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.

  • 2
    $\begingroup$ 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

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

  • 1
    $\begingroup$ $z$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.