why is the density of a standard normal random variable $z$ such that $z$ is greater than $c$ defined as it is? $$ E(z\mid z > c) = \dfrac{\varphi(c)}{1 - \Phi(c)}$$
This says the expected value of a (continuous) standard normal random variable $z$ given that $z$ is greater than some constant $c$ is given by the PDF of $z$ at $c$ divided by $1$ less the CDF of $z$ at $c$.
The numerator $\varphi(c)$ is the probability that $z$ is 'near' the constant $c$ and the denominator $1 - \Phi(c)$ is the probability that $z$ is higher than $c$.
This seems to be saying that the expected value of standard normal variable $z$ given that $z > c$ is the probability that $z$ is near $c$ relative to the probability $z$ is greater than $c.$
Is this correct? If so why is the $\Pr(z$ the bottom of its constrained distribution ${}\mid z > c)$ divided by $\Pr(z >c)$ give us the expected value of $z \mid z > c$?
 A: Suppose $A\subseteq [c,\infty).$ Then
$$
\Pr(Z\in A\mid Z\ge c) = \frac{\Pr(Z\in A)}{\Pr(Z\ge c)} = \frac{\int_A \varphi(u)\,du}{\Pr(Z\ge c)} = \int_A \frac{\varphi(u)}{1-\Phi(c)} \, du,
$$
and therefore the conditional density of $Z$ given that $Z\ge c$ is
$$
u \mapsto f(u) = \frac{\varphi(u)}{1-\Phi(c)} \text{ for } u \ge c.
$$
Therefore the conditional expected value is
\begin{align}
\int_c^\infty uf(u)\,du = \int_c^\infty u \frac{\varphi(u)}{1-\Phi(u)} \,du & = \frac 1 {1-\Phi(c)} \cdot \frac 1 {\sqrt{2\pi}} \int_c^\infty e^{-u^2/2} \Big(u \, du\Big) \\[10pt]
& = \frac 1 {1-\Phi(c)} \cdot \frac 1 {\sqrt{2\pi}} \int _{c^2/2}^\infty e^{-w}\,dw \\[10pt]
& =  \frac 1 {1-\Phi(c)} \cdot \frac 1 {\sqrt{2\pi}} e^{-c^2/2} \\[10pt]
& = \frac 1 {1-\Phi(c)} \varphi(c).
\end{align}
A: $E[Z \mid Z > c]$ is the conditional expected value of $Z$ given that $Z > c$. Obviously, $E[Z \mid Z > c]$ must be larger than $c$ because the values that you are averaging are all larger than $c$. So, you need to find $f_{Z\mid Z > c}(z\mid Z > c)$, the conditional probability density function of $Z$ given that $Z > c$ and then find the conditional expected value of $Z$ using $f_{Z\mid Z > c}(z\mid Z > c)$ as
$$E[Z \mid Z > c] = \int_{-\infty}^\infty z 
\cdot f_{Z\mid Z > c}(z\mid Z > c) \, \mathrm dz = \int_c^\infty z 
\cdot f_{Z\mid Z > c}(z\mid Z > c) \, \mathrm dz.$$
Do you know how to find $f_{Z\mid Z > c}(z\mid Z > c)$? Do you know why the second integral follows from the first (and the known properties of $f_{Z\mid Z > c}(z\mid Z > c)$?)
A: Note that $\int_c^\infty z e^{-z^2/2}\,dz = e^{-c^2/2},$ so $$\int_c^\infty z \,\phi(z)\,dz = \phi(c).$$
This, put together with the formula for the  density for $Z$ conditional on $Z\ge c$ is $$ f(z) = \begin{cases}0&z<c\\\phi(z)/(1-\Phi(c))&z\ge c\end{cases}$$ gives the desired result.
