Mixing
Logistic Regression is convex proof
$begingroup$
I am trying to make sense of this paper qwone.com/~jason/writing/convexLR.pdf
"Regularized Logistic Regression is Strictly Convex" by Jason D. M. Rennie.
I am following the proof and formula (1) is a given:
$$
-ln(P(vec{y}mid X,vec{w})) = sum_{i=1}^N ln(1+e^{(-y_{i}vec{w}^Tvec{x}_i)}
$$
Assuming:
$$
g(z) = frac{1}{1+e^{-z}}
$$
I also see how
$$
1-g(z) = frac{e^{-z}}{1+e^{-z}}
$$
However, I don't follow how
$$
frac{partial g(z)}{partial z} = -g(z)(1-g(z))
$$
If I differentiate g(z) w.r.t. z I get:
$$
frac{partial g(z)}{partial z} = frac{e^{-z}}{(1+e^{-z})^2}
$$
which is $g(z)(1-g(z))$ not $-g(z)(1-g(z))$
Also, when doing (2) I get the negative of what is expressed there (taking into account it is performing the partial differential of - L.H.S. of (1)):
$$
frac{partial (-text{L.H.S. (1)} )}{partial w_j}
$$
Thanks in advance!
convex-optimization
edited Feb 21 '15 at 17:16
Michael Hardy
1
asked Feb 21 '15 at 17:14
user1064285user1064285
334
$endgroup$
add a comment |
$begingroup$
I am trying to make sense of this paper qwone.com/~jason/writing/convexLR.pdf
"Regularized Logistic Regression is Strictly Convex" by Jason D. M. Rennie.
I am following the proof and formula (1) is a given:
$$
-ln(P(vec{y}mid X,vec{w})) = sum_{i=1}^N ln(1+e^{(-y_{i}vec{w}^Tvec{x}_i)}
$$
Assuming:
$$
g(z) = frac{1}{1+e^{-z}}
$$
I also see how
$$
1-g(z) = frac{e^{-z}}{1+e^{-z}}
$$
However, I don't follow how
$$
frac{partial g(z)}{partial z} = -g(z)(1-g(z))
$$
If I differentiate g(z) w.r.t. z I get:
$$
frac{partial g(z)}{partial z} = frac{e^{-z}}{(1+e^{-z})^2}
$$
which is $g(z)(1-g(z))$ not $-g(z)(1-g(z))$
Also, when doing (2) I get the negative of what is expressed there (taking into account it is performing the partial differential of - L.H.S. of (1)):
$$
frac{partial (-text{L.H.S. (1)} )}{partial w_j}
$$
Thanks in advance!
convex-optimization
edited Feb 21 '15 at 17:16
Michael Hardy
1
asked Feb 21 '15 at 17:14
user1064285user1064285
334
$endgroup$
$begingroup$
your calculations are correct, not the paper you were reading.
$endgroup$
– Math-fun
Feb 21 '15 at 17:22
1
$begingroup$
Someone else can check? I have seen other successful proofs that Logistic Regression optimization is a convex problem, I'm just wondering if there is anything I'm not seeing...
$endgroup$
– user1064285
Feb 22 '15 at 14:29
add a comment |
$begingroup$
I am trying to make sense of this paper qwone.com/~jason/writing/convexLR.pdf
"Regularized Logistic Regression is Strictly Convex" by Jason D. M. Rennie.
I am following the proof and formula (1) is a given:
$$
-ln(P(vec{y}mid X,vec{w})) = sum_{i=1}^N ln(1+e^{(-y_{i}vec{w}^Tvec{x}_i)}
$$
Assuming:
$$
g(z) = frac{1}{1+e^{-z}}
$$
I also see how
$$
1-g(z) = frac{e^{-z}}{1+e^{-z}}
$$
However, I don't follow how
$$
frac{partial g(z)}{partial z} = -g(z)(1-g(z))
$$
If I differentiate g(z) w.r.t. z I get:
$$
frac{partial g(z)}{partial z} = frac{e^{-z}}{(1+e^{-z})^2}
$$
which is $g(z)(1-g(z))$ not $-g(z)(1-g(z))$
Also, when doing (2) I get the negative of what is expressed there (taking into account it is performing the partial differential of - L.H.S. of (1)):
$$
frac{partial (-text{L.H.S. (1)} )}{partial w_j}
$$
Thanks in advance!
convex-optimization
edited Feb 21 '15 at 17:16
Michael Hardy
1
asked Feb 21 '15 at 17:14
user1064285user1064285
334
$endgroup$
I am trying to make sense of this paper qwone.com/~jason/writing/convexLR.pdf
"Regularized Logistic Regression is Strictly Convex" by Jason D. M. Rennie.
I am following the proof and formula (1) is a given:
$$
-ln(P(vec{y}mid X,vec{w})) = sum_{i=1}^N ln(1+e^{(-y_{i}vec{w}^Tvec{x}_i)}
$$
Assuming:
$$
g(z) = frac{1}{1+e^{-z}}
$$
I also see how
$$
1-g(z) = frac{e^{-z}}{1+e^{-z}}
$$
However, I don't follow how
$$
frac{partial g(z)}{partial z} = -g(z)(1-g(z))
$$
If I differentiate g(z) w.r.t. z I get:
$$
frac{partial g(z)}{partial z} = frac{e^{-z}}{(1+e^{-z})^2}
$$
which is $g(z)(1-g(z))$ not $-g(z)(1-g(z))$
Also, when doing (2) I get the negative of what is expressed there (taking into account it is performing the partial differential of - L.H.S. of (1)):
$$
frac{partial (-text{L.H.S. (1)} )}{partial w_j}
$$
Thanks in advance!
convex-optimization
convex-optimization
edited Feb 21 '15 at 17:16
Michael Hardy
1
asked Feb 21 '15 at 17:14
user1064285user1064285
334
edited Feb 21 '15 at 17:16
Michael Hardy
1
asked Feb 21 '15 at 17:14
user1064285user1064285
334
edited Feb 21 '15 at 17:16
Michael Hardy
1
edited Feb 21 '15 at 17:16
Michael Hardy
1
edited Feb 21 '15 at 17:16
Michael Hardy
1
1
asked Feb 21 '15 at 17:14
user1064285user1064285
334
asked Feb 21 '15 at 17:14
user1064285user1064285
334
asked Feb 21 '15 at 17:14
user1064285user1064285
334
334
$begingroup$
your calculations are correct, not the paper you were reading.
$endgroup$
– Math-fun
Feb 21 '15 at 17:22
1
$begingroup$
Someone else can check? I have seen other successful proofs that Logistic Regression optimization is a convex problem, I'm just wondering if there is anything I'm not seeing...
$endgroup$
– user1064285
Feb 22 '15 at 14:29
add a comment |
$begingroup$
your calculations are correct, not the paper you were reading.
$endgroup$
– Math-fun
Feb 21 '15 at 17:22
1
$begingroup$
Someone else can check? I have seen other successful proofs that Logistic Regression optimization is a convex problem, I'm just wondering if there is anything I'm not seeing...
$endgroup$
– user1064285
Feb 22 '15 at 14:29
$begingroup$
your calculations are correct, not the paper you were reading.
$endgroup$
– Math-fun
Feb 21 '15 at 17:22
$begingroup$
your calculations are correct, not the paper you were reading.
$endgroup$
– Math-fun
Feb 21 '15 at 17:22
1
1
$begingroup$
Someone else can check? I have seen other successful proofs that Logistic Regression optimization is a convex problem, I'm just wondering if there is anything I'm not seeing...
$endgroup$
– user1064285
Feb 22 '15 at 14:29
$begingroup$
Someone else can check? I have seen other successful proofs that Logistic Regression optimization is a convex problem, I'm just wondering if there is anything I'm not seeing...
$endgroup$
– user1064285
Feb 22 '15 at 14:29
add a comment |
1 Answer
1
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$begingroup$
Here is the graph of $displaystyle g(z)=frac{1}{1+e^{-z}}$:
which is clearly increasing, hence the derivative should be positive and as your calculations show $g'(z)=g(z)(1-g(z))>0$. This could be a typo in the paper you mentioned. Further equation (2) in the "paper" is correct since it is the derivative of $displaystylelog P(.)$ not $displaystyle-log P(.)$. The second derivative in the paper is also correct.
answered Feb 22 '15 at 15:45
Math-funMath-fun
7,0061425
$endgroup$
add a comment |
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$begingroup$
Here is the graph of $displaystyle g(z)=frac{1}{1+e^{-z}}$:
which is clearly increasing, hence the derivative should be positive and as your calculations show $g'(z)=g(z)(1-g(z))>0$. This could be a typo in the paper you mentioned. Further equation (2) in the "paper" is correct since it is the derivative of $displaystylelog P(.)$ not $displaystyle-log P(.)$. The second derivative in the paper is also correct.
answered Feb 22 '15 at 15:45
Math-funMath-fun
7,0061425
$endgroup$
add a comment |
$begingroup$
Here is the graph of $displaystyle g(z)=frac{1}{1+e^{-z}}$:
which is clearly increasing, hence the derivative should be positive and as your calculations show $g'(z)=g(z)(1-g(z))>0$. This could be a typo in the paper you mentioned. Further equation (2) in the "paper" is correct since it is the derivative of $displaystylelog P(.)$ not $displaystyle-log P(.)$. The second derivative in the paper is also correct.
answered Feb 22 '15 at 15:45
Math-funMath-fun
7,0061425
$endgroup$
add a comment |
$begingroup$
Here is the graph of $displaystyle g(z)=frac{1}{1+e^{-z}}$:
which is clearly increasing, hence the derivative should be positive and as your calculations show $g'(z)=g(z)(1-g(z))>0$. This could be a typo in the paper you mentioned. Further equation (2) in the "paper" is correct since it is the derivative of $displaystylelog P(.)$ not $displaystyle-log P(.)$. The second derivative in the paper is also correct.
answered Feb 22 '15 at 15:45
Math-funMath-fun
7,0061425
$endgroup$
Here is the graph of $displaystyle g(z)=frac{1}{1+e^{-z}}$:
which is clearly increasing, hence the derivative should be positive and as your calculations show $g'(z)=g(z)(1-g(z))>0$. This could be a typo in the paper you mentioned. Further equation (2) in the "paper" is correct since it is the derivative of $displaystylelog P(.)$ not $displaystyle-log P(.)$. The second derivative in the paper is also correct.
answered Feb 22 '15 at 15:45
Math-funMath-fun
7,0061425
answered Feb 22 '15 at 15:45
Math-funMath-fun
7,0061425
answered Feb 22 '15 at 15:45
Math-funMath-fun
7,0061425
answered Feb 22 '15 at 15:45
Math-funMath-fun
7,0061425
7,0061425
add a comment |
add a comment |
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$begingroup$
your calculations are correct, not the paper you were reading.
$endgroup$
– Math-fun
Feb 21 '15 at 17:22
1
$begingroup$
Someone else can check? I have seen other successful proofs that Logistic Regression optimization is a convex problem, I'm just wondering if there is anything I'm not seeing...
$endgroup$
– user1064285
Feb 22 '15 at 14:29