Expectation of a censored variable. My problem is to find $E(Y)$, if $Y=max(X,k)$, for $X \thicksim N(,²)$ and $k \in ℝ$.
I've come across $ \int_k^{\infty} x. \phi(x) dx$, where $\phi(x)$ is the pmf of the normal distribution.
Can I find a closed form solution for $E(Y)$?
 A: First observe that
$$\int_k^{\infty}x \phi(x)dx$$
is not your $E(Y)$ (but sure you already know this)

Example: $X\sim N(0;1)$ and $k=4$
Without any calculation, $E(Y)\approx 4$ because the probability  $\mathbb{P}(X>4)\approx 0$

If the problem is only to evaluate the integral, it is very easy to answer:
$$\int_k^{\infty}x \phi(x)dx=\int_k^{\infty}x \frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/(2\sigma^2)}dx$$
Set $\frac{x-\mu}{\sigma}=t$ and get
$$\int_{h}^{\infty}(\sigma t+\mu)\frac{1}{\sqrt{2\pi}}e^{-t^2/2}dt$$
Now split the integral in two separate integrals, one is quite immediate and the other one can be expressed in funciton  of $\Phi(t)$ the CDF of the Standard Gaussian, tabulated everywhere, on paper tables as given by any calculator

Notes:

*

*Following what you wrote, I used $\phi(x)$ indicating the pfd of a Gaussian rv but it is used to indicate with $\phi(x)$ the density of a STANDARD gaussian...


*The normal rv does not hav a pmf: it has a pdf; pmf is for discrete rv's


*your $Y=max(X,k)$ for a fixed $k$, is not a censored rv but it is a "mixed" rv.
