For a standard normal distribution, how to prove $$E(Z|Z\geq z)=\frac{f(z)}{1-F(z)}$$

$f(z)$ is the pdf and $F(z)$ is the cdf.

This is how I did it. But I got stuck here. Should I use integration by part? I couldn't get the result using integration by part.

$$E(Z|Z\geq z)=\frac{\int_{z}^{\infty} u f(u) du}{1-F(z)}$$

  • $\begingroup$ Take a look at How to ask a good question at Math.SE. To avoid downvotes and closing you should add your own efforts to the question by means of an edit (not a comment), and tell us where you got stuck. $\endgroup$
    – drhab
    Sep 11 '21 at 12:19
  • $\begingroup$ I just edited the question. I hope it's clearer now. $\endgroup$ Sep 11 '21 at 12:26
  • $\begingroup$ This answer might help. $\endgroup$
    – drhab
    Sep 11 '21 at 12:37
  • $\begingroup$ Thank you! It helped. $\endgroup$ Sep 11 '21 at 12:47

So if I understand correctly, you are having trouble in computing the integral right?. Otherwise your steps seem fine to me.


substitute $u^{2}=t$ You get.

$$\frac{1}{2\sqrt{2\pi}}\int_{z^{2}}^{\infty}e^{\frac{-t}{2}}dt=\frac{1}{\sqrt{2\pi}} e^{\frac{-z^{2}}{2}}=f(z)$$.

So you have the required answer $$\frac{f(z)}{1-F(z)}$$

(Procedure to derive the integral expression):-

$$\mathbb{E}(Z|Z\geq z) = \frac{\mathbb{E}(Z\cdot I_{\{Z\geq z\}})}{\mathbb{P}(Z\geq z)}$$

Where $I_{A}$ is the random variable such that $I_{A}=1$ if $Z(\omega)\in A$ and $I_{A}=0$ if $Z(\omega)\in \mathbb{R-}A$ .

$I_{A}$ is most commonly known as the indicator random variable.

Here $A=[z,\infty)$

So $\int_{-\infty}^{\infty}z.I_{\{Z\geq z\}}f(z)dz=\int_{z}^{\infty}zf(z)dz$.

And $\mathbb{P}(Z\geq z) = 1- F(z)$

Now it is the same as I had done before.

  • $\begingroup$ Thank you. It helped! $\endgroup$ Sep 11 '21 at 12:49
  • $\begingroup$ @crazystats Consider upvoting and marking as answer if a particular answer helped you $\endgroup$ Sep 11 '21 at 12:49
  • $\begingroup$ I marked it as the answer. I'm having trouble with $E(Z^2|Z>=z)$ now, do you have any idea? $\endgroup$ Sep 11 '21 at 13:10
  • $\begingroup$ @kellyohhhh . You will not get a closed form expression . Sorry for my initial comment. I thought you wanted $E(Z|Z^{2}\leq z)$ $\endgroup$ Sep 11 '21 at 17:30
  • $\begingroup$ No No it's fine. Your solution helped. $\endgroup$ Sep 11 '21 at 23:54

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