Mixing
Finding the distribution of a random variable made of a random vector
Firstly let's explain some signs used later:
$y_1, ldots, y_2$ are random variables,
$mathbb{E}(y_i) = mu_i$,
$text{Cov}(y_i, y_j) = sigma_{ij}$,
$Y = (y_1 ldots y_n)^{T}$,
$mathbb{E}(Y) = mu = (mu_1 ldots mu_n)^{T}$,
$text{Cov}(Y) = mathbb{E}[(Y- mu)(Y - mu)^{T}]$.
Now we can define a random vector $Y$ with normal distribution - $N(mu, Sigma)$.
Let's consider a random variable:
$$Z = (Y-mu)^{T} Sigma^{-1}(Y-mu).$$
What is the distribution of $Z$? How can it be found?
I know that one method would be to find expected value and covariance but I don't know how. Are there any other possibilities?
probability random-variables
asked Nov 20 '18 at 21:46
Hendrra
1,079516
add a comment |
Firstly let's explain some signs used later:
$y_1, ldots, y_2$ are random variables,
$mathbb{E}(y_i) = mu_i$,
$text{Cov}(y_i, y_j) = sigma_{ij}$,
$Y = (y_1 ldots y_n)^{T}$,
$mathbb{E}(Y) = mu = (mu_1 ldots mu_n)^{T}$,
$text{Cov}(Y) = mathbb{E}[(Y- mu)(Y - mu)^{T}]$.
Now we can define a random vector $Y$ with normal distribution - $N(mu, Sigma)$.
Let's consider a random variable:
$$Z = (Y-mu)^{T} Sigma^{-1}(Y-mu).$$
What is the distribution of $Z$? How can it be found?
I know that one method would be to find expected value and covariance but I don't know how. Are there any other possibilities?
probability random-variables
asked Nov 20 '18 at 21:46
Hendrra
1,079516
isn't it just a Chi-squared distribution? since Z is as sort of "square" of standard normals?
– Ramiro Scorolli
Nov 20 '18 at 22:47
@RamiroScorolli pretty much! Typically you diagonalise the quadratic form and then rewrite it as a sum of independent non-central Chi-squared random variables
– Nadiels
Nov 20 '18 at 23:21
add a comment |
Firstly let's explain some signs used later:
$y_1, ldots, y_2$ are random variables,
$mathbb{E}(y_i) = mu_i$,
$text{Cov}(y_i, y_j) = sigma_{ij}$,
$Y = (y_1 ldots y_n)^{T}$,
$mathbb{E}(Y) = mu = (mu_1 ldots mu_n)^{T}$,
$text{Cov}(Y) = mathbb{E}[(Y- mu)(Y - mu)^{T}]$.
Now we can define a random vector $Y$ with normal distribution - $N(mu, Sigma)$.
Let's consider a random variable:
$$Z = (Y-mu)^{T} Sigma^{-1}(Y-mu).$$
What is the distribution of $Z$? How can it be found?
I know that one method would be to find expected value and covariance but I don't know how. Are there any other possibilities?
probability random-variables
asked Nov 20 '18 at 21:46
Hendrra
1,079516
Firstly let's explain some signs used later:
$y_1, ldots, y_2$ are random variables,
$mathbb{E}(y_i) = mu_i$,
$text{Cov}(y_i, y_j) = sigma_{ij}$,
$Y = (y_1 ldots y_n)^{T}$,
$mathbb{E}(Y) = mu = (mu_1 ldots mu_n)^{T}$,
$text{Cov}(Y) = mathbb{E}[(Y- mu)(Y - mu)^{T}]$.
Now we can define a random vector $Y$ with normal distribution - $N(mu, Sigma)$.
Let's consider a random variable:
$$Z = (Y-mu)^{T} Sigma^{-1}(Y-mu).$$
What is the distribution of $Z$? How can it be found?
I know that one method would be to find expected value and covariance but I don't know how. Are there any other possibilities?
probability random-variables
probability random-variables
asked Nov 20 '18 at 21:46
Hendrra
1,079516
asked Nov 20 '18 at 21:46
Hendrra
1,079516
asked Nov 20 '18 at 21:46
Hendrra
1,079516
asked Nov 20 '18 at 21:46
Hendrra
1,079516
asked Nov 20 '18 at 21:46
Hendrra
1,079516
1,079516
isn't it just a Chi-squared distribution? since Z is as sort of "square" of standard normals?
– Ramiro Scorolli
Nov 20 '18 at 22:47
@RamiroScorolli pretty much! Typically you diagonalise the quadratic form and then rewrite it as a sum of independent non-central Chi-squared random variables
– Nadiels
Nov 20 '18 at 23:21
add a comment |
isn't it just a Chi-squared distribution? since Z is as sort of "square" of standard normals?
– Ramiro Scorolli
Nov 20 '18 at 22:47
@RamiroScorolli pretty much! Typically you diagonalise the quadratic form and then rewrite it as a sum of independent non-central Chi-squared random variables
– Nadiels
Nov 20 '18 at 23:21
isn't it just a Chi-squared distribution? since Z is as sort of "square" of standard normals?
– Ramiro Scorolli
Nov 20 '18 at 22:47
isn't it just a Chi-squared distribution? since Z is as sort of "square" of standard normals?
– Ramiro Scorolli
Nov 20 '18 at 22:47
@RamiroScorolli pretty much! Typically you diagonalise the quadratic form and then rewrite it as a sum of independent non-central Chi-squared random variables
– Nadiels
Nov 20 '18 at 23:21
@RamiroScorolli pretty much! Typically you diagonalise the quadratic form and then rewrite it as a sum of independent non-central Chi-squared random variables
– Nadiels
Nov 20 '18 at 23:21
add a comment |
1 Answer
1
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votes
You have that $X$ is distributed $N(mu,Sigma)$ so:
$(X-mu)sim N(0,Sigma)$
and
$Sigma^{-1/2}(X-mu)sim N(0,Sigma^{-1/2}cdotSigmacdot{Sigma^{-1/2}}^{'})$
$Sigma^{-1/2}(X-mu)sim N(0,Sigma^{-1/2}cdotSigma^{-1/2}({Sigma^{1/2}}^{'}cdotSigma^{1/2})cdot{Sigma^{-1/2}}^{'})$ Since $Sigma^{-1/2}$ is symmetric by construction (use an orthogonal eigendecomposition) this leads to:
$Sigma^{-1/2}(X-mu)sim N(0,I)$ i.e. a standard normal.
Calculating the square of this new random variable gives us:
$(Sigma^{-1/2}(X-mu))^{'}cdot Sigma^{-1/2}(X-mu)=(X-mu)^{'}Sigma^{-1}(X-mu)$ Which is precisely $Z$
Since it's the square of a Standard Normal, the distribution will be Chi-Squared with k (dimension of the vector $y$) df
answered Nov 20 '18 at 23:34
Ramiro Scorolli
655113
Thank you very much! Your solution is very smart and elegant :)
– Hendrra
Nov 21 '18 at 7:57
1
Your welcome! Glad you found it useful
– Ramiro Scorolli
Nov 21 '18 at 8:06
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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You have that $X$ is distributed $N(mu,Sigma)$ so:
$(X-mu)sim N(0,Sigma)$
and
$Sigma^{-1/2}(X-mu)sim N(0,Sigma^{-1/2}cdotSigmacdot{Sigma^{-1/2}}^{'})$
$Sigma^{-1/2}(X-mu)sim N(0,Sigma^{-1/2}cdotSigma^{-1/2}({Sigma^{1/2}}^{'}cdotSigma^{1/2})cdot{Sigma^{-1/2}}^{'})$ Since $Sigma^{-1/2}$ is symmetric by construction (use an orthogonal eigendecomposition) this leads to:
$Sigma^{-1/2}(X-mu)sim N(0,I)$ i.e. a standard normal.
Calculating the square of this new random variable gives us:
$(Sigma^{-1/2}(X-mu))^{'}cdot Sigma^{-1/2}(X-mu)=(X-mu)^{'}Sigma^{-1}(X-mu)$ Which is precisely $Z$
Since it's the square of a Standard Normal, the distribution will be Chi-Squared with k (dimension of the vector $y$) df
answered Nov 20 '18 at 23:34
Ramiro Scorolli
655113
Thank you very much! Your solution is very smart and elegant :)
– Hendrra
Nov 21 '18 at 7:57
1
Your welcome! Glad you found it useful
– Ramiro Scorolli
Nov 21 '18 at 8:06
add a comment |
You have that $X$ is distributed $N(mu,Sigma)$ so:
$(X-mu)sim N(0,Sigma)$
and
$Sigma^{-1/2}(X-mu)sim N(0,Sigma^{-1/2}cdotSigmacdot{Sigma^{-1/2}}^{'})$
$Sigma^{-1/2}(X-mu)sim N(0,Sigma^{-1/2}cdotSigma^{-1/2}({Sigma^{1/2}}^{'}cdotSigma^{1/2})cdot{Sigma^{-1/2}}^{'})$ Since $Sigma^{-1/2}$ is symmetric by construction (use an orthogonal eigendecomposition) this leads to:
$Sigma^{-1/2}(X-mu)sim N(0,I)$ i.e. a standard normal.
Calculating the square of this new random variable gives us:
$(Sigma^{-1/2}(X-mu))^{'}cdot Sigma^{-1/2}(X-mu)=(X-mu)^{'}Sigma^{-1}(X-mu)$ Which is precisely $Z$
Since it's the square of a Standard Normal, the distribution will be Chi-Squared with k (dimension of the vector $y$) df
answered Nov 20 '18 at 23:34
Ramiro Scorolli
655113
Thank you very much! Your solution is very smart and elegant :)
– Hendrra
Nov 21 '18 at 7:57
1
Your welcome! Glad you found it useful
– Ramiro Scorolli
Nov 21 '18 at 8:06
add a comment |
You have that $X$ is distributed $N(mu,Sigma)$ so:
$(X-mu)sim N(0,Sigma)$
and
$Sigma^{-1/2}(X-mu)sim N(0,Sigma^{-1/2}cdotSigmacdot{Sigma^{-1/2}}^{'})$
$Sigma^{-1/2}(X-mu)sim N(0,Sigma^{-1/2}cdotSigma^{-1/2}({Sigma^{1/2}}^{'}cdotSigma^{1/2})cdot{Sigma^{-1/2}}^{'})$ Since $Sigma^{-1/2}$ is symmetric by construction (use an orthogonal eigendecomposition) this leads to:
$Sigma^{-1/2}(X-mu)sim N(0,I)$ i.e. a standard normal.
Calculating the square of this new random variable gives us:
$(Sigma^{-1/2}(X-mu))^{'}cdot Sigma^{-1/2}(X-mu)=(X-mu)^{'}Sigma^{-1}(X-mu)$ Which is precisely $Z$
Since it's the square of a Standard Normal, the distribution will be Chi-Squared with k (dimension of the vector $y$) df
answered Nov 20 '18 at 23:34
Ramiro Scorolli
655113
You have that $X$ is distributed $N(mu,Sigma)$ so:
$(X-mu)sim N(0,Sigma)$
and
$Sigma^{-1/2}(X-mu)sim N(0,Sigma^{-1/2}cdotSigmacdot{Sigma^{-1/2}}^{'})$
$Sigma^{-1/2}(X-mu)sim N(0,Sigma^{-1/2}cdotSigma^{-1/2}({Sigma^{1/2}}^{'}cdotSigma^{1/2})cdot{Sigma^{-1/2}}^{'})$ Since $Sigma^{-1/2}$ is symmetric by construction (use an orthogonal eigendecomposition) this leads to:
$Sigma^{-1/2}(X-mu)sim N(0,I)$ i.e. a standard normal.
Calculating the square of this new random variable gives us:
$(Sigma^{-1/2}(X-mu))^{'}cdot Sigma^{-1/2}(X-mu)=(X-mu)^{'}Sigma^{-1}(X-mu)$ Which is precisely $Z$
Since it's the square of a Standard Normal, the distribution will be Chi-Squared with k (dimension of the vector $y$) df
answered Nov 20 '18 at 23:34
Ramiro Scorolli
655113
answered Nov 20 '18 at 23:34
Ramiro Scorolli
655113
answered Nov 20 '18 at 23:34
Ramiro Scorolli
655113
answered Nov 20 '18 at 23:34
Ramiro Scorolli
655113
655113
Thank you very much! Your solution is very smart and elegant :)
– Hendrra
Nov 21 '18 at 7:57
1
Your welcome! Glad you found it useful
– Ramiro Scorolli
Nov 21 '18 at 8:06
add a comment |
Thank you very much! Your solution is very smart and elegant :)
– Hendrra
Nov 21 '18 at 7:57
1
Your welcome! Glad you found it useful
– Ramiro Scorolli
Nov 21 '18 at 8:06
Thank you very much! Your solution is very smart and elegant :)
– Hendrra
Nov 21 '18 at 7:57
Thank you very much! Your solution is very smart and elegant :)
– Hendrra
Nov 21 '18 at 7:57
1
1
Your welcome! Glad you found it useful
– Ramiro Scorolli
Nov 21 '18 at 8:06
Your welcome! Glad you found it useful
– Ramiro Scorolli
Nov 21 '18 at 8:06
add a comment |
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isn't it just a Chi-squared distribution? since Z is as sort of "square" of standard normals?
– Ramiro Scorolli
Nov 20 '18 at 22:47
@RamiroScorolli pretty much! Typically you diagonalise the quadratic form and then rewrite it as a sum of independent non-central Chi-squared random variables
– Nadiels
Nov 20 '18 at 23:21