# Thread: change of variable proof...

1. ## change of variable proof...

Let X~Uniform[L,R]. Let Y=cX+d, where c<0. Prove that Y~Uniform[cL+d,cR+d]. (In particular if L=0 and R=1 and c=-1 and d=1, then X~Uniform[0,1] and also
Y=1-X~Uniform[0,1])

my attempt
Y=cX+d
X= (Y-d)/c since c<0
(Y-d)/c ~ Uniform[L,R] since c<0
isolating for Y gets you
L <= (Y-d)/c <= R since c<0
cL <= Y-d <= cR
cL+d <= Y <= cR+d
Y~Uniform[cL+d,cR+d]

Y~Uniform[(-1)0+1,(-1)1+1]
Y~Uniform[1, 0]
1 <= -X+1 <= 0
0 <= -X <= -1
0 <= X <= 1
X~Uniform[0,1]

Y=1-X~Uniform[0,1]
cX+d=1-X~Uniform[0,1]
X~Uniform[0,1] = 1-(cX+d)
X~Uniform[0,1] = 1-cX-d
X~Uniform[0,1] = 1+X-1
X~Uniform[0,1] = X

is all this correct?

also
another question
Let X~exponential(lambda). Let Y=cX where c>0. prove that Y~Exponential(lanbda/c)

i don't understand how to do this proof.

2. Originally Posted by Sneaky
Let X~Uniform[L,R]. Let Y=cX+d, where c<0. Prove that Y~Uniform[cL+d,cR+d]. (In particular if L=0 and R=1 and c=-1 and d=1, then X~Uniform[0,1] and also
Y=1-X~Uniform[0,1])

my attempt
Y=cX+d
X= (Y-d)/c since c<0
(Y-d)/c ~ Uniform[L,R] since c<0
isolating for Y gets you
L <= (Y-d)/c <= R since c<0
cL <= Y-d <= cR
cL+d <= Y <= cR+d
Y~Uniform[cL+d,cR+d]
You can't do that!

What you need to do is show that $\displaystyle P(a<Y<b)=\frac{b-a}{cR-cL}$ when $\displaystyle a,b \in (cL+d, cR+d)$

(that the probability for any interval outside $\displaystyle (cL+d, cR+d)$ will be zero should follow)

CB

3. how do i do that?

4. Originally Posted by Sneaky
how do i do that?
It is simple to do both questions using the same approach as I explained here: http://www.mathhelpforum.com/math-he...ge-159842.html

Note that for the first question, note that the support of Y is $\displaystyle cL+d \leq y \leq cR + d$ and that the cdf of Y is $\displaystyle \displaystyle G(y) = R - \left( \frac{y - d}{c}\right)$ since c < 0 (all working is left for you to do).