PDF of the minimum of a geometric random variable and a constant I have $X \sim Geo(p)$, such that
$$p(x) = (1-p)^{k-1}p, \ \ x = 1,2,3, \ldots$$
and Y is a constant random variable which assumes the value of the constant integer $t$, such that
$$P(Y=t) = 1, \ \ t>0$$
And now I am considering a random variable $Z$, where 
$$Z = min(X, Y)$$
I am attempting to compute the PDF of Z, and have followed something similar to the answer posted here i.e.
$\begin{align}
f_Z(z) & = \frac{\mathrm d\;}{\mathrm d z} \Bbb P(\min(X,Y)\leq z)
\\[1ex] & = \frac{\mathrm d\;}{\mathrm d z} (1 - \Bbb P(\min(X,Y)\gt z))
\\[1ex] & =  -\frac{\mathrm d\;}{\mathrm d z}\Big( \Bbb P(X>z)\,\Bbb P(Y>z) \Big)
\\[1ex] & =  -\frac{\mathrm d\;}{\mathrm d z} \Big(\big(1-F_X(z)\big)\big(1-F_Y(z)\big)\Big)
\\[1ex] & = f_X(z)\Big(1-F_Y(z)\Big) + \Big(1-F_X(z)\Big)f_Y(z)
\\[1ex] & = (1-p)^{z-1}p\Big(1 - 1_{z \geq t}\Big) - (1-p)^z1_{z=t}
\end{align}$
Am I on the right path? I kind of expected a more compact kind of expression, one I could relate to the exponential family?
 A: 
Proposition.
  Let $X$ be a random variable with cumulative distribution function $F_X$ (i.e.,
  $$
F_X(x)
= P(X \leq x)
$$
  for all $x \in \mathbb{R}$, and let $t \in \mathbb{R}$ be a constant.
  Define $Z = \min\{X, t\}$.
  Then the cumulative distribution function $F_Z$ of $Z$ is
  $$
F_Z(z)
= \begin{cases}
F_X(z) & \text{if $z < t$} \\
1 & \text{if $z \geq t$}
\end{cases}
$$
  for all $z \in \mathbb{R}$.

Proof.
Let $Y$ be a constant random variable with value $t$.
Note that $Y$ is necessarily independent of $X$ and that
$$
P(Y > z)
= \mathbf{1}_{(-\infty, t)}(z)
= \begin{cases}
1 & \text{if $z < t$,} \\
0 & \text{if $z \geq t$.}
\end{cases}
$$
Then we have
$$
\begin{aligned}
F_Z(z)
&= P(Z \leq z) \\
&= 1 - P(Z > z) \\
&= 1 - P(\min\{X, Y\} > z) \\
&= 1 - P(X > z, Y > z) \\
&= 1 - P(X > z) P(Y > z) \\
&= 1 - (1 - F_X(z)) \mathbf{1}_{(-\infty, t)}(z) \\
&= \begin{cases}
F_X(z) & \text{if $z < t$,} \\
1 & \text{if $z \geq t$.}
\end{cases}
\end{aligned}
$$

You can use this Proposition to figure out the probability mass function (not the probability density function!) of $Z = \min\{X, Y\}$ where $X \sim \operatorname{Geometric}(p)$ and $Y = t$ almost surely, as in your question.
More concretely, your random variable $Z$ will be a discrete random variable supported on $\{1, 2, \ldots, t\}$ satisfying
$$
\begin{aligned}
P(Z = z)
&= P(Z \leq z) - P(Z \leq z - 1) \\
&= F_Z(z) - F_Z(z - 1)
\end{aligned}
$$
for all $z \in \{1, 2, \ldots, t\}$.
The cumulative distribution function needed for the above computation can be determined using the Proposition above.
