For a non-negative real number $x$, write $Int(x)$ for the largest integer that is less than or equal to $x$. Let $U$ be a uniform random variable on $(0,1)$ and $n \geq 1$ an integer. Find the probability mass function of the (discrete) random variable $X = Int(nU) + 1$.
I don't really know how to start on this question, can someone please give me some hints?
We analyze in detail the case $n=3$. We will use the standard notation $\lfloor x\rfloor$ for what in the OP is called $\text{Int}(x)$.
(i) For $0\lt u\lt \frac{1}{3}$, we have $\lfloor 3u\rfloor +1=1$. So for these $u$, $X=1$.
(ii) For $\frac{1}{3}\le u\lt \frac{2}{3}$, we have $\lfloor 3u\rfloor +1=2$. So for these $u$, $X=2$.
(iii) Finally, for $\frac{2}{3}\le u\lt 1$, we have $\lfloor 3u\rfloor +1=3$. For these $u$, $X=3$.
Since the range of $U$ is $(0,1)$, $X$ can take on no values other than $1$, $2$, or $3$.
The probability that $X=1$ is the probability that $0\lt U\lt \frac{1}{3}$. Since $U$ is uniform on $(0,1)$, we conclude that $\Pr(X=1)=\frac{1}{3}$. The same sort of calculation shows that $\Pr(X=2)=\Pr(X=3)=\frac{1}{3}$.
Thus $X$ has the discrete uniform distribution on the set $\{1,2,3\}$.
A very similar analysis shows that for general $n$, your random variable $X$ has the discrete uniform distribution on the set $\{1,2,3,\dots,n\}$.
For general $n$, the random variable $X$ takes on the value $1$ if $0\lt u\lt \frac{1}{n}$, the value $2$ if $\frac{1}{n}\le u\lt \frac{2}{n}$, the value $3$ when $\frac{2}{n}\le u\lt \frac{3}{n}$, and so on. Each of these intervals has length $\frac{1}{n}$. In general, because $U$ has continuous uniform distribution on $(0,1)$, $U$ lands in any such interval with probability $\frac{1}{n}$.
Remark: The typical pseudo-random number generator simulates a continuous uniform distribution on $(0,1)$. By using the method described in this problem, we can use the pseudo-random number generator to simulate certain discrete uniform distributions. For example, by taking $n=6$, we can simulate the result of throwing a fair die.
