Question

The random variable X is said to be a discrete uniform random variable on the integers 1, 2, . . . , n if P{X = i } = 1 n i = 1, 2, . . , n For any nonnegative real number x, let In t(x) (sometimes written as [x]) be the largest integer that is less than or equal to x. Show that if U is a uniform random variable on (0, 1), then X = In t (n U) + 1 is a discrete uniform random variable on 1, . . . , n.

Short answer

To get the claimed, look at the relationship between X and continuous uniform distribution.

Step-by-step solution

Step:1 Given Information

If $PX=I=1nI=1,2,...,n,$the random variable X is said to be a discrete uniform random variable on the integers $1,2,...,n$. Let In t(x) (often written as [x]) be the largest integer that is less than or equal to x for any nonnegative real number x. Demonstrate that if U is a uniform random variable on $(0,1),X=Int(nU)+1$is a discrete uniform random variable on 1,..., n.

Step:2 Definition

A uniform probability distribution is a continuous opportunity distribution this is related to occurrences which are equally probably to occur. It has two parameters, x and y, with x denoting the minimal fee and y denoting the most value. The letter u is commonly used to represent it (x, y)

Step:3 Explanation of the solution

We are given that $U~Unif(0,1)$. That means that $nU\in (0,n)$which implies that $Int\left(nU\right)\in \{0,\dots ,n-1\}$. Finally, we have that $X=Int\left(nU\right)+1\in \{1,\dots ,n\}$. So, take any $k\in \{1,\dots ,n\}$. We have that

$\begin{array}{r}P(X=k)=P(\mathrm{Int}(nU)+1=k)=P(\mathrm{Int}(nU)=k-1)\\ =P(nU\in [k-1,k\left)\right)=P\left(U\in \left[\frac{k-1}{n},\frac{k}{n}\right)\right)\\ ={\int }_{\frac{k-1}{n}}^{\frac{k}{n}} 1dx=\frac{k}{n}-\frac{k-1}{n}=\frac{1}{n}\end{array}$

Hence, we have proved that $X~\mathrm{DUnif}(1,\dots ,n)$