Truncated normal random variable Find the cdf and quantile function for the truncated (at a) normal random variable given that $$\frac{\varphi(x) I_{x>a}}{1-\Phi(a)}$$ where $\varphi(x)$ is the density for standard normal and $\Phi(x)$ is the cdf for standard normal distribution. Express answers in terms of $\varphi(x)$ and $\Phi(x)$.
Appreciate your help, thank you!
 A: Perhaps the most common use of truncated normal distributions is in applied situations where the distribution has a normal shape, but negative values
are not logically possible. This can be an issue when $\mu > 0$ is less
then a few standard deviations $\sigma$, so that a substantial amount of
probability under the normal curve would be in 'negative territory'.
For example, if positive test scores have $\mu = 10$ and $\sigma = 5$. Then if
we try to model scores as $X \sim \mathsf{Norm}(\mu = 10,\, \sigma = 5),$ we have
$P(X < 0) = 0.023.$ [The computation is shown below using R statistical software,
but printed normal tables would give a similar result.]
pnorm(0, 10, 5)
## 0.02275013

It may be best to find a non-normal distribution to
model the scores more precisely (without need for adjustment), or it may be good enough to truncate the normal distribution at 0.
For a general discussion of the 'truncated normal distribution', see
the Wikipedia article.

You are truncating a standard normal distribution to consider only values
above $a$. Then, as in the answer of @Any, the cdf of the truncated random variable $T$ is
$$F_T(u) = \frac{\int_a^u \varphi(z)\,dz}{1 - \Phi(a)} = 
\frac{\Phi(u) - \Phi(a)}{1 - \Phi(a)}.$$ 
The quantile function is the inverse of the CDF.
If you want $x$ such that $F_T(u),$ then solve $x = F_T(u)$ to get $u = F_T^{-1}(x),$
remembering that $K = 1 - \Phi(a)$ is a constant. Thus,
$Kx + \Phi(a) = \Phi(u)$ and 
$$u = F_T^{-1}(x) = \Phi^{-1}(Kx + \Phi(a)),$$
where $\Phi^{-1}$ is the standard normal quantile function (inverse CDF).
Here is an example in R statistical software, in which pnorm is $\Phi$
and qnorm is $\Phi^{-1}.$  We truncate the standard normal distribution
to ignore values below -1, finding $K = 0.84134$ and $P(T < 1) = 0.81143$. Then, going in reverse as a check, we ask what $x$ on the truncated scale has probability 0.81143 below it,
that is $F_T^{-1}(.81143).$ Of course, the answer is 1. 
a = -1;  k = 1 - pnorm(a);  k
## 0.8413447                      # K
(pnorm(1) - pnorm(a))/k
## 0.8114266                      # P(T < 1)
x = .81143
qnorm(k*x+pnorm(a))
## 1.000012                       # T-quantile of x

In the graph below, the dashed black density curve is the standard normal PDF $\varphi$, the solid blue density curve is for standard normal, truncated at $-1$; it has been 'inflated' by $1/K$ so that it includes total area 1 (between
-1 and $\infty$). The area under the truncated density curve between -1 and 1 is 0.81143.
A: Since the cdf is $F_t (u)=\int_a^u\frac{\varphi(x) }{1-\Phi(a)} dx$, it can be expressed in terms of normal and hence standard normal (using the substitution $v=x-a $ to 'evaluate' the integral).
