If $X\sim U(0,n) ; n \in \mathbb N$ , how can I show that the distribution of $Y =X-[X]$ is $U(0,1)$?
Any hint will also help me...
2026-03-26 09:19:41.1774516781
If $X\sim U(0,n)$, how can I show that $X-[X]\sim U(0,1)$?
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For $Y=X-[X]$, and $0<y<1$ $P(Y<y)=P(0<X<y,1<X<1+y,\ldots n-1<X<n-1+y)=P(0<X<y)+P(1<X<1+y)+\ldots +P(n-1<X<n-1+y)=n\times y/n=y.$ Hence $Y\sim U(0,1).$