# 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?

• You seem to be on the right track if the problem specifies that $X$ and $Y$ are independent. If not, you might have to deal with their joint PDF. – rgk Mar 18 '19 at 16:05
• There is no PDF anywhere. And since you are assuming independence of $X,Y$, you should mention that in your post. – StubbornAtom Mar 18 '19 at 16:10

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.