup$ Dec 2 '16 at 15:48
  • $\begingroup$ ... and $f$ is the density of $Z$. $\endgroup$ Dec 2 '16 at 15:50
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This is S. Catterall's hint: \begin{align*} E\{ZΦ(Z)\} &= \int_{\mathbb{R}} z \Phi(z) \phi(z) dz. \end{align*} And an extra hint: let $u=\Phi(z)$ and $v' = z\phi(z)$.

One more hint: $v = \int_{-\infty}^z \phi(s)sds = - \frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}$.

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    $\begingroup$ $z$ should go into $v'$ rather than $u$ $\endgroup$ Dec 2 '16 at 16:27
  • $\begingroup$ and v = ∫zϕ(z)dz . It is expected value of z? which is zero $\endgroup$
    – May
    Dec 2 '16 at 16:35
  • $\begingroup$ The expected value of $Z$ is the integral of $v'$ over the whole real line (a so-called definite integral). What you're looking for is the 'indefinite integral' $\int_{-\infty}^z v'(t) dt$. In other words you need to find a function that differentiates to give $v'$. $\endgroup$ Dec 2 '16 at 17:14
  • $\begingroup$ My v = -ϕ(z). Is it correct? $\endgroup$
    – May
    Dec 2 '16 at 17:42
  • $\begingroup$ When you take the derivative of that is it equal to $\phi(z)z$? Try using the $v(z)$ @S.Catterall gave you $\endgroup$
    – Taylor
    Dec 2 '16 at 17:44
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I'm going to offer an alternative that uses a special property of the Gaussian distribution - but it really comes from applying integration by parts, let $f$ be a function (I won't say much about what this function class is) then it is the case that if $Z$ is a standard Gaussian random variable we have $$ \begin{align} \mathbb{E} \left[ Z f(Z) \right] = \mathbb{E} \left[ f^{\prime}(Z)\right], \end{align} $$ so applying that to this problem we get $$ \begin{align} \mathbb{E}\left[ Z \Phi(Z) \right] &= \mathbb{E}\left[ \phi(Z) \right] \\ &= \frac{1}{2\pi} \int e^{-z^2}dz \\ &= \frac{\sqrt{\pi}}{2\pi} = \frac{1}{2\sqrt{\pi}}. \end{align} $$ The identity used above is the starting point of what is called Stein's method in probability and statistics.

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  • $\begingroup$ Thank you for showing your method. Does Stein's method use to solve E[Z^2Φ(Z)] ? $\endgroup$
    – May
    Dec 2 '16 at 22:59
  • $\begingroup$ You are very welcome, Stein's method is really a way of showing that if $\mathbb{E} f^{\prime}(Z) - Z f(Z) = 0$ then the distribution is a standard normal. Therefore it is not going to be immediately applicable in this case. As I mentioned there really is integration by parts going on to derive this result and so you would do well to tackle it directly using the method suggested in other answers - this is just a fun result that I thought I would share! $\endgroup$
    – Nadiels
    Dec 2 '16 at 23:03
  • $\begingroup$ A bit more reflection and of course it still applies in this case! Using the same result we just say $\mathbb{E} \left[ x^2 \Phi(x) \right] = \mathbb{E}\left[ x (x \Phi(x) )^{\prime} \right] = \mathbb{E} \left[ \Phi(x) \right] + \mathbb{E}\left[ x \phi(x) \right] = \mathbb{E}\left[ \Phi(x) \right] = 1/2$. $\endgroup$
    – Nadiels
    Dec 4 '16 at 12:47
  • $\begingroup$ Little mistake that should read $\mathbb{E} \left[ x(x\Phi(x)) \right] = \mathbb{E}\left[ (x\Phi(x))^{\prime} \right]$, and that will work with any monomial in the same way $\endgroup$
    – Nadiels
    Dec 4 '16 at 12:49

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