MHB Distribution of Y: cX+d | X~uniform(0,1)

[oyth94]

Let X~ uniform(0,1), ie f(x) = 1/(R-L) for L<x<R and c<0. Let Y= cX+d. What is the distribution of Y?

 

[Barioth]

Let X~ uniform(0,1), ie f(x) = 1/(R-L) for L<x<R and c<0. Let Y= cX+d. What is the distribution of Y?

Think about what happen to X when you add d.

Or

Try to think about what would happen if you only have c*X

What would happen if d was equal to 1?

 

[oyth94]

I rearranged Y=cX+d to get X=(Y-d)/c and substitute it into L<x<R
So it's L< (Y-d)/c < R => cL + d < Y< cR+d
So is that the dist. of Y?

 

[I like Serena]

I rearranged Y=cX+d to get X=(Y-d)/c and substitute it into L<x<R
So it's L< (Y-d)/c < R => cL + d < Y< cR+d
So is that the dist. of Y?

Hi oyth94! :)

Yes, that is the case.
Btw, you can make your result a little more specific.
Note that your X distribution is U(0,1).
Those 0 and 1 mean that your L=0 and your R=1...
It also means that you can write Y ~ U(a,b) for some a and b.