How is $\min\{X,Y\}$ defined for $X, Y$ random variables?
Is it defined as the $\min$ of their probability functions for a constant value they get?
For example:
$X\sim \operatorname{Geom}\left(\frac16\right), Y\sim \operatorname{Geom}\left(\frac12\right)$
We have $P(X=k)= \frac {5^{k-1}} {6^k} $, $P(Y=k)= \frac 1 {2^k} $, so when comparing them we get: $$\min\{X,Y\}=\begin{cases} X, & 1\leq k\leq 3 \\ Y, & k>3 \end{cases}$$
Is this true or is the definition different?
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$\begingroup$No, $M:=\min\{X,Y\}$ is a random variable itself that "records" the lowest value of $X,Y$. You do not compare the probabilities but the values of the random variables. For example first dice $X$, you roll and you get a $5$, second dice $Y$ you roll and you get a $3$ then $\min\{X,Y\}=3$.
As a random variable $M$ has its own distribution that in the special case $X,Y$ are independent is known and is given by
\begin{align}F_M(t)&=P(\min\{X,Y\}\le t)=1-P(\min\{X,Y\}>t)=1-P(X>t, Y>t)\\&=1-P(X>t)P(Y>t)=1-(1-P(X\le t))(1-P(Y\le t))\\&=1-(1-F_X(t))(1-F_Y(t))\end{align}
$\endgroup$ 1 $\begingroup$If $X\sim\text{Geom}(1/6), Y\sim\text{Geom}(1/2)$ be independent, where each is over $\{1,2,3,\dotsc\}$, and if $M = \min\{X,Y\}$, then for $m = 1,2,3,\dotsc$, \begin{align*} P(M = m) &= P(M>m-1)-P(M>m) \\ &=P(X>m-1,Y>m-1)-P(X>m,Y>m)\\ &=P(X>m-1)P(Y>m-1)-P(X>m)P(Y>m)\tag{1}\\ &= \left(\frac{5}{6}\right)^{m-1} \left(\frac{1}{2}\right)^{m-1}-\left(\frac{5}{6}\right)^{m}\left(\frac{1}{2}\right)^{m}, \end{align*} where you can split the probabilities in $(1)$ since $X,Y$ are independent. Notice that we are interested in the probability that the minimum is equal to some value $m$, not whether $X$ or $Y$ was the minimum. If you imagine $X$ as a fair die and and $Y$ as a fair coin, then I would just be flipping the coin (until I get a tail) and rolling the die (until I get a 6 for example) behind a wall or inside a box and simply report to you that the minimum is 2, or 12, or 100, not that it came from the coin or the die.
$\endgroup$ 1 $\begingroup$Recall that a random variable is a function on a sample space. Accordingly the minimum is taken pointwise: for each $\omega$ in the sample space, $\min\{X, Y\}$ is defined as the function
$$\omega \mapsto \min\{X(\omega), Y(\omega)\}.$$
An important remark is that information on each of $X$ and $Y$ is not enough to determine $\min\{X, Y\}$. You need to know how they very jointly. One the best situation is that $X$ and $Y$ are independent, but other dependent cases also arise in many meaningful situations.
Here are two examples such that $X \sim \operatorname{Geom}(1/6)$ and $Y \sim \operatorname{Geom}(1/2)$, but $\min\{X, Y\}$ have different distributions.
Example 1. Let $X \sim \operatorname{Geom}(1/6)$ and $Y \sim \operatorname{Geom}(1/2)$ be independent. Then the CDF of $M = \min\{X, Y\}$ is easily computed by exploiting the independence: \begin{align*} F_M(z) = \Bbb{P}(M \leq z) &= 1 - \Bbb{P}(M > z) \\ & = 1 - \Bbb{P}(X > z, Y > z) \\ & = 1 - \Bbb{P}(X > z)\Bbb{P}(Y > z) \\ & = 1 - (1 - F_X(z))(1 - F_Y(z)). \end{align*}
$\endgroup$ 1 $\begingroup$Example 2. We construct an example where $X \sim \operatorname{Geom}(1/6)$ and $Y \sim \operatorname{Geom}(1/2)$, but $M = \min\{X, Y\}$ has a distribution which is different from Example 1.
Let $Y_1, Y_2, \cdots$ be i.i.d. and $Y_i \sim \operatorname{Geom}(1/2)$. Also let $N$ be independent of all of $(Y_i)$ and $N \sim \operatorname{Geom}(1/3)$. (Here, we use the convention that any geometric distribution is supported on the set $\{1, 2, \cdots\}$.) Then it is well-known that $Y$ defined by $$ X = Y_1 + Y_2 + \cdots + Y_N $$ is also a geometric random variable with $X \sim \operatorname{Geom}(1/6)$. But if we let $Y = Y_1$, then we always have $Y \leq X$ and hence $M = Y$. Consequently $$ F_M(z) = F_Y(z). $$
It is defined as the minimum of two functions in the normal way. Think of $$\min(f(x),g(x))$$
How is this defined? Well for every $x$, compute the value of $f(x)$, the value of $g(x)$, and take the minimum of them. Same thing for random variables
Actually determining the law of the minimum of two random variables is not always easy
$\endgroup$ $\begingroup$If it's an explicit formula that you're looking for, then $$\min\{X,Y\}=\frac12(X+Y)-\frac12\lvert Y-X\rvert.$$ In other words, we start at the midpoint of $X$'s value and $Y$'s value, then subtract half of the distance between $X$'s value and $Y$'s value. Had we added half the distance, we'd instead obtain the random variable $\max\{X,Y\}.$
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