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

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The distribution of Y, defined as Y = cX + d where X follows a uniform distribution U(0,1) and c < 0, results in Y being uniformly distributed between the bounds cL + d and cR + d. Given that L = 0 and R = 1, the distribution simplifies to Y ~ U(d, c + d). This conclusion is reached by rearranging the equation and substituting the limits of X into the derived inequalities.

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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?
 
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oyth94 said:
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?
 
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?
 
oyth94 said:
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.
 

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