91影视

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.

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)

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

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

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