Mixing
Under what condition are $U$ and $V$ uncorrelated?
$begingroup$
Let $X$ and $Y$ be independent random variables with finite variances, and let $U = X + Y$ and $V = XY$. Under what condition are $U$ and $V$ uncorrelated?
MY ATTEMPT
We say that two variables are uncorrelated if their covariance equals zero. Bearing this in mind, we have
begin{align*}
textbf{Cov}(U,V) & = textbf{E}(UV) - textbf{E}(U)textbf{E}(V)\\
& = textbf{E}(X^{2}Y + XY^{2}) - textbf{E}(X+Y)textbf{E}(XY)\\
& = textbf{E}(X^{2})textbf{E}(Y) + textbf{E}(X)textbf{
E}(Y^{2}) - textbf{E}(X)^{2}textbf{E}(Y) - textbf{E}(X)textbf{E}(Y)^{2}\\
& = textbf{E}(Y)[textbf{E}(X^{2}) - textbf{E}(X)^{2}] + textbf{E}(X)[textbf{E}(Y^{2}) - textbf{E}(Y)^{2}]\\
& = textbf{E}(Y)textbf{Var}(X) + textbf{E}(X)textbf{Var}(Y) = 0
end{align*}
Is there a specific name for this last expression? Any contribution is appreciated. Thanks :)
probability probability-theory independence covariance
asked Jan 30 at 17:35
user1337user1337
47210
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be independent random variables with finite variances, and let $U = X + Y$ and $V = XY$. Under what condition are $U$ and $V$ uncorrelated?
MY ATTEMPT
We say that two variables are uncorrelated if their covariance equals zero. Bearing this in mind, we have
begin{align*}
textbf{Cov}(U,V) & = textbf{E}(UV) - textbf{E}(U)textbf{E}(V)\\
& = textbf{E}(X^{2}Y + XY^{2}) - textbf{E}(X+Y)textbf{E}(XY)\\
& = textbf{E}(X^{2})textbf{E}(Y) + textbf{E}(X)textbf{
E}(Y^{2}) - textbf{E}(X)^{2}textbf{E}(Y) - textbf{E}(X)textbf{E}(Y)^{2}\\
& = textbf{E}(Y)[textbf{E}(X^{2}) - textbf{E}(X)^{2}] + textbf{E}(X)[textbf{E}(Y^{2}) - textbf{E}(Y)^{2}]\\
& = textbf{E}(Y)textbf{Var}(X) + textbf{E}(X)textbf{Var}(Y) = 0
end{align*}
Is there a specific name for this last expression? Any contribution is appreciated. Thanks :)
probability probability-theory independence covariance
asked Jan 30 at 17:35
user1337user1337
47210
$endgroup$
$begingroup$
"Is there a specific name for this last expression?" No.
$endgroup$
– Did
Jan 30 at 17:37
$begingroup$
Looks like you've found the condition. There is no specific name for this.
$endgroup$
– Ben
Jan 31 at 9:16
add a comment |
$begingroup$
Let $X$ and $Y$ be independent random variables with finite variances, and let $U = X + Y$ and $V = XY$. Under what condition are $U$ and $V$ uncorrelated?
MY ATTEMPT
We say that two variables are uncorrelated if their covariance equals zero. Bearing this in mind, we have
begin{align*}
textbf{Cov}(U,V) & = textbf{E}(UV) - textbf{E}(U)textbf{E}(V)\\
& = textbf{E}(X^{2}Y + XY^{2}) - textbf{E}(X+Y)textbf{E}(XY)\\
& = textbf{E}(X^{2})textbf{E}(Y) + textbf{E}(X)textbf{
E}(Y^{2}) - textbf{E}(X)^{2}textbf{E}(Y) - textbf{E}(X)textbf{E}(Y)^{2}\\
& = textbf{E}(Y)[textbf{E}(X^{2}) - textbf{E}(X)^{2}] + textbf{E}(X)[textbf{E}(Y^{2}) - textbf{E}(Y)^{2}]\\
& = textbf{E}(Y)textbf{Var}(X) + textbf{E}(X)textbf{Var}(Y) = 0
end{align*}
Is there a specific name for this last expression? Any contribution is appreciated. Thanks :)
probability probability-theory independence covariance
asked Jan 30 at 17:35
user1337user1337
47210
$endgroup$
Let $X$ and $Y$ be independent random variables with finite variances, and let $U = X + Y$ and $V = XY$. Under what condition are $U$ and $V$ uncorrelated?
MY ATTEMPT
We say that two variables are uncorrelated if their covariance equals zero. Bearing this in mind, we have
begin{align*}
textbf{Cov}(U,V) & = textbf{E}(UV) - textbf{E}(U)textbf{E}(V)\\
& = textbf{E}(X^{2}Y + XY^{2}) - textbf{E}(X+Y)textbf{E}(XY)\\
& = textbf{E}(X^{2})textbf{E}(Y) + textbf{E}(X)textbf{
E}(Y^{2}) - textbf{E}(X)^{2}textbf{E}(Y) - textbf{E}(X)textbf{E}(Y)^{2}\\
& = textbf{E}(Y)[textbf{E}(X^{2}) - textbf{E}(X)^{2}] + textbf{E}(X)[textbf{E}(Y^{2}) - textbf{E}(Y)^{2}]\\
& = textbf{E}(Y)textbf{Var}(X) + textbf{E}(X)textbf{Var}(Y) = 0
end{align*}
Is there a specific name for this last expression? Any contribution is appreciated. Thanks :)
probability probability-theory independence covariance
probability probability-theory independence covariance
asked Jan 30 at 17:35
user1337user1337
47210
asked Jan 30 at 17:35
user1337user1337
47210
asked Jan 30 at 17:35
user1337user1337
47210
asked Jan 30 at 17:35
user1337user1337
47210
asked Jan 30 at 17:35
user1337user1337
47210
47210
$begingroup$
"Is there a specific name for this last expression?" No.
$endgroup$
– Did
Jan 30 at 17:37
$begingroup$
Looks like you've found the condition. There is no specific name for this.
$endgroup$
– Ben
Jan 31 at 9:16
add a comment |
$begingroup$
"Is there a specific name for this last expression?" No.
$endgroup$
– Did
Jan 30 at 17:37
$begingroup$
Looks like you've found the condition. There is no specific name for this.
$endgroup$
– Ben
Jan 31 at 9:16
$begingroup$
"Is there a specific name for this last expression?" No.
$endgroup$
– Did
Jan 30 at 17:37
$begingroup$
"Is there a specific name for this last expression?" No.
$endgroup$
– Did
Jan 30 at 17:37
$begingroup$
Looks like you've found the condition. There is no specific name for this.
$endgroup$
– Ben
Jan 31 at 9:16
$begingroup$
Looks like you've found the condition. There is no specific name for this.
$endgroup$
– Ben
Jan 31 at 9:16
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
First of all, there is no specific name of what you've got. Second of all, assuming r.v. $X$ is not degenerated we have the following condition
$$
mathbf{E} Y = - mathbf{E} X frac{mathbf{Var} Y}{mathbf{Var} X}.
$$
answered Jan 30 at 17:41
pointguard0pointguard0
1,55211121
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First of all, there is no specific name of what you've got. Second of all, assuming r.v. $X$ is not degenerated we have the following condition
$$
mathbf{E} Y = - mathbf{E} X frac{mathbf{Var} Y}{mathbf{Var} X}.
$$
answered Jan 30 at 17:41
pointguard0pointguard0
1,55211121
$endgroup$
add a comment |
$begingroup$
First of all, there is no specific name of what you've got. Second of all, assuming r.v. $X$ is not degenerated we have the following condition
$$
mathbf{E} Y = - mathbf{E} X frac{mathbf{Var} Y}{mathbf{Var} X}.
$$
answered Jan 30 at 17:41
pointguard0pointguard0
1,55211121
$endgroup$
add a comment |
$begingroup$
First of all, there is no specific name of what you've got. Second of all, assuming r.v. $X$ is not degenerated we have the following condition
$$
mathbf{E} Y = - mathbf{E} X frac{mathbf{Var} Y}{mathbf{Var} X}.
$$
answered Jan 30 at 17:41
pointguard0pointguard0
1,55211121
$endgroup$
First of all, there is no specific name of what you've got. Second of all, assuming r.v. $X$ is not degenerated we have the following condition
$$
mathbf{E} Y = - mathbf{E} X frac{mathbf{Var} Y}{mathbf{Var} X}.
$$
answered Jan 30 at 17:41
pointguard0pointguard0
1,55211121
answered Jan 30 at 17:41
pointguard0pointguard0
1,55211121
answered Jan 30 at 17:41
pointguard0pointguard0
1,55211121
answered Jan 30 at 17:41
pointguard0pointguard0
1,55211121
1,55211121
add a comment |
add a comment |
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$begingroup$
"Is there a specific name for this last expression?" No.
$endgroup$
– Did
Jan 30 at 17:37
$begingroup$
Looks like you've found the condition. There is no specific name for this.
$endgroup$
– Ben
Jan 31 at 9:16