Normal Random Variable less than Uniform If X is a normally distributed RV and Y is a uniform RV on a to b, is there a distribution of X given it is less than Y?
Intuitively it would be similar to the density of a truncated RV truncated at b, but with slightly less mass in the region a to b than usual. 
Is there a way to express this distribution as a function of Phi functions (standard normals) if it exists?
Cheers
Angus
 A: Given $X \sim \mathcal{N}(\mu, \sigma^2), Y \sim \mathcal{U}(a, b)$ and they are independent, consider the CDF of $X|X < Y$:
$$ \begin{align} F_{X|X<Y}(x) &= \Pr\{X \leq x|X < Y\} \\
&= \frac {\Pr\{X \leq x, X < Y\}} {\Pr\{X < Y\}} \\
&= \frac {\displaystyle \frac {1} {b - a} \int_a^b\Pr\{X \leq x, X < y\}dy} 
{\displaystyle \frac {1} {b - a}  \int_a^b \Pr\{X < y\}dy}  \\
\end{align}$$
Consider the numerator. The split the range of $x$ into $3$ cases:
When $x \leq a$,
$$ \int_a^b\Pr\{X \leq x, X < y\}dy = \int_a^b\Pr\{X \leq x\}dy = (b - a)\Phi\left(\frac {x - \mu} {\sigma}\right)$$
When $x \geq b$,
$$ \int_a^b\Pr\{X \leq x, X < y\}dy = \int_a^b\Pr\{X < y\}dy = \int_a^b \Phi\left(\frac {y - \mu} {\sigma}\right)dy$$
When $a < x < b$,
$$ \begin{align} \int_a^b\Pr\{X \leq x, X < y\}dy &= \int_a^x\Pr\{X < y\}dy + \int_x^b\Pr\{X < x\}dy \\
& = \int_a^x \Phi\left(\frac {y - \mu} {\sigma}\right)dy + (b - x)\Phi\left(\frac {x - \mu} {\sigma}\right) \end{align}$$
So obviously for the case $x \geq b$, we have $F_{X|X<Y}(x) = 1$ (the denominator. And there is not much to simplify for the remaining. You may also try to differentiate and obtain the pdf.
A: Comment: I have not checked the details, but it seems @BGM's Answer 
has a method for an analytic solution to your Problem. (+1)
But in your question, you seemed to be
searching for an intuitive view of what is happening. Below is a histogram for 
a million values of $Z = X|X < Y,$ where $X$ is standard normal and
$Y \sim \mathsf{Unif}(0,.5).$ [I chose this particular case because it ought to be easy to plug numbers into analytic answers to see what happens.] 
The histogram below suggests that the normal is not sharply truncated, but
perhaps one might say 'gradually nearly-linearly truncated' between 0 and .5.
Green lines indicate the interval $(0,.5).$ The red curve is the standard
normal density for negative values. Of course, the total area under the
histogram approximates 1.

Skeletal R code, if you want it:
m = 10^6;  x = rnorm(m);  y = runif(m, 0,.5)
z = x[x < y];  hist(z, prob=T, br=100, col="skyblue")

A: Here is some R code I wrote to calculate the density. It uses the truncnorm package in R. Plotting this over the hist suggested by BruceET seems to confirm it is correct. 
dens<-function(x,a,b,mean,sd)
  {
  #first find the normalising cosntant.
  etruncab<-etruncnorm(a=a,b=b,mean=mean,sd=sd)
  normc<-1/(1-(1-ptruncnorm(a,-Inf,b,mean,sd))*punif(etruncab,a,b)) 

  out<-numeric(length=length(x))
  out<- dtruncnorm(x=x,a=-Inf,b=b,mean=mean,sd=sd)*normc
  out[x>a]<-out[x>a]*(1-punif(x[x>a],a,b))

  out
}

