I am trying to find $E[X | X> 0]$, and in doing so, I know (according to the book solution below) I should integrate $$ \int_{0}^{\infty} x N(0,1) dx $$
where you observe the lower limit is 0. But is this suggesting that the conditional pdf of $X|X>0$ is just the right half of $N(0,1)$? But this doesn't seem to make sense because this pdf only integrates to 0.5.
Generally, I know that
$$ E[X | Y = y] = \int_{X_{\min}}^{X_{\max}} x f_{X|Y=y}(x)dx $$
So I am trying to find how this general definition relates to the one above.
Book solution:
The requested density is
$$\frac{1}{1-\Phi_X(0)}\phi_X(x)=\frac{1}{\frac{1}{2}}\phi(x)=2\phi(x)$$
...the integral to calculate the conditional expectation is trivial
Further details of the explanation.
As per definition, with continuous distributions
$$F(x|x>0)=\frac{\mathbb{P}[X \leq x;X>0]}{\mathbb{P}[X>0]}=\frac{F_X(x)-F_X(0)}{1-F_X(0)}$$
Derivating the conditional CDF you find the conditional density...
$$f(x|x>0)=\frac{1}{1-F(0)}f(x)$$
As in the standard normal law $1-F(0)=\frac{1}{2}$ the conditional density is
$$f(x|x>0)=\sqrt{\frac{2}{\pi}}e^{-\frac{x^2}{2}}$$
$x>0$
In order to calculate the expectation you have to solve
$$\mathbb{E}=\sqrt{\frac{2}{\pi}}\underbrace{\int_{0}^{+\infty}{xe^{-\frac{x^2}{2}}dx}}_{=1}=\sqrt{\frac{2}{\pi}}$$
the integral is 1 without any calculation because the integrand is a Rayleigh distribution. If you want to calculate it....it is trivial. In fact
$$\int_{0}^{+\infty}{xe^{-\frac{x^2}{2}}dx}=-e^{-\frac{x^2}{2}}\Bigg|_{0}^{+\infty}=1$$
...that's all