Mixing
Interpreting a matrix calculation
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I recently came across this problem, although trivial to compute by hand - is a little challenging for me to interpret. Notably, we have three matrices:
$$vecc=
beginbmatrix
0.5 \
0.5
endbmatrix,hspace0.2in
vecx=
beginbmatrix
2 \
3
endbmatrix,hspace0.2in
vecmu=
beginbmatrix
1 \
2
endbmatrix, hspace0.2in
mathbfSigma =
beginbmatrix
2 & 1 \
1 & 4
endbmatrix
$$
We then have the following calculation, which I have solved below.
$vecc^,T(vecx-vecmu) (vecc^ TmathbfSigmavecc)^-1$
$=beginbmatrix
0.5
endbmatrix$
The calculation is trivial. However, how do I interpret this solution - assuming $vecx$ is a data vector, $vecmu$ is a mean vector, and $mathbfSigma$ is a covariance matrix?
linear-algebra
asked yesterday
Workhorse
1262
add a comment |Â
up vote
5
down vote
favorite
I recently came across this problem, although trivial to compute by hand - is a little challenging for me to interpret. Notably, we have three matrices:
$$vecc=
beginbmatrix
0.5 \
0.5
endbmatrix,hspace0.2in
vecx=
beginbmatrix
2 \
3
endbmatrix,hspace0.2in
vecmu=
beginbmatrix
1 \
2
endbmatrix, hspace0.2in
mathbfSigma =
beginbmatrix
2 & 1 \
1 & 4
endbmatrix
$$
We then have the following calculation, which I have solved below.
$vecc^,T(vecx-vecmu) (vecc^ TmathbfSigmavecc)^-1$
$=beginbmatrix
0.5
endbmatrix$
The calculation is trivial. However, how do I interpret this solution - assuming $vecx$ is a data vector, $vecmu$ is a mean vector, and $mathbfSigma$ is a covariance matrix?
linear-algebra
asked yesterday
Workhorse
1262
1
Hint: think of standardization and consider the variance of $vecc^prime vecx.$
– whuber♦
yesterday
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I recently came across this problem, although trivial to compute by hand - is a little challenging for me to interpret. Notably, we have three matrices:
$$vecc=
beginbmatrix
0.5 \
0.5
endbmatrix,hspace0.2in
vecx=
beginbmatrix
2 \
3
endbmatrix,hspace0.2in
vecmu=
beginbmatrix
1 \
2
endbmatrix, hspace0.2in
mathbfSigma =
beginbmatrix
2 & 1 \
1 & 4
endbmatrix
$$
We then have the following calculation, which I have solved below.
$vecc^,T(vecx-vecmu) (vecc^ TmathbfSigmavecc)^-1$
$=beginbmatrix
0.5
endbmatrix$
The calculation is trivial. However, how do I interpret this solution - assuming $vecx$ is a data vector, $vecmu$ is a mean vector, and $mathbfSigma$ is a covariance matrix?
linear-algebra
asked yesterday
Workhorse
1262
I recently came across this problem, although trivial to compute by hand - is a little challenging for me to interpret. Notably, we have three matrices:
$$vecc=
beginbmatrix
0.5 \
0.5
endbmatrix,hspace0.2in
vecx=
beginbmatrix
2 \
3
endbmatrix,hspace0.2in
vecmu=
beginbmatrix
1 \
2
endbmatrix, hspace0.2in
mathbfSigma =
beginbmatrix
2 & 1 \
1 & 4
endbmatrix
$$
We then have the following calculation, which I have solved below.
$vecc^,T(vecx-vecmu) (vecc^ TmathbfSigmavecc)^-1$
$=beginbmatrix
0.5
endbmatrix$
The calculation is trivial. However, how do I interpret this solution - assuming $vecx$ is a data vector, $vecmu$ is a mean vector, and $mathbfSigma$ is a covariance matrix?
linear-algebra
linear-algebra
asked yesterday
Workhorse
1262
asked yesterday
Workhorse
1262
edited yesterday
asked yesterday
Workhorse
1262
asked yesterday
Workhorse
1262
asked yesterday
Workhorse
1262
1262
1
Hint: think of standardization and consider the variance of $vecc^prime vecx.$
– whuber♦
yesterday
add a comment |Â
1
Hint: think of standardization and consider the variance of $vecc^prime vecx.$
– whuber♦
yesterday
1
1
Hint: think of standardization and consider the variance of $vecc^prime vecx.$
– whuber♦
yesterday
Hint: think of standardization and consider the variance of $vecc^prime vecx.$
– whuber♦
yesterday
add a comment |Â
1 Answer
1
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6
down vote
Consider the random variable $$y = c^T x$$Then the mean of $y$ is
$$E(y) = c^T E(x) = c^T mu$$
and variance of $y$ is
$$var(y) = E(y - c^T mu)^2 = E(y^2) - (c^T mu)^2 = E(y^2) + c^T mumu^T c tag*$$
But
$$y^2 = (c^Tx)^2 = c^Tx c^Tx= c^Txx^Tc$$
So $$E(y^2) = c^TE(xx^T)c = c^T(Sigma - mumu^T)c tag**$$
Replace $(**)$ is $(*)$, we get
$$var(y) = c^TSigma c$$
Let $z$ be a standard version of $y$, i.e.
$$z = fracy - E(y)var(y) = fracc^Tx - c^Tmuc^TSigma c = c^T(x - mu)(c^TSigma c)^-1 $$
So, you have computed the $z$ here.
answered yesterday
Ahmad Bazzi
1815
New contributor
Ahmad Bazzi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
Consider the random variable $$y = c^T x$$Then the mean of $y$ is
$$E(y) = c^T E(x) = c^T mu$$
and variance of $y$ is
$$var(y) = E(y - c^T mu)^2 = E(y^2) - (c^T mu)^2 = E(y^2) + c^T mumu^T c tag*$$
But
$$y^2 = (c^Tx)^2 = c^Tx c^Tx= c^Txx^Tc$$
So $$E(y^2) = c^TE(xx^T)c = c^T(Sigma - mumu^T)c tag**$$
Replace $(**)$ is $(*)$, we get
$$var(y) = c^TSigma c$$
Let $z$ be a standard version of $y$, i.e.
$$z = fracy - E(y)var(y) = fracc^Tx - c^Tmuc^TSigma c = c^T(x - mu)(c^TSigma c)^-1 $$
So, you have computed the $z$ here.
answered yesterday
Ahmad Bazzi
1815
New contributor
Ahmad Bazzi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
6
down vote
Consider the random variable $$y = c^T x$$Then the mean of $y$ is
$$E(y) = c^T E(x) = c^T mu$$
and variance of $y$ is
$$var(y) = E(y - c^T mu)^2 = E(y^2) - (c^T mu)^2 = E(y^2) + c^T mumu^T c tag*$$
But
$$y^2 = (c^Tx)^2 = c^Tx c^Tx= c^Txx^Tc$$
So $$E(y^2) = c^TE(xx^T)c = c^T(Sigma - mumu^T)c tag**$$
Replace $(**)$ is $(*)$, we get
$$var(y) = c^TSigma c$$
Let $z$ be a standard version of $y$, i.e.
$$z = fracy - E(y)var(y) = fracc^Tx - c^Tmuc^TSigma c = c^T(x - mu)(c^TSigma c)^-1 $$
So, you have computed the $z$ here.
answered yesterday
Ahmad Bazzi
1815
New contributor
Ahmad Bazzi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Consider the random variable $$y = c^T x$$Then the mean of $y$ is
$$E(y) = c^T E(x) = c^T mu$$
and variance of $y$ is
$$var(y) = E(y - c^T mu)^2 = E(y^2) - (c^T mu)^2 = E(y^2) + c^T mumu^T c tag*$$
But
$$y^2 = (c^Tx)^2 = c^Tx c^Tx= c^Txx^Tc$$
So $$E(y^2) = c^TE(xx^T)c = c^T(Sigma - mumu^T)c tag**$$
Replace $(**)$ is $(*)$, we get
$$var(y) = c^TSigma c$$
Let $z$ be a standard version of $y$, i.e.
$$z = fracy - E(y)var(y) = fracc^Tx - c^Tmuc^TSigma c = c^T(x - mu)(c^TSigma c)^-1 $$
So, you have computed the $z$ here.
answered yesterday
Ahmad Bazzi
1815
New contributor
Ahmad Bazzi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Consider the random variable $$y = c^T x$$Then the mean of $y$ is
$$E(y) = c^T E(x) = c^T mu$$
and variance of $y$ is
$$var(y) = E(y - c^T mu)^2 = E(y^2) - (c^T mu)^2 = E(y^2) + c^T mumu^T c tag*$$
But
$$y^2 = (c^Tx)^2 = c^Tx c^Tx= c^Txx^Tc$$
So $$E(y^2) = c^TE(xx^T)c = c^T(Sigma - mumu^T)c tag**$$
Replace $(**)$ is $(*)$, we get
$$var(y) = c^TSigma c$$
Let $z$ be a standard version of $y$, i.e.
$$z = fracy - E(y)var(y) = fracc^Tx - c^Tmuc^TSigma c = c^T(x - mu)(c^TSigma c)^-1 $$
So, you have computed the $z$ here.
answered yesterday
Ahmad Bazzi
1815
New contributor
Ahmad Bazzi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
answered yesterday
Ahmad Bazzi
1815
New contributor
Ahmad Bazzi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Ahmad Bazzi
1815
answered yesterday
Ahmad Bazzi
1815
1815
New contributor
Ahmad Bazzi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ahmad Bazzi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ahmad Bazzi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
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1
Hint: think of standardization and consider the variance of $vecc^prime vecx.$
– whuber♦
yesterday