Prove standard normal distribution conditional expectation [duplicate]

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.

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

• 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. Sep 11 '21 at 12:19
• I just edited the question. I hope it's clearer now. Sep 11 '21 at 12:26
• This answer might help. Sep 11 '21 at 12:37
• Thank you! It helped. 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.

$$\frac{1}{\sqrt{2\pi}}\int_{z}^{\infty}ue^{\frac{-u^{2}}{2}}du$$.

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.

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