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