• Support PF! Buy your school textbooks, materials and every day products Here!

Sum of independent uniform distribution conditional on uniform

  • #1

Homework Statement


Let X and Y be independent and normal, then we know that
It must be the case that X+Y and X are jointly normal
Therefore we can apply the projection theorem:
which states that if A and B are jointly normal then VAR(A|B)=VAR(B)-[tex]\rho[/tex]^2VAR(B) , apply the theorem to A=X+Y, B=Y to find
VAR(X+Y|X)

There is a similar procedure of finding E(X+Y|X)



I know how to do the above. However, what I don't know is what if X and Y are independent but each are UNIFORMLY distributed on [-1,1]
What is:
1.VAR(X+Y|X)
2.E(X+Y|X)


Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 
Last edited:

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
2
this may help (or may not), with bits borrowed form wiki
http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables
though it should be independent of the probabilty distribution used

let Z = X + Y, and consider the joint probability density [itex] f_{X,Y,Z}(x,y,z) [/itex] the probability density of z is given by
[tex]f_{Z}(z) = \int f_{X,Y,Z}(x,y,z)dxdy [/tex]

you can re-write the joint distribution using conditional probabilities & the independence of X & Y, following that through leads to the convolution
[tex]f_{Z}(z) = \int f_{X,Y,Z}(x,y,z)dxdy = \int f_{X}(x)f_{Y}(y|x)f_{Z}(z|x,y)dxdy = \int f_{X}(x)f_{Y}(y)\delta(z-x-y)dxdy = \int f_{X}(x)f_{Y}(z-x)dx[/tex]

The conditional probability given [itex] X = x_0 [/itex] will be
[tex]f_{Z}(z|x_0) = \int f_{X}(x|x_0)f_{Y}(y|x_0)f_{Z}(z|x_0,y)dxdy = \int \delta(x-x_0)f_{Y}(y)\delta(z-x_0-y)dxdy=\int f_{Y}(y)\delta(z-x_0-y)dy= f_{Y}(z-x_0)[/tex]

Then take the expectation
[tex] E(Z|x_0) = \int z f(z|x_0)dz = \int z f_{Y}(z-x_0)dz [/tex]

and similar for the variance
 
  • #3
Thank you for your reply.
Unfortunately I am already familiar with this topic with regard to normally distibuted variables.
What I need is the mean and variance of sum of two independent Uniformly distributed variables conditional on one of the uniformly distributed variable.
I know that X+Y should be a triangular distributed, if X and Y are independent uniformly distributed.
However, what is
E(x+y|x)
Var(x+y|x)
 
Last edited:
  • #4
lanedance
Homework Helper
3,304
2
the derivation does not assume normal variables & should work for any distribution

the convolution integral will give you the triangular distribution

the last integral should give you the expectation with conditional X=x_0 (try it, it should be pretty simple)

then you should be able to use the conditional distiribution for the finding the variance as well

Thats rigorous... but if you just think about what is happening when X=x_0 is known you should be able to guess the mean & variance based on the distribution of Y easily enough
 
Last edited:
  • #5
lanedance
Homework Helper
3,304
2
PS, hint - as X is a set value, the only random variable is Y, so in effect Z = constant + Y...
 
Last edited:
  • #6
Thank you very much for your reply. I mostly understand what you wrote except 3 things.
1. Why is the conditional pdf of (z given x,y) equal to dirac's delta function of z-(x+y). Wikipedia just says they are trivially equal. Yes it's true that z needs to be equal to x+y, so you can say there is zero probabiltiy z not equal to x+y. But somehow I feel this is just too much intuition and not rigorous enough.
2. When the double integral regard to x y become a single integral to x, why does this work out to be fx(x)fy(z-x) ?

Yes y=z-x but don't we need
integral[fx(x)integral{fy(z-x)d(-x)}dx]. The d(-x) is due to the fact that dy=d(z-x)=-dx
3. How would we find E(X|X+Y)
 
Last edited:
  • #7
lanedance
Homework Helper
3,304
2
Thank you very much for your reply. I mostly understand what you wrote except 3 things.
1. Why is the conditional pdf of (z given x,y) equal to dirac's delta function of z-(x+y). Wikipedia just says they are trivially equal. Yes it's true that z needs to be equal to x+y, so you can say there is zero probabiltiy z not equal to x+y. But somehow I feel this is just too much intuition and not rigorous enough.
it can only be that, given X=x & Y=y, Z =X+Y is totally determined to be Z=x+y, hence
[tex] f_Z(z=x+y|X=x,Y=y) = \delta(z-x+y) [/tex]
note this intgrates to one as it must and satifies the constraint

2. When the double integral regard to x y become a single integral to x, why does this work out to be fx(x)fy(z-x) ? Yes y=z-x but don't we need integral[fx(x)integral{fy(z-x)d(-x)}dx]. The d(-x) is due to the fact that dy=d(z-x)=-dx
the following integral is contracted by integrating over all y, integrating over the delta function sets y = z-x
[tex] = \int \int f_{X}(x)f_{Y}(y)\delta(z-x-y)dxdy = \int f_{X}(x)f_{Y}(z-x)dx[/tex]

you could similarly choose to integrate over x first & get
[tex] = \int \int f_{X}(x)f_{Y}(y)\delta(z-x-y)dxdy = \int \int f_{X}(x)f_{Y}(y)\delta(z-x-y)dydx = \int f_{X}(z-y)f_{Y}(y)dy[/tex]

the results are equivalent as they must be, we choose to integrate over Y first as it makes life easier later on with the constrant on x

3. How would we find E(X|X+Y)
this takes a little more thought as i assume you mean E(X|Z), with Z = X+Y? you would need to find the conditional probability distribution

for the previous questions, say you have a random variable X with
[tex] E(X) = \mu [/tex]
[tex] VAR(X) = \sigma^2 [/tex]

note that for a constant a
[tex] E(X+a) = \mu +a [/tex]
[tex] VAR(X+a) = \sigma^2 [/tex]
 
  • #8
I am able to completely understand your post now. Thank you. But I tried to derive the entire thing in your early post but with different objective:
Give X and Y are indpendent uniform
Find
E(X|Z=X+Y)
Var(X|Z=X+Y)
and I am stuck in the imitation of your proof. Simply because fx(x|z) is no longer fx(x).
Would you please show me how to redo your early post in this scenario?
 

Related Threads on Sum of independent uniform distribution conditional on uniform

Replies
1
Views
10K
Replies
2
Views
1K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
8
Views
1K
Replies
2
Views
3K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
3
Views
694
  • Last Post
Replies
7
Views
592
Top