Find variable that is uncorrelated but not independent
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I am given PMF of random variable X. P(X=0) = P(x=1) =0.5. Now there is another RV Y such that Y = XZ. I have to find Z independent of X such that X and Y are uncorrelated but not independent. My first intuition is that the PMF of Z should be 1/z^3. And Z can take values form 1 to n. This way each time it is multiplied with X, U gets reduced. So, even if X is increased to 1 U won't increase.
Is this correct?
correlation
asked Nov 15 at 0:48
puffles
669
add a comment |
up vote
0
down vote
favorite
I am given PMF of random variable X. P(X=0) = P(x=1) =0.5. Now there is another RV Y such that Y = XZ. I have to find Z independent of X such that X and Y are uncorrelated but not independent. My first intuition is that the PMF of Z should be 1/z^3. And Z can take values form 1 to n. This way each time it is multiplied with X, U gets reduced. So, even if X is increased to 1 U won't increase.
Is this correct?
correlation
asked Nov 15 at 0:48
puffles
669
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am given PMF of random variable X. P(X=0) = P(x=1) =0.5. Now there is another RV Y such that Y = XZ. I have to find Z independent of X such that X and Y are uncorrelated but not independent. My first intuition is that the PMF of Z should be 1/z^3. And Z can take values form 1 to n. This way each time it is multiplied with X, U gets reduced. So, even if X is increased to 1 U won't increase.
Is this correct?
correlation
asked Nov 15 at 0:48
puffles
669
I am given PMF of random variable X. P(X=0) = P(x=1) =0.5. Now there is another RV Y such that Y = XZ. I have to find Z independent of X such that X and Y are uncorrelated but not independent. My first intuition is that the PMF of Z should be 1/z^3. And Z can take values form 1 to n. This way each time it is multiplied with X, U gets reduced. So, even if X is increased to 1 U won't increase.
Is this correct?
correlation
correlation
asked Nov 15 at 0:48
puffles
669
asked Nov 15 at 0:48
puffles
669
asked Nov 15 at 0:48
puffles
669
asked Nov 15 at 0:48
puffles
669
asked Nov 15 at 0:48
puffles
669
669
add a comment |
add a comment |
1 Answer
1
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0
down vote
You've mentioned both U and Y as RVs. I'll assume that they both refer to the same variable so U=XZ.
Note that Cov(U,X)=0
E(UX) - E(U)E(X)
Since Z and X are independent we have E(U) = E(XZ) = E(X).E(Z)
Substituting that in the equation above we get:
= E(X^2).E(Z) - E(Z).E(X)^2
=E(Z).(E(X^2) - E(X)^2)
=E(Z).var(X)
Now we know that var(X) is not 0. So just define Z such that E(Z) = 0.
That's pretty simple to do. Just define an RV symmetric around 0. So +-1 both with probability 0.5 and you get your Z.
answered Nov 18 at 18:40
helloworld
477
I defined RV Z that takes value 1 with probability 1. And I guess that worked too. Thanks anyways
– puffles
Nov 18 at 19:04
That way you'll get E(Z) = 1. Since var(X) ≠ 0, you won't get cov(u,x)=0
– helloworld
Nov 18 at 20:05
E(Z) = 1 and E(X) = 0.5 and E(U) = 0.5 and E(UX) = 0.25. If I plug into equation E(UX) - E(X)E(U) = 0.25 - 0.5*0.5.
– puffles
Nov 19 at 6:59
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You've mentioned both U and Y as RVs. I'll assume that they both refer to the same variable so U=XZ.
Note that Cov(U,X)=0
E(UX) - E(U)E(X)
Since Z and X are independent we have E(U) = E(XZ) = E(X).E(Z)
Substituting that in the equation above we get:
= E(X^2).E(Z) - E(Z).E(X)^2
=E(Z).(E(X^2) - E(X)^2)
=E(Z).var(X)
Now we know that var(X) is not 0. So just define Z such that E(Z) = 0.
That's pretty simple to do. Just define an RV symmetric around 0. So +-1 both with probability 0.5 and you get your Z.
answered Nov 18 at 18:40
helloworld
477
I defined RV Z that takes value 1 with probability 1. And I guess that worked too. Thanks anyways
– puffles
Nov 18 at 19:04
That way you'll get E(Z) = 1. Since var(X) ≠ 0, you won't get cov(u,x)=0
– helloworld
Nov 18 at 20:05
E(Z) = 1 and E(X) = 0.5 and E(U) = 0.5 and E(UX) = 0.25. If I plug into equation E(UX) - E(X)E(U) = 0.25 - 0.5*0.5.
– puffles
Nov 19 at 6:59
add a comment |
up vote
0
down vote
You've mentioned both U and Y as RVs. I'll assume that they both refer to the same variable so U=XZ.
Note that Cov(U,X)=0
E(UX) - E(U)E(X)
Since Z and X are independent we have E(U) = E(XZ) = E(X).E(Z)
Substituting that in the equation above we get:
= E(X^2).E(Z) - E(Z).E(X)^2
=E(Z).(E(X^2) - E(X)^2)
=E(Z).var(X)
Now we know that var(X) is not 0. So just define Z such that E(Z) = 0.
That's pretty simple to do. Just define an RV symmetric around 0. So +-1 both with probability 0.5 and you get your Z.
answered Nov 18 at 18:40
helloworld
477
I defined RV Z that takes value 1 with probability 1. And I guess that worked too. Thanks anyways
– puffles
Nov 18 at 19:04
That way you'll get E(Z) = 1. Since var(X) ≠ 0, you won't get cov(u,x)=0
– helloworld
Nov 18 at 20:05
E(Z) = 1 and E(X) = 0.5 and E(U) = 0.5 and E(UX) = 0.25. If I plug into equation E(UX) - E(X)E(U) = 0.25 - 0.5*0.5.
– puffles
Nov 19 at 6:59
add a comment |
up vote
0
down vote
up vote
0
down vote
You've mentioned both U and Y as RVs. I'll assume that they both refer to the same variable so U=XZ.
Note that Cov(U,X)=0
E(UX) - E(U)E(X)
Since Z and X are independent we have E(U) = E(XZ) = E(X).E(Z)
Substituting that in the equation above we get:
= E(X^2).E(Z) - E(Z).E(X)^2
=E(Z).(E(X^2) - E(X)^2)
=E(Z).var(X)
Now we know that var(X) is not 0. So just define Z such that E(Z) = 0.
That's pretty simple to do. Just define an RV symmetric around 0. So +-1 both with probability 0.5 and you get your Z.
answered Nov 18 at 18:40
helloworld
477
You've mentioned both U and Y as RVs. I'll assume that they both refer to the same variable so U=XZ.
Note that Cov(U,X)=0
E(UX) - E(U)E(X)
Since Z and X are independent we have E(U) = E(XZ) = E(X).E(Z)
Substituting that in the equation above we get:
= E(X^2).E(Z) - E(Z).E(X)^2
=E(Z).(E(X^2) - E(X)^2)
=E(Z).var(X)
Now we know that var(X) is not 0. So just define Z such that E(Z) = 0.
That's pretty simple to do. Just define an RV symmetric around 0. So +-1 both with probability 0.5 and you get your Z.
answered Nov 18 at 18:40
helloworld
477
answered Nov 18 at 18:40
helloworld
477
answered Nov 18 at 18:40
helloworld
477
answered Nov 18 at 18:40
helloworld
477
477
I defined RV Z that takes value 1 with probability 1. And I guess that worked too. Thanks anyways
– puffles
Nov 18 at 19:04
That way you'll get E(Z) = 1. Since var(X) ≠ 0, you won't get cov(u,x)=0
– helloworld
Nov 18 at 20:05
E(Z) = 1 and E(X) = 0.5 and E(U) = 0.5 and E(UX) = 0.25. If I plug into equation E(UX) - E(X)E(U) = 0.25 - 0.5*0.5.
– puffles
Nov 19 at 6:59
add a comment |
I defined RV Z that takes value 1 with probability 1. And I guess that worked too. Thanks anyways
– puffles
Nov 18 at 19:04
That way you'll get E(Z) = 1. Since var(X) ≠ 0, you won't get cov(u,x)=0
– helloworld
Nov 18 at 20:05
E(Z) = 1 and E(X) = 0.5 and E(U) = 0.5 and E(UX) = 0.25. If I plug into equation E(UX) - E(X)E(U) = 0.25 - 0.5*0.5.
– puffles
Nov 19 at 6:59
I defined RV Z that takes value 1 with probability 1. And I guess that worked too. Thanks anyways
– puffles
Nov 18 at 19:04
I defined RV Z that takes value 1 with probability 1. And I guess that worked too. Thanks anyways
– puffles
Nov 18 at 19:04
That way you'll get E(Z) = 1. Since var(X) ≠ 0, you won't get cov(u,x)=0
– helloworld
Nov 18 at 20:05
That way you'll get E(Z) = 1. Since var(X) ≠ 0, you won't get cov(u,x)=0
– helloworld
Nov 18 at 20:05
E(Z) = 1 and E(X) = 0.5 and E(U) = 0.5 and E(UX) = 0.25. If I plug into equation E(UX) - E(X)E(U) = 0.25 - 0.5*0.5.
– puffles
Nov 19 at 6:59
E(Z) = 1 and E(X) = 0.5 and E(U) = 0.5 and E(UX) = 0.25. If I plug into equation E(UX) - E(X)E(U) = 0.25 - 0.5*0.5.
– puffles
Nov 19 at 6:59
add a comment |
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