Find CDF of $Z:=\min\{X,Y\}$

Question

Let be $X,Y$ independent random variables with $X\sim\mathcal{U}(\{0,1\})$ and $Y\sim\mathcal{U}(\{0,1,2\})$. Compute the CDF of $Z:=\min\{X,Y\}$.

My idea:

$$ \begin{align*} P(\min\{X,Y\}\leq x) &= 1-P(\min\{X,Y\}>x)\\ &= 1-P(\{X>x\}\cap\{Y>x\})\\ &=1-P(X>x)\cdot P(Y>x)\\ &= 1-(1-P(X\leq x))(1-P(Y\leq x))\\ &=1-(1-F_X(x))(1-F_Y(x)), \end{align*} $$ while

$$F_X(x)=\begin{cases} 0 &,x<a\\ \frac{x-a}{b-a} &, x\in[a,b]\\ 1&, x>b \end{cases}.$$

What sould I do next?

Answer 1

You gave a generic definition of the CDF of uniform continuous distributions.

You have two specific uniform discrete distributions.   These will have step functions.

The CDF of the minimum of those RV will also be a step function, although the steps will not be uniform .   We also know the support of $Z$ since: $\min\{X, Y\}\in \{0,1,2\}$

$$\begin{align}F_X (x) & = \begin{cases} 0 & x < 0 \\ \tfrac 12 & : 0\leq x < 1 \\ 1 & : 1\leq x\end{cases} \\[2ex] F_Y (y) & = \begin{cases} 0 & y < 0 \\ \tfrac 13 & : 0\leq y < 1 \\ \tfrac 2 3 & : 1\leq y < 2 \\ 1 & : 2\leq y\end{cases} \\[3ex] F_Z(z) & = 1-(1-F_X(z))(1-F_Y(z)) \\[1ex] & = \begin{cases} 0 & z < 0 \\ ? & : 0\leq z < 1 \\ ? & : 1\leq z < 2 \\ 1 & : 2\leq z\end{cases} \end{align}$$

Can you complete?

Answer 2

Hint: If $X$ takes on two possible values and $Y$ takes on three possible values, then the pair $(X,Y)$ has six possible values, each one equally likely (since $X$ and $Y$ are independent and each one has uniform distribution). This means you can enumerate the possible values for $Z:=\min(X,Y)$ and determine their probabilities (construct a table!). You'll find that $Z$ takes value $0$ with probability $2/3$ and value $1$ with probability $1/3$, which is enough information to determine the CDF.