Cumulative probability of minimum function Let $X$ be an exponential random variable such that $F_{X}(x) = 1 - e^{-x/\tau}$. What is the cdf of $Y=\min(X, \tau)$ where $x \ge 0$? I think the reasoning is something like this: 
$F_Y(y) = P(\min(X, \tau) \le y) = 1 - P(X>y, y>\tau)$. Since $\tau$ is a parameter for the distribution, not a random variable, I don't know how to work around with $P(X > Y, y > \tau)$.
 A: I think you have some inequalities reversed.
Let $X \sim \mathsf{EXP}(rate = 1/\tau) = \mathsf{EXP}(mean=\tau).$ 
Then $Y$ is just $X$ right-censored
at time $\tau.$ You might want to google 'right censoring', and look
at the Wikipedia article among others.
The distribution of $Y$ is a mixed distribution (partly continuous and partly discrete) with
$P(Y \le x) = P(X \le x),$ for $x < \tau,$ and 
$P(Y = \tau) = P(X > \tau) = e^{-\tau/\tau} = e^{-1}.$ 
A graph of the CDF is shown below for the case $\tau = 5.$ The vertical
red line is at $y = 5$ and the horizontal green line at $1 - e^{-1}.$

Note: The following simulation in R statistical software is an easy way to make the plot.
mu = 5;  x = rexp(10^4, 1/mu);  y = pmin(x, mu)
plot(ecdf(y), lwd=2, ylab="CDF", main="CDF of EXP(rate = 1/5) Right-Censored at 5")
abline(v=5, col="red");  abline(h=1-exp(-1), col="green2")

A: $Y  = \min(X,\tau)$ is a truncated exponential variable truncated to the right at $\tau$.
$Y = X,\, X<\tau$
$Y = \tau,\, X \ge \tau$
This translates into a truncated exponential distribution.
Suppose that $X$ is a random variable with
exponential Probability Density Function (PDF) of
mean $\tau,$ then the PDF of the random variable $Y,$ the
truncated version of $X$ truncated on the right at $\tau,$ is
given by:
$f(y,\tau) = \frac{1}{\tau}e^{-\frac{y}{\tau}}(1-\frac{1}{e})^{-1},\, 0\le y\le\tau$
$f(y,\tau) = 0$, elsewhere
$F_{Y}(y) = \int_0^y  \frac{1}{\tau}e^{-\frac{y}{\tau}}(1-\frac{1}{e})^{-1},\, 0\le y\le\tau$
$F_{Y}(y) = 1,\, y\gt \tau$
