0
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ You should show your work/share your thoughts in your questions. $\endgroup$ – StubbornAtom Dec 20 '18 at 14:46
1
$\begingroup$

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 $Y\sim U(0,1).$

$\endgroup$
  • 1
    $\begingroup$ It probably would be helpful to use align so that the key step is better readable. $\endgroup$ – Just_to_Answer Dec 29 '18 at 2:37
0
$\begingroup$

Hint :

If for all $\phi$ continuous bounded functions :

$$E[\phi(Y)]=\int_{\mathbb{R}} \phi(y) g(y) dy.$$

Then $g$ is the pdf of $Y$.

First step :

$$E[\phi(Y)]=\int_{\mathbb{R}} \phi(x-[x]) f_X(x) dx.$$

With $f_X$ the pdf of $X$.

Can you finish ?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.