3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – rgk Mar 18 '19 at 16:05
  • $\begingroup$ There is no PDF anywhere. And since you are assuming independence of $X,Y$, you should mention that in your post. $\endgroup$ – StubbornAtom Mar 18 '19 at 16:10
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.