Variance of the gradient function
$begingroup$
I was reading the Xavier initialization technique used in deep learning. But I'm struct in the basic math used in the paper. For example for a dense artificial neural network using symmetric activation function $f$ with unit derivative at $0$ (i.e. $f′(0) = 1$), if we write $z^i$ for the activation vector of layer $i$, and $s^i$ the argument vector of the activation function at layer $i$, we have $s^i = z^i W^i + b^i$ and $z^{i+1} = f(s^i)$. From these definitions we obtain the following:
$$
dfrac{partial Cost}{partial s^i_k}
=
f'(s_k^i) W^{i+1}_{k^{i'}} dfrac{partial Cost}{partial s^{i+1}}
$$
$$
dfrac{partial Cost}{partial w^i_{l,k}}
=
z^i_l dfrac{partial Cost}{partial s^{i}_k}
$$
Consider the hypothesis that we are in a linear regime at the initialization, that the weights are initialized independently and that the inputs features variances are the same ($= Var[x]$). Then we can say that, with $n_i$ the size of layer $i$ and $x$ the network input,
$$
f'(s_k^i) approx 1
$$
$$
Varleft[ z^i right]
=
Varleft[ x right]
prod_{i'=0}^{i-1} n_{i'}Varleft[W^{i'}right]
$$
We write $Var[W^{i′} ]$ for the shared scalar variance of all weights at layer $i′$. Then for a network with $d$ layers,
$$
Varleft[dfrac{partial Cost}{partial s^i}right]
=
Varleft[dfrac{partial Cost}{partial s^d}right]
prod_{i'=i}^{d}n_{i' +1}Varleft[W^{i'}right]
$$
$$
Varleft[dfrac{partial Cost}{partial w^i}right]
=
prod_{i'=0}^{i-1}n_{i'}Varleft[W^{i'}right]
prod_{i'=i}^{d-1}n_{i' +1}Varleft[W^{i'}right] times
Var[x]
Varleft[dfrac{partial Cost}{partial s^d}right]
$$
My question is, how were the last two equations derived from the first two equations ?
calculus statistics machine-learning
asked Dec 9 '18 at 16:46
Ansh KumarAnsh Kumar
156111
$endgroup$
add a comment |
$begingroup$
I was reading the Xavier initialization technique used in deep learning. But I'm struct in the basic math used in the paper. For example for a dense artificial neural network using symmetric activation function $f$ with unit derivative at $0$ (i.e. $f′(0) = 1$), if we write $z^i$ for the activation vector of layer $i$, and $s^i$ the argument vector of the activation function at layer $i$, we have $s^i = z^i W^i + b^i$ and $z^{i+1} = f(s^i)$. From these definitions we obtain the following:
$$
dfrac{partial Cost}{partial s^i_k}
=
f'(s_k^i) W^{i+1}_{k^{i'}} dfrac{partial Cost}{partial s^{i+1}}
$$
$$
dfrac{partial Cost}{partial w^i_{l,k}}
=
z^i_l dfrac{partial Cost}{partial s^{i}_k}
$$
Consider the hypothesis that we are in a linear regime at the initialization, that the weights are initialized independently and that the inputs features variances are the same ($= Var[x]$). Then we can say that, with $n_i$ the size of layer $i$ and $x$ the network input,
$$
f'(s_k^i) approx 1
$$
$$
Varleft[ z^i right]
=
Varleft[ x right]
prod_{i'=0}^{i-1} n_{i'}Varleft[W^{i'}right]
$$
We write $Var[W^{i′} ]$ for the shared scalar variance of all weights at layer $i′$. Then for a network with $d$ layers,
$$
Varleft[dfrac{partial Cost}{partial s^i}right]
=
Varleft[dfrac{partial Cost}{partial s^d}right]
prod_{i'=i}^{d}n_{i' +1}Varleft[W^{i'}right]
$$
$$
Varleft[dfrac{partial Cost}{partial w^i}right]
=
prod_{i'=0}^{i-1}n_{i'}Varleft[W^{i'}right]
prod_{i'=i}^{d-1}n_{i' +1}Varleft[W^{i'}right] times
Var[x]
Varleft[dfrac{partial Cost}{partial s^d}right]
$$
My question is, how were the last two equations derived from the first two equations ?
calculus statistics machine-learning
asked Dec 9 '18 at 16:46
Ansh KumarAnsh Kumar
156111
$endgroup$
add a comment |
$begingroup$
I was reading the Xavier initialization technique used in deep learning. But I'm struct in the basic math used in the paper. For example for a dense artificial neural network using symmetric activation function $f$ with unit derivative at $0$ (i.e. $f′(0) = 1$), if we write $z^i$ for the activation vector of layer $i$, and $s^i$ the argument vector of the activation function at layer $i$, we have $s^i = z^i W^i + b^i$ and $z^{i+1} = f(s^i)$. From these definitions we obtain the following:
$$
dfrac{partial Cost}{partial s^i_k}
=
f'(s_k^i) W^{i+1}_{k^{i'}} dfrac{partial Cost}{partial s^{i+1}}
$$
$$
dfrac{partial Cost}{partial w^i_{l,k}}
=
z^i_l dfrac{partial Cost}{partial s^{i}_k}
$$
Consider the hypothesis that we are in a linear regime at the initialization, that the weights are initialized independently and that the inputs features variances are the same ($= Var[x]$). Then we can say that, with $n_i$ the size of layer $i$ and $x$ the network input,
$$
f'(s_k^i) approx 1
$$
$$
Varleft[ z^i right]
=
Varleft[ x right]
prod_{i'=0}^{i-1} n_{i'}Varleft[W^{i'}right]
$$
We write $Var[W^{i′} ]$ for the shared scalar variance of all weights at layer $i′$. Then for a network with $d$ layers,
$$
Varleft[dfrac{partial Cost}{partial s^i}right]
=
Varleft[dfrac{partial Cost}{partial s^d}right]
prod_{i'=i}^{d}n_{i' +1}Varleft[W^{i'}right]
$$
$$
Varleft[dfrac{partial Cost}{partial w^i}right]
=
prod_{i'=0}^{i-1}n_{i'}Varleft[W^{i'}right]
prod_{i'=i}^{d-1}n_{i' +1}Varleft[W^{i'}right] times
Var[x]
Varleft[dfrac{partial Cost}{partial s^d}right]
$$
My question is, how were the last two equations derived from the first two equations ?
calculus statistics machine-learning
asked Dec 9 '18 at 16:46
Ansh KumarAnsh Kumar
156111
$endgroup$
I was reading the Xavier initialization technique used in deep learning. But I'm struct in the basic math used in the paper. For example for a dense artificial neural network using symmetric activation function $f$ with unit derivative at $0$ (i.e. $f′(0) = 1$), if we write $z^i$ for the activation vector of layer $i$, and $s^i$ the argument vector of the activation function at layer $i$, we have $s^i = z^i W^i + b^i$ and $z^{i+1} = f(s^i)$. From these definitions we obtain the following:
$$
dfrac{partial Cost}{partial s^i_k}
=
f'(s_k^i) W^{i+1}_{k^{i'}} dfrac{partial Cost}{partial s^{i+1}}
$$
$$
dfrac{partial Cost}{partial w^i_{l,k}}
=
z^i_l dfrac{partial Cost}{partial s^{i}_k}
$$
Consider the hypothesis that we are in a linear regime at the initialization, that the weights are initialized independently and that the inputs features variances are the same ($= Var[x]$). Then we can say that, with $n_i$ the size of layer $i$ and $x$ the network input,
$$
f'(s_k^i) approx 1
$$
$$
Varleft[ z^i right]
=
Varleft[ x right]
prod_{i'=0}^{i-1} n_{i'}Varleft[W^{i'}right]
$$
We write $Var[W^{i′} ]$ for the shared scalar variance of all weights at layer $i′$. Then for a network with $d$ layers,
$$
Varleft[dfrac{partial Cost}{partial s^i}right]
=
Varleft[dfrac{partial Cost}{partial s^d}right]
prod_{i'=i}^{d}n_{i' +1}Varleft[W^{i'}right]
$$
$$
Varleft[dfrac{partial Cost}{partial w^i}right]
=
prod_{i'=0}^{i-1}n_{i'}Varleft[W^{i'}right]
prod_{i'=i}^{d-1}n_{i' +1}Varleft[W^{i'}right] times
Var[x]
Varleft[dfrac{partial Cost}{partial s^d}right]
$$
My question is, how were the last two equations derived from the first two equations ?
calculus statistics machine-learning
calculus statistics machine-learning
asked Dec 9 '18 at 16:46
Ansh KumarAnsh Kumar
156111
asked Dec 9 '18 at 16:46
Ansh KumarAnsh Kumar
156111
edited Dec 9 '18 at 16:52
Ansh Kumar
asked Dec 9 '18 at 16:46
Ansh KumarAnsh Kumar
156111
asked Dec 9 '18 at 16:46
Ansh KumarAnsh Kumar
156111
asked Dec 9 '18 at 16:46
Ansh KumarAnsh Kumar
156111
156111
add a comment |
add a comment |
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