Matrix differentiation involving exponential function
The function of interest is $textbf{X}'exp[textbf{X} boldsymbol{beta}]$. $textbf{X}$ is a $n times K$ matrix. The columns of $textbf{X}$ contain $K$ variables each with $n$ observations. That is, $textbf{x}_{k} = [x_{ik}, ldots, x_{nk}]'$ is a column in $textbf{X}$. $boldsymbol{beta}$ is a $K times 1$ parameter vector such that $boldsymbol{beta} = [beta_{1}, ldots, beta_{K}]'$. I need to differentiate this function with respect to $boldsymbol{beta}$. Since the function $textbf{X}'exp[textbf{X} boldsymbol{beta}]$ is $K times 1$ and parameter vector $boldsymbol{beta}$ is $K times 1$, the resulting derivative in matrix form will be $K times K$. I propose that the derivative uses the denominator layout. I happen to end up using the Hadamard product but struggle to get the final result.
matrices derivatives
asked Nov 25 at 15:26
Snoopy
235
add a comment |
The function of interest is $textbf{X}'exp[textbf{X} boldsymbol{beta}]$. $textbf{X}$ is a $n times K$ matrix. The columns of $textbf{X}$ contain $K$ variables each with $n$ observations. That is, $textbf{x}_{k} = [x_{ik}, ldots, x_{nk}]'$ is a column in $textbf{X}$. $boldsymbol{beta}$ is a $K times 1$ parameter vector such that $boldsymbol{beta} = [beta_{1}, ldots, beta_{K}]'$. I need to differentiate this function with respect to $boldsymbol{beta}$. Since the function $textbf{X}'exp[textbf{X} boldsymbol{beta}]$ is $K times 1$ and parameter vector $boldsymbol{beta}$ is $K times 1$, the resulting derivative in matrix form will be $K times K$. I propose that the derivative uses the denominator layout. I happen to end up using the Hadamard product but struggle to get the final result.
matrices derivatives
asked Nov 25 at 15:26
Snoopy
235
add a comment |
The function of interest is $textbf{X}'exp[textbf{X} boldsymbol{beta}]$. $textbf{X}$ is a $n times K$ matrix. The columns of $textbf{X}$ contain $K$ variables each with $n$ observations. That is, $textbf{x}_{k} = [x_{ik}, ldots, x_{nk}]'$ is a column in $textbf{X}$. $boldsymbol{beta}$ is a $K times 1$ parameter vector such that $boldsymbol{beta} = [beta_{1}, ldots, beta_{K}]'$. I need to differentiate this function with respect to $boldsymbol{beta}$. Since the function $textbf{X}'exp[textbf{X} boldsymbol{beta}]$ is $K times 1$ and parameter vector $boldsymbol{beta}$ is $K times 1$, the resulting derivative in matrix form will be $K times K$. I propose that the derivative uses the denominator layout. I happen to end up using the Hadamard product but struggle to get the final result.
matrices derivatives
asked Nov 25 at 15:26
Snoopy
235
The function of interest is $textbf{X}'exp[textbf{X} boldsymbol{beta}]$. $textbf{X}$ is a $n times K$ matrix. The columns of $textbf{X}$ contain $K$ variables each with $n$ observations. That is, $textbf{x}_{k} = [x_{ik}, ldots, x_{nk}]'$ is a column in $textbf{X}$. $boldsymbol{beta}$ is a $K times 1$ parameter vector such that $boldsymbol{beta} = [beta_{1}, ldots, beta_{K}]'$. I need to differentiate this function with respect to $boldsymbol{beta}$. Since the function $textbf{X}'exp[textbf{X} boldsymbol{beta}]$ is $K times 1$ and parameter vector $boldsymbol{beta}$ is $K times 1$, the resulting derivative in matrix form will be $K times K$. I propose that the derivative uses the denominator layout. I happen to end up using the Hadamard product but struggle to get the final result.
matrices derivatives
matrices derivatives
asked Nov 25 at 15:26
Snoopy
235
asked Nov 25 at 15:26
Snoopy
235
asked Nov 25 at 15:26
Snoopy
235
asked Nov 25 at 15:26
Snoopy
235
asked Nov 25 at 15:26
Snoopy
235
235
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Define some new variables
$$eqalign{
y &= Xbeta &implies dy = X,dbeta cr
e &= exp(y) &implies de = eodot dy cr
E &= {rm Diag}(e) &implies de = E,dy cr
}$$
Write the function of interest in terms of these variables.
Then find its differential and gradient.
$$eqalign{
f &= X^Te cr
df &= X^Tde = X^TE,dy = X^TEX,dbeta cr
frac{partial f}{partialbeta} &= X^TEX cr
}$$
The trick is to use a Diag operation to eliminate the Hadamard product.
answered Nov 26 at 3:23
greg
7,5001721
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
Define some new variables
$$eqalign{
y &= Xbeta &implies dy = X,dbeta cr
e &= exp(y) &implies de = eodot dy cr
E &= {rm Diag}(e) &implies de = E,dy cr
}$$
Write the function of interest in terms of these variables.
Then find its differential and gradient.
$$eqalign{
f &= X^Te cr
df &= X^Tde = X^TE,dy = X^TEX,dbeta cr
frac{partial f}{partialbeta} &= X^TEX cr
}$$
The trick is to use a Diag operation to eliminate the Hadamard product.
answered Nov 26 at 3:23
greg
7,5001721
add a comment |
Define some new variables
$$eqalign{
y &= Xbeta &implies dy = X,dbeta cr
e &= exp(y) &implies de = eodot dy cr
E &= {rm Diag}(e) &implies de = E,dy cr
}$$
Write the function of interest in terms of these variables.
Then find its differential and gradient.
$$eqalign{
f &= X^Te cr
df &= X^Tde = X^TE,dy = X^TEX,dbeta cr
frac{partial f}{partialbeta} &= X^TEX cr
}$$
The trick is to use a Diag operation to eliminate the Hadamard product.
answered Nov 26 at 3:23
greg
7,5001721
add a comment |
Define some new variables
$$eqalign{
y &= Xbeta &implies dy = X,dbeta cr
e &= exp(y) &implies de = eodot dy cr
E &= {rm Diag}(e) &implies de = E,dy cr
}$$
Write the function of interest in terms of these variables.
Then find its differential and gradient.
$$eqalign{
f &= X^Te cr
df &= X^Tde = X^TE,dy = X^TEX,dbeta cr
frac{partial f}{partialbeta} &= X^TEX cr
}$$
The trick is to use a Diag operation to eliminate the Hadamard product.
answered Nov 26 at 3:23
greg
7,5001721
Define some new variables
$$eqalign{
y &= Xbeta &implies dy = X,dbeta cr
e &= exp(y) &implies de = eodot dy cr
E &= {rm Diag}(e) &implies de = E,dy cr
}$$
Write the function of interest in terms of these variables.
Then find its differential and gradient.
$$eqalign{
f &= X^Te cr
df &= X^Tde = X^TE,dy = X^TEX,dbeta cr
frac{partial f}{partialbeta} &= X^TEX cr
}$$
The trick is to use a Diag operation to eliminate the Hadamard product.
answered Nov 26 at 3:23
greg
7,5001721
answered Nov 26 at 3:23
greg
7,5001721
answered Nov 26 at 3:23
greg
7,5001721
answered Nov 26 at 3:23
greg
7,5001721
7,5001721
add a comment |
add a comment |
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