Making Random variable from other Random variables but keep them independent
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I am trying to solve the following question.
Random Variable X has the following PMF:
P(X = 0) = 0.5
P(X = 1) = 0.5
We define another random variable U = XZ.
Construct random variable Z , independent of X such that Cov(U,X) = 0 i.e. U and X are uncorrelated but not
independent.
As we can see that Z = U/X, U and X are also dependent then how can we construct a random variable Z which is independent of X. One way i was thinking is that write it in term of U but in turn U is also dependent on X.
How to approach this problem
probability probability-theory probability-distributions random-variables covariance
asked Nov 16 at 19:03
chuffles
203
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up vote
0
down vote
favorite
I am trying to solve the following question.
Random Variable X has the following PMF:
P(X = 0) = 0.5
P(X = 1) = 0.5
We define another random variable U = XZ.
Construct random variable Z , independent of X such that Cov(U,X) = 0 i.e. U and X are uncorrelated but not
independent.
As we can see that Z = U/X, U and X are also dependent then how can we construct a random variable Z which is independent of X. One way i was thinking is that write it in term of U but in turn U is also dependent on X.
How to approach this problem
probability probability-theory probability-distributions random-variables covariance
asked Nov 16 at 19:03
chuffles
203
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to solve the following question.
Random Variable X has the following PMF:
P(X = 0) = 0.5
P(X = 1) = 0.5
We define another random variable U = XZ.
Construct random variable Z , independent of X such that Cov(U,X) = 0 i.e. U and X are uncorrelated but not
independent.
As we can see that Z = U/X, U and X are also dependent then how can we construct a random variable Z which is independent of X. One way i was thinking is that write it in term of U but in turn U is also dependent on X.
How to approach this problem
probability probability-theory probability-distributions random-variables covariance
asked Nov 16 at 19:03
chuffles
203
I am trying to solve the following question.
Random Variable X has the following PMF:
P(X = 0) = 0.5
P(X = 1) = 0.5
We define another random variable U = XZ.
Construct random variable Z , independent of X such that Cov(U,X) = 0 i.e. U and X are uncorrelated but not
independent.
As we can see that Z = U/X, U and X are also dependent then how can we construct a random variable Z which is independent of X. One way i was thinking is that write it in term of U but in turn U is also dependent on X.
How to approach this problem
probability probability-theory probability-distributions random-variables covariance
probability probability-theory probability-distributions random-variables covariance
asked Nov 16 at 19:03
chuffles
203
asked Nov 16 at 19:03
chuffles
203
asked Nov 16 at 19:03
chuffles
203
asked Nov 16 at 19:03
chuffles
203
asked Nov 16 at 19:03
chuffles
203
203
add a comment |
add a comment |
1 Answer
1
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up vote
1
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${Cov(UX)=E(UX)-E(U)E(X)={E(X^2)E(Z)-E(Z)(E(X))^2,}}$
since $X$ and $Z$ are independent. For $Cov(UX)=0$, either $E(Z)=0$ or $E(X^2)-(E(X))^2=0$. We know the latter cannot be $0$, since it is the variance of $X$. To make $E(Z)=0$, let $Z$ be any random variable symmetric around $0$, for example $Z=pm 1$, each with probability $0.5$.
answered Nov 16 at 20:42
herb steinberg
2,2132310
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
${Cov(UX)=E(UX)-E(U)E(X)={E(X^2)E(Z)-E(Z)(E(X))^2,}}$
since $X$ and $Z$ are independent. For $Cov(UX)=0$, either $E(Z)=0$ or $E(X^2)-(E(X))^2=0$. We know the latter cannot be $0$, since it is the variance of $X$. To make $E(Z)=0$, let $Z$ be any random variable symmetric around $0$, for example $Z=pm 1$, each with probability $0.5$.
answered Nov 16 at 20:42
herb steinberg
2,2132310
add a comment |
up vote
1
down vote
accepted
${Cov(UX)=E(UX)-E(U)E(X)={E(X^2)E(Z)-E(Z)(E(X))^2,}}$
since $X$ and $Z$ are independent. For $Cov(UX)=0$, either $E(Z)=0$ or $E(X^2)-(E(X))^2=0$. We know the latter cannot be $0$, since it is the variance of $X$. To make $E(Z)=0$, let $Z$ be any random variable symmetric around $0$, for example $Z=pm 1$, each with probability $0.5$.
answered Nov 16 at 20:42
herb steinberg
2,2132310
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
${Cov(UX)=E(UX)-E(U)E(X)={E(X^2)E(Z)-E(Z)(E(X))^2,}}$
since $X$ and $Z$ are independent. For $Cov(UX)=0$, either $E(Z)=0$ or $E(X^2)-(E(X))^2=0$. We know the latter cannot be $0$, since it is the variance of $X$. To make $E(Z)=0$, let $Z$ be any random variable symmetric around $0$, for example $Z=pm 1$, each with probability $0.5$.
answered Nov 16 at 20:42
herb steinberg
2,2132310
${Cov(UX)=E(UX)-E(U)E(X)={E(X^2)E(Z)-E(Z)(E(X))^2,}}$
since $X$ and $Z$ are independent. For $Cov(UX)=0$, either $E(Z)=0$ or $E(X^2)-(E(X))^2=0$. We know the latter cannot be $0$, since it is the variance of $X$. To make $E(Z)=0$, let $Z$ be any random variable symmetric around $0$, for example $Z=pm 1$, each with probability $0.5$.
answered Nov 16 at 20:42
herb steinberg
2,2132310
answered Nov 16 at 20:42
herb steinberg
2,2132310
answered Nov 16 at 20:42
herb steinberg
2,2132310
answered Nov 16 at 20:42
herb steinberg
2,2132310
2,2132310
add a comment |
add a comment |
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