Matrix calculus for statistics
I want to calculate the variance of a sum of random variables $ (X_1+X_2) $.
Doing statistics, I have to learn maths starting from the end and it is quite difficult, yet very interesting (please consider that I only have the very basic skills in maths).
For now I am doing this calculus manually with the formula $ mathrm{var}(X_1+X_2) = mathrm{var}(X_1) + mathrm{var}(X_2) + 2mathrm{cov}(X_1,X_2)$.
But I am now facing much larger sums (with some minus) and being able to do so with matrix calculation would save me a lot of time (and would be very satisfying too).
I searched every resource in matrix calculus but couldn't find anything usable with my knowledge.
How can I do this calculus from the variance-covariance matrix
$$
begin{pmatrix}
mathrm{var}(X_1) & mathrm{cov}(X_1,X_2) \
mathrm{cov}(X_1,X_2) & mathrm{var}(X_2) \
end{pmatrix}
$$
[preferentially extended to substractions and n terms, like $ (X_1+X_2-X_3) $]?
NB: this is not a statistic question and doesn't belong to stats.stackexchange. I want to understand the thought process of turning scalar calculation to matrix.
matrices descriptive-statistics
add a comment |
I want to calculate the variance of a sum of random variables $ (X_1+X_2) $.
Doing statistics, I have to learn maths starting from the end and it is quite difficult, yet very interesting (please consider that I only have the very basic skills in maths).
For now I am doing this calculus manually with the formula $ mathrm{var}(X_1+X_2) = mathrm{var}(X_1) + mathrm{var}(X_2) + 2mathrm{cov}(X_1,X_2)$.
But I am now facing much larger sums (with some minus) and being able to do so with matrix calculation would save me a lot of time (and would be very satisfying too).
I searched every resource in matrix calculus but couldn't find anything usable with my knowledge.
How can I do this calculus from the variance-covariance matrix
$$
begin{pmatrix}
mathrm{var}(X_1) & mathrm{cov}(X_1,X_2) \
mathrm{cov}(X_1,X_2) & mathrm{var}(X_2) \
end{pmatrix}
$$
[preferentially extended to substractions and n terms, like $ (X_1+X_2-X_3) $]?
NB: this is not a statistic question and doesn't belong to stats.stackexchange. I want to understand the thought process of turning scalar calculation to matrix.
matrices descriptive-statistics
add a comment |
I want to calculate the variance of a sum of random variables $ (X_1+X_2) $.
Doing statistics, I have to learn maths starting from the end and it is quite difficult, yet very interesting (please consider that I only have the very basic skills in maths).
For now I am doing this calculus manually with the formula $ mathrm{var}(X_1+X_2) = mathrm{var}(X_1) + mathrm{var}(X_2) + 2mathrm{cov}(X_1,X_2)$.
But I am now facing much larger sums (with some minus) and being able to do so with matrix calculation would save me a lot of time (and would be very satisfying too).
I searched every resource in matrix calculus but couldn't find anything usable with my knowledge.
How can I do this calculus from the variance-covariance matrix
$$
begin{pmatrix}
mathrm{var}(X_1) & mathrm{cov}(X_1,X_2) \
mathrm{cov}(X_1,X_2) & mathrm{var}(X_2) \
end{pmatrix}
$$
[preferentially extended to substractions and n terms, like $ (X_1+X_2-X_3) $]?
NB: this is not a statistic question and doesn't belong to stats.stackexchange. I want to understand the thought process of turning scalar calculation to matrix.
matrices descriptive-statistics
I want to calculate the variance of a sum of random variables $ (X_1+X_2) $.
Doing statistics, I have to learn maths starting from the end and it is quite difficult, yet very interesting (please consider that I only have the very basic skills in maths).
For now I am doing this calculus manually with the formula $ mathrm{var}(X_1+X_2) = mathrm{var}(X_1) + mathrm{var}(X_2) + 2mathrm{cov}(X_1,X_2)$.
But I am now facing much larger sums (with some minus) and being able to do so with matrix calculation would save me a lot of time (and would be very satisfying too).
I searched every resource in matrix calculus but couldn't find anything usable with my knowledge.
How can I do this calculus from the variance-covariance matrix
$$
begin{pmatrix}
mathrm{var}(X_1) & mathrm{cov}(X_1,X_2) \
mathrm{cov}(X_1,X_2) & mathrm{var}(X_2) \
end{pmatrix}
$$
[preferentially extended to substractions and n terms, like $ (X_1+X_2-X_3) $]?
NB: this is not a statistic question and doesn't belong to stats.stackexchange. I want to understand the thought process of turning scalar calculation to matrix.
matrices descriptive-statistics
matrices descriptive-statistics
edited Nov 22 '18 at 15:10
Jean-Claude Arbaut
14.7k63464
14.7k63464
asked Nov 22 '18 at 12:31
Dan Chaltiel
1134
1134
add a comment |
add a comment |
2 Answers
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The variance-covariance matrix of $X$ is $frac1n(X-bar X)^T(X-bar X)$.
Now, you want to compute the variance of the vector $u=Xbeta$ for some vector $beta$. This variance is
$$Var(u)=frac1n(u-bar u)^T(u-bar u)=frac1n(Xbeta-bar Xbeta)^T(Xbeta-bar Xbeta)\
=frac1nbeta^T(X-bar X)^T(X-bar X)beta=beta^TVar(X)beta$$
+1 Elegant solution
– caverac
Nov 22 '18 at 13:12
Great answer, thanks ! Learned a lot with this !
– Dan Chaltiel
Nov 22 '18 at 13:28
add a comment |
The key point here is that
$$
mathbb{V}{rm ar}[X] = mathbb{C}{rm ov}[X, X]
$$
so that you can express your first expression as
begin{eqnarray}
mathbb{V}{rm ar}[a_1 X_1 + a_2 X_2] &=& a_1^2mathbb{V}{rm ar}[X_1] + a_2^2mathbb{V}{rm ar}[X_2] + 2 a_1a_2 mathbb{C}{rm ov}[X_1, X_2] \
&=& a_1^2 mathbb{C}{rm ov}[X_1, X_1] + a_2^2 mathbb{C}{rm ov}[X_2, X_2] + 2a_1a_2 mathbb{C}{rm ov}[X_1, X_2] \
&=& a_1^2 mathbb{C}{rm ov}[X_1, X_1] + a_2^2 mathbb{C}{rm ov}[X_2, X_2] + a_1a_2 mathbb{C}{rm ov}[X_1, X_2] + a_2a_1 mathbb{C}{rm ov}[X_2, X_1] \
&=& sum_{i=1}^2 sum_{j=1}^2 a_i a_j mathbb{C}{rm ov}[X_i, X_j]
end{eqnarray}
In general
begin{eqnarray}
mathbb{V}{rm ar}[a_1 X_1 + cdots a_n X_n] &=& sum_{i=1}^n sum_{j=1}^n a_i a_j mathbb{C}{rm ov}[X_i, X_j]
end{eqnarray}
1
+1, but I think the OP wanted to see how this is done with matrices, that is how to write it as $a^TVar(X)a$.
– Jean-Claude Arbaut
Nov 22 '18 at 12:57
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The variance-covariance matrix of $X$ is $frac1n(X-bar X)^T(X-bar X)$.
Now, you want to compute the variance of the vector $u=Xbeta$ for some vector $beta$. This variance is
$$Var(u)=frac1n(u-bar u)^T(u-bar u)=frac1n(Xbeta-bar Xbeta)^T(Xbeta-bar Xbeta)\
=frac1nbeta^T(X-bar X)^T(X-bar X)beta=beta^TVar(X)beta$$
+1 Elegant solution
– caverac
Nov 22 '18 at 13:12
Great answer, thanks ! Learned a lot with this !
– Dan Chaltiel
Nov 22 '18 at 13:28
add a comment |
The variance-covariance matrix of $X$ is $frac1n(X-bar X)^T(X-bar X)$.
Now, you want to compute the variance of the vector $u=Xbeta$ for some vector $beta$. This variance is
$$Var(u)=frac1n(u-bar u)^T(u-bar u)=frac1n(Xbeta-bar Xbeta)^T(Xbeta-bar Xbeta)\
=frac1nbeta^T(X-bar X)^T(X-bar X)beta=beta^TVar(X)beta$$
+1 Elegant solution
– caverac
Nov 22 '18 at 13:12
Great answer, thanks ! Learned a lot with this !
– Dan Chaltiel
Nov 22 '18 at 13:28
add a comment |
The variance-covariance matrix of $X$ is $frac1n(X-bar X)^T(X-bar X)$.
Now, you want to compute the variance of the vector $u=Xbeta$ for some vector $beta$. This variance is
$$Var(u)=frac1n(u-bar u)^T(u-bar u)=frac1n(Xbeta-bar Xbeta)^T(Xbeta-bar Xbeta)\
=frac1nbeta^T(X-bar X)^T(X-bar X)beta=beta^TVar(X)beta$$
The variance-covariance matrix of $X$ is $frac1n(X-bar X)^T(X-bar X)$.
Now, you want to compute the variance of the vector $u=Xbeta$ for some vector $beta$. This variance is
$$Var(u)=frac1n(u-bar u)^T(u-bar u)=frac1n(Xbeta-bar Xbeta)^T(Xbeta-bar Xbeta)\
=frac1nbeta^T(X-bar X)^T(X-bar X)beta=beta^TVar(X)beta$$
edited Nov 22 '18 at 13:25
answered Nov 22 '18 at 13:00
Jean-Claude Arbaut
14.7k63464
14.7k63464
+1 Elegant solution
– caverac
Nov 22 '18 at 13:12
Great answer, thanks ! Learned a lot with this !
– Dan Chaltiel
Nov 22 '18 at 13:28
add a comment |
+1 Elegant solution
– caverac
Nov 22 '18 at 13:12
Great answer, thanks ! Learned a lot with this !
– Dan Chaltiel
Nov 22 '18 at 13:28
+1 Elegant solution
– caverac
Nov 22 '18 at 13:12
+1 Elegant solution
– caverac
Nov 22 '18 at 13:12
Great answer, thanks ! Learned a lot with this !
– Dan Chaltiel
Nov 22 '18 at 13:28
Great answer, thanks ! Learned a lot with this !
– Dan Chaltiel
Nov 22 '18 at 13:28
add a comment |
The key point here is that
$$
mathbb{V}{rm ar}[X] = mathbb{C}{rm ov}[X, X]
$$
so that you can express your first expression as
begin{eqnarray}
mathbb{V}{rm ar}[a_1 X_1 + a_2 X_2] &=& a_1^2mathbb{V}{rm ar}[X_1] + a_2^2mathbb{V}{rm ar}[X_2] + 2 a_1a_2 mathbb{C}{rm ov}[X_1, X_2] \
&=& a_1^2 mathbb{C}{rm ov}[X_1, X_1] + a_2^2 mathbb{C}{rm ov}[X_2, X_2] + 2a_1a_2 mathbb{C}{rm ov}[X_1, X_2] \
&=& a_1^2 mathbb{C}{rm ov}[X_1, X_1] + a_2^2 mathbb{C}{rm ov}[X_2, X_2] + a_1a_2 mathbb{C}{rm ov}[X_1, X_2] + a_2a_1 mathbb{C}{rm ov}[X_2, X_1] \
&=& sum_{i=1}^2 sum_{j=1}^2 a_i a_j mathbb{C}{rm ov}[X_i, X_j]
end{eqnarray}
In general
begin{eqnarray}
mathbb{V}{rm ar}[a_1 X_1 + cdots a_n X_n] &=& sum_{i=1}^n sum_{j=1}^n a_i a_j mathbb{C}{rm ov}[X_i, X_j]
end{eqnarray}
1
+1, but I think the OP wanted to see how this is done with matrices, that is how to write it as $a^TVar(X)a$.
– Jean-Claude Arbaut
Nov 22 '18 at 12:57
add a comment |
The key point here is that
$$
mathbb{V}{rm ar}[X] = mathbb{C}{rm ov}[X, X]
$$
so that you can express your first expression as
begin{eqnarray}
mathbb{V}{rm ar}[a_1 X_1 + a_2 X_2] &=& a_1^2mathbb{V}{rm ar}[X_1] + a_2^2mathbb{V}{rm ar}[X_2] + 2 a_1a_2 mathbb{C}{rm ov}[X_1, X_2] \
&=& a_1^2 mathbb{C}{rm ov}[X_1, X_1] + a_2^2 mathbb{C}{rm ov}[X_2, X_2] + 2a_1a_2 mathbb{C}{rm ov}[X_1, X_2] \
&=& a_1^2 mathbb{C}{rm ov}[X_1, X_1] + a_2^2 mathbb{C}{rm ov}[X_2, X_2] + a_1a_2 mathbb{C}{rm ov}[X_1, X_2] + a_2a_1 mathbb{C}{rm ov}[X_2, X_1] \
&=& sum_{i=1}^2 sum_{j=1}^2 a_i a_j mathbb{C}{rm ov}[X_i, X_j]
end{eqnarray}
In general
begin{eqnarray}
mathbb{V}{rm ar}[a_1 X_1 + cdots a_n X_n] &=& sum_{i=1}^n sum_{j=1}^n a_i a_j mathbb{C}{rm ov}[X_i, X_j]
end{eqnarray}
1
+1, but I think the OP wanted to see how this is done with matrices, that is how to write it as $a^TVar(X)a$.
– Jean-Claude Arbaut
Nov 22 '18 at 12:57
add a comment |
The key point here is that
$$
mathbb{V}{rm ar}[X] = mathbb{C}{rm ov}[X, X]
$$
so that you can express your first expression as
begin{eqnarray}
mathbb{V}{rm ar}[a_1 X_1 + a_2 X_2] &=& a_1^2mathbb{V}{rm ar}[X_1] + a_2^2mathbb{V}{rm ar}[X_2] + 2 a_1a_2 mathbb{C}{rm ov}[X_1, X_2] \
&=& a_1^2 mathbb{C}{rm ov}[X_1, X_1] + a_2^2 mathbb{C}{rm ov}[X_2, X_2] + 2a_1a_2 mathbb{C}{rm ov}[X_1, X_2] \
&=& a_1^2 mathbb{C}{rm ov}[X_1, X_1] + a_2^2 mathbb{C}{rm ov}[X_2, X_2] + a_1a_2 mathbb{C}{rm ov}[X_1, X_2] + a_2a_1 mathbb{C}{rm ov}[X_2, X_1] \
&=& sum_{i=1}^2 sum_{j=1}^2 a_i a_j mathbb{C}{rm ov}[X_i, X_j]
end{eqnarray}
In general
begin{eqnarray}
mathbb{V}{rm ar}[a_1 X_1 + cdots a_n X_n] &=& sum_{i=1}^n sum_{j=1}^n a_i a_j mathbb{C}{rm ov}[X_i, X_j]
end{eqnarray}
The key point here is that
$$
mathbb{V}{rm ar}[X] = mathbb{C}{rm ov}[X, X]
$$
so that you can express your first expression as
begin{eqnarray}
mathbb{V}{rm ar}[a_1 X_1 + a_2 X_2] &=& a_1^2mathbb{V}{rm ar}[X_1] + a_2^2mathbb{V}{rm ar}[X_2] + 2 a_1a_2 mathbb{C}{rm ov}[X_1, X_2] \
&=& a_1^2 mathbb{C}{rm ov}[X_1, X_1] + a_2^2 mathbb{C}{rm ov}[X_2, X_2] + 2a_1a_2 mathbb{C}{rm ov}[X_1, X_2] \
&=& a_1^2 mathbb{C}{rm ov}[X_1, X_1] + a_2^2 mathbb{C}{rm ov}[X_2, X_2] + a_1a_2 mathbb{C}{rm ov}[X_1, X_2] + a_2a_1 mathbb{C}{rm ov}[X_2, X_1] \
&=& sum_{i=1}^2 sum_{j=1}^2 a_i a_j mathbb{C}{rm ov}[X_i, X_j]
end{eqnarray}
In general
begin{eqnarray}
mathbb{V}{rm ar}[a_1 X_1 + cdots a_n X_n] &=& sum_{i=1}^n sum_{j=1}^n a_i a_j mathbb{C}{rm ov}[X_i, X_j]
end{eqnarray}
answered Nov 22 '18 at 12:56
caverac
13.9k21130
13.9k21130
1
+1, but I think the OP wanted to see how this is done with matrices, that is how to write it as $a^TVar(X)a$.
– Jean-Claude Arbaut
Nov 22 '18 at 12:57
add a comment |
1
+1, but I think the OP wanted to see how this is done with matrices, that is how to write it as $a^TVar(X)a$.
– Jean-Claude Arbaut
Nov 22 '18 at 12:57
1
1
+1, but I think the OP wanted to see how this is done with matrices, that is how to write it as $a^TVar(X)a$.
– Jean-Claude Arbaut
Nov 22 '18 at 12:57
+1, but I think the OP wanted to see how this is done with matrices, that is how to write it as $a^TVar(X)a$.
– Jean-Claude Arbaut
Nov 22 '18 at 12:57
add a comment |
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