Probability mass function of $ \min(X, Y)$ where $X,Y$ are i.i.d discrete uniform

Question

Let $X$ and $Y$ be two discrete uniform i.i.d random variables distributed over $\{0, 1, 2,\ldots, N\}$. Find the pmf of $Z = \min(X, Y)$.

From what I understand, I have to find the joint $pmf$ first, which is just $1/(N+1)^2$ by independence. Now, I have to find the probability function $P(X_{(1)})$. Is this right? If so, how do I determine $P(X_{(1)})$ for $X_i$ discrete?

From two textbooks I have, I only found $P(X_{(1)} = x_i)$. Is $P(X_{(1)})$ = $P(X_{(1)} = x_i)$ for a generic $x_i$?

Thanks.

Answer 1

$P(Z\le n)=P(X\le n\ or\ Y\le n)=P(X\le n)+P(Y\le n)-P(X\le n \ and\ Y\le n) =\frac{2(n+!)}{N+1}-(\frac{n+1}{N+1})^2$

The last step uses the independence of $X$ and $Y$.

Since the random variables are discrete I am not sure what you would want for a density function.

Answer 2

$P(\min(X,Y) = k) = \sum_{j=k}^N P(X = k)P(Y = j) + \sum_{j=k}^N P(Y = k)P(X = j) - P(Y=k,X=k) $

$= 2\sum_{j=k}^N P(X = k)P(Y = j)-\frac{1}{(N+1)^2}$ by symmetry.

$ = 2\sum_{j=k}^N \frac{1}{N+1}\frac{1}{N+1} = \frac{2(N+1-k)}{(N+1)^2}-\frac{1}{(N+1)^2}$

$ = \frac{2(N+\frac{1}{2}-k)}{(N+1)^2}$