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Finding argmax of an integral — can I take logarithm of one of the terms?
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Suppose I need to solve for the argmax of $x$ for an integral that can be factored as follows:
$$
A = mathrmargmax_x int_Y f(x,y)g(y)dy
$$
Where both $f()$ and $g()$ are strictly positive, and the form of $f(x,y)$ is sufficiently horrible that I can't solve the integral directly. However, what is easy to solve is this different integral
$$
int_Y logleft[f(x,y)right]g(y)dy
$$
Since $x$ only appears in the $f()$ function, and the integrand is factored like so, it seems to me that the argmax of the second integral is equal to the argmax of the first integral, because the logarithm is continuous and strictly increasing. So, am I allowed to do:
$$
A = mathrmargmax_x int_Y logleft[f(x,y)right]g(y)dy
$$
?
I think I can, but I want to double check to see if there are any subtleties to this.
(If it helps to give stronger conditions / motivate what I'm trying to do: $f()$ and $g()$ are both probability density functions, and I need to marginalize over $y$ and get the MLE for $x$. Assume that the functions are continuous and otherwise obey enough regularity conditions for all of this to make sense).
integration optimization logarithms
asked Aug 2 at 19:53
dain
1084
add a comment |Â
up vote
0
down vote
favorite
Suppose I need to solve for the argmax of $x$ for an integral that can be factored as follows:
$$
A = mathrmargmax_x int_Y f(x,y)g(y)dy
$$
Where both $f()$ and $g()$ are strictly positive, and the form of $f(x,y)$ is sufficiently horrible that I can't solve the integral directly. However, what is easy to solve is this different integral
$$
int_Y logleft[f(x,y)right]g(y)dy
$$
Since $x$ only appears in the $f()$ function, and the integrand is factored like so, it seems to me that the argmax of the second integral is equal to the argmax of the first integral, because the logarithm is continuous and strictly increasing. So, am I allowed to do:
$$
A = mathrmargmax_x int_Y logleft[f(x,y)right]g(y)dy
$$
?
I think I can, but I want to double check to see if there are any subtleties to this.
(If it helps to give stronger conditions / motivate what I'm trying to do: $f()$ and $g()$ are both probability density functions, and I need to marginalize over $y$ and get the MLE for $x$. Assume that the functions are continuous and otherwise obey enough regularity conditions for all of this to make sense).
integration optimization logarithms
asked Aug 2 at 19:53
dain
1084
1
No you cannot, just like you cannot take the logarithm of terms of a sum.
– LinAlg
Aug 2 at 22:02
I know that it changes the value of the integral, but I don't care about the value of the integral. All I care about is whether the operation preserves the argmax of $x$, and I can't see why it doesn't.
– dain
Aug 2 at 23:43
you can take the logarithm of a sum and keep the same maximizer but you cannot take the logarithm of the individual terms (which is essentially what you are doing with the integral).
– LinAlg
Aug 3 at 1:28
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose I need to solve for the argmax of $x$ for an integral that can be factored as follows:
$$
A = mathrmargmax_x int_Y f(x,y)g(y)dy
$$
Where both $f()$ and $g()$ are strictly positive, and the form of $f(x,y)$ is sufficiently horrible that I can't solve the integral directly. However, what is easy to solve is this different integral
$$
int_Y logleft[f(x,y)right]g(y)dy
$$
Since $x$ only appears in the $f()$ function, and the integrand is factored like so, it seems to me that the argmax of the second integral is equal to the argmax of the first integral, because the logarithm is continuous and strictly increasing. So, am I allowed to do:
$$
A = mathrmargmax_x int_Y logleft[f(x,y)right]g(y)dy
$$
?
I think I can, but I want to double check to see if there are any subtleties to this.
(If it helps to give stronger conditions / motivate what I'm trying to do: $f()$ and $g()$ are both probability density functions, and I need to marginalize over $y$ and get the MLE for $x$. Assume that the functions are continuous and otherwise obey enough regularity conditions for all of this to make sense).
integration optimization logarithms
asked Aug 2 at 19:53
dain
1084
Suppose I need to solve for the argmax of $x$ for an integral that can be factored as follows:
$$
A = mathrmargmax_x int_Y f(x,y)g(y)dy
$$
Where both $f()$ and $g()$ are strictly positive, and the form of $f(x,y)$ is sufficiently horrible that I can't solve the integral directly. However, what is easy to solve is this different integral
$$
int_Y logleft[f(x,y)right]g(y)dy
$$
Since $x$ only appears in the $f()$ function, and the integrand is factored like so, it seems to me that the argmax of the second integral is equal to the argmax of the first integral, because the logarithm is continuous and strictly increasing. So, am I allowed to do:
$$
A = mathrmargmax_x int_Y logleft[f(x,y)right]g(y)dy
$$
?
I think I can, but I want to double check to see if there are any subtleties to this.
(If it helps to give stronger conditions / motivate what I'm trying to do: $f()$ and $g()$ are both probability density functions, and I need to marginalize over $y$ and get the MLE for $x$. Assume that the functions are continuous and otherwise obey enough regularity conditions for all of this to make sense).
integration optimization logarithms
asked Aug 2 at 19:53
dain
1084
edited Aug 3 at 11:29
asked Aug 2 at 19:53
dain
1084
asked Aug 2 at 19:53
dain
1084
asked Aug 2 at 19:53
dain
1084
1084
1
No you cannot, just like you cannot take the logarithm of terms of a sum.
– LinAlg
Aug 2 at 22:02
I know that it changes the value of the integral, but I don't care about the value of the integral. All I care about is whether the operation preserves the argmax of $x$, and I can't see why it doesn't.
– dain
Aug 2 at 23:43
you can take the logarithm of a sum and keep the same maximizer but you cannot take the logarithm of the individual terms (which is essentially what you are doing with the integral).
– LinAlg
Aug 3 at 1:28
add a comment |Â
1
No you cannot, just like you cannot take the logarithm of terms of a sum.
– LinAlg
Aug 2 at 22:02
I know that it changes the value of the integral, but I don't care about the value of the integral. All I care about is whether the operation preserves the argmax of $x$, and I can't see why it doesn't.
– dain
Aug 2 at 23:43
you can take the logarithm of a sum and keep the same maximizer but you cannot take the logarithm of the individual terms (which is essentially what you are doing with the integral).
– LinAlg
Aug 3 at 1:28
1
1
No you cannot, just like you cannot take the logarithm of terms of a sum.
– LinAlg
Aug 2 at 22:02
No you cannot, just like you cannot take the logarithm of terms of a sum.
– LinAlg
Aug 2 at 22:02
I know that it changes the value of the integral, but I don't care about the value of the integral. All I care about is whether the operation preserves the argmax of $x$, and I can't see why it doesn't.
– dain
Aug 2 at 23:43
I know that it changes the value of the integral, but I don't care about the value of the integral. All I care about is whether the operation preserves the argmax of $x$, and I can't see why it doesn't.
– dain
Aug 2 at 23:43
you can take the logarithm of a sum and keep the same maximizer but you cannot take the logarithm of the individual terms (which is essentially what you are doing with the integral).
– LinAlg
Aug 3 at 1:28
you can take the logarithm of a sum and keep the same maximizer but you cannot take the logarithm of the individual terms (which is essentially what you are doing with the integral).
– LinAlg
Aug 3 at 1:28
add a comment |Â
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1
No you cannot, just like you cannot take the logarithm of terms of a sum.
– LinAlg
Aug 2 at 22:02
I know that it changes the value of the integral, but I don't care about the value of the integral. All I care about is whether the operation preserves the argmax of $x$, and I can't see why it doesn't.
– dain
Aug 2 at 23:43
you can take the logarithm of a sum and keep the same maximizer but you cannot take the logarithm of the individual terms (which is essentially what you are doing with the integral).
– LinAlg
Aug 3 at 1:28