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

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Discussion Overview

The discussion focuses on determining the distribution of the random variable Y defined as Y = cX + d, where X is uniformly distributed over the interval (0, 1) and c is a negative constant. Participants explore the implications of the transformation on the distribution of Y, considering various scenarios involving the parameters c and d.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • Some participants propose that the transformation Y = cX + d, with X uniformly distributed, leads to a specific distribution for Y that can be derived from rearranging the equation.
  • One participant suggests considering the effect of adding d to X and the implications of varying c, particularly when c is negative.
  • Another participant derives inequalities based on the transformation, concluding that Y must lie within a certain range defined by cL + d and cR + d.
  • A later reply confirms the earlier derivation and encourages further specification of the result, noting the uniform nature of X and suggesting that Y can be expressed as a uniform distribution over a new interval.

Areas of Agreement / Disagreement

Participants generally agree on the approach to derive the distribution of Y from the transformation of X, but there is no consensus on the final form or specific parameters of the distribution.

Contextual Notes

Limitations include the dependence on the values of c and d, as well as the assumptions regarding the uniform distribution of X. The discussion does not resolve the specific parameters of the resulting distribution for Y.

oyth94
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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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