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Jacobian of linear map, with variable matrix
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Let $beta in mathbbR^k, x in mathbbR^q$, and $A: mathbbR^k to mathbbR^k times q$ a matrix function of $beta$. Define
$$
h(beta) = A(beta) x,
$$
so that $h:mathbbR^k to mathbbR^k$. I would like the Jacobian of $h$ with respect to $beta$, but it seems like that would require some notion of a tensor derivative that I am unfamiliar with. I am having a hard time figuring out where to even begin looking, since any search results involving
"linear map" assumes $A$ is constant.
If it helps, $A$ has some structure. Specifically, $A(beta) = mathbbE[g(beta, w)z'] B$ where $z in mathbbR^q$, $w$ are random variables and $B$ is a constant $mathbbR^q times q$ matrix. The function $g : mathbbR^k times mathcalW to mathbbR^k$, where $mathcalW$ is the space the random variable $w$ lies in. It is reasonable to assume that the derivative can "pass through" the expectation.
If it helps further, we can assume that $g$ is affine/linear in $beta$.
calculus
asked Jul 25 at 15:40
Matt
1566
add a comment |Â
up vote
0
down vote
favorite
Let $beta in mathbbR^k, x in mathbbR^q$, and $A: mathbbR^k to mathbbR^k times q$ a matrix function of $beta$. Define
$$
h(beta) = A(beta) x,
$$
so that $h:mathbbR^k to mathbbR^k$. I would like the Jacobian of $h$ with respect to $beta$, but it seems like that would require some notion of a tensor derivative that I am unfamiliar with. I am having a hard time figuring out where to even begin looking, since any search results involving
"linear map" assumes $A$ is constant.
If it helps, $A$ has some structure. Specifically, $A(beta) = mathbbE[g(beta, w)z'] B$ where $z in mathbbR^q$, $w$ are random variables and $B$ is a constant $mathbbR^q times q$ matrix. The function $g : mathbbR^k times mathcalW to mathbbR^k$, where $mathcalW$ is the space the random variable $w$ lies in. It is reasonable to assume that the derivative can "pass through" the expectation.
If it helps further, we can assume that $g$ is affine/linear in $beta$.
calculus
asked Jul 25 at 15:40
Matt
1566
It might help to write $A(beta)$ as a matrix with rows of the form $a_i(beta)$ and then perform the derivation step by step starting from $h_i(beta)=a_i(beta)xin mathbbR$.
– WalterJ
Jul 25 at 16:09
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $beta in mathbbR^k, x in mathbbR^q$, and $A: mathbbR^k to mathbbR^k times q$ a matrix function of $beta$. Define
$$
h(beta) = A(beta) x,
$$
so that $h:mathbbR^k to mathbbR^k$. I would like the Jacobian of $h$ with respect to $beta$, but it seems like that would require some notion of a tensor derivative that I am unfamiliar with. I am having a hard time figuring out where to even begin looking, since any search results involving
"linear map" assumes $A$ is constant.
If it helps, $A$ has some structure. Specifically, $A(beta) = mathbbE[g(beta, w)z'] B$ where $z in mathbbR^q$, $w$ are random variables and $B$ is a constant $mathbbR^q times q$ matrix. The function $g : mathbbR^k times mathcalW to mathbbR^k$, where $mathcalW$ is the space the random variable $w$ lies in. It is reasonable to assume that the derivative can "pass through" the expectation.
If it helps further, we can assume that $g$ is affine/linear in $beta$.
calculus
asked Jul 25 at 15:40
Matt
1566
Let $beta in mathbbR^k, x in mathbbR^q$, and $A: mathbbR^k to mathbbR^k times q$ a matrix function of $beta$. Define
$$
h(beta) = A(beta) x,
$$
so that $h:mathbbR^k to mathbbR^k$. I would like the Jacobian of $h$ with respect to $beta$, but it seems like that would require some notion of a tensor derivative that I am unfamiliar with. I am having a hard time figuring out where to even begin looking, since any search results involving
"linear map" assumes $A$ is constant.
If it helps, $A$ has some structure. Specifically, $A(beta) = mathbbE[g(beta, w)z'] B$ where $z in mathbbR^q$, $w$ are random variables and $B$ is a constant $mathbbR^q times q$ matrix. The function $g : mathbbR^k times mathcalW to mathbbR^k$, where $mathcalW$ is the space the random variable $w$ lies in. It is reasonable to assume that the derivative can "pass through" the expectation.
If it helps further, we can assume that $g$ is affine/linear in $beta$.
calculus
asked Jul 25 at 15:40
Matt
1566
edited Jul 25 at 15:50
asked Jul 25 at 15:40
Matt
1566
asked Jul 25 at 15:40
Matt
1566
asked Jul 25 at 15:40
Matt
1566
1566
It might help to write $A(beta)$ as a matrix with rows of the form $a_i(beta)$ and then perform the derivation step by step starting from $h_i(beta)=a_i(beta)xin mathbbR$.
– WalterJ
Jul 25 at 16:09
add a comment |Â
It might help to write $A(beta)$ as a matrix with rows of the form $a_i(beta)$ and then perform the derivation step by step starting from $h_i(beta)=a_i(beta)xin mathbbR$.
– WalterJ
Jul 25 at 16:09
It might help to write $A(beta)$ as a matrix with rows of the form $a_i(beta)$ and then perform the derivation step by step starting from $h_i(beta)=a_i(beta)xin mathbbR$.
– WalterJ
Jul 25 at 16:09
It might help to write $A(beta)$ as a matrix with rows of the form $a_i(beta)$ and then perform the derivation step by step starting from $h_i(beta)=a_i(beta)xin mathbbR$.
– WalterJ
Jul 25 at 16:09
add a comment |Â
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It might help to write $A(beta)$ as a matrix with rows of the form $a_i(beta)$ and then perform the derivation step by step starting from $h_i(beta)=a_i(beta)xin mathbbR$.
– WalterJ
Jul 25 at 16:09