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Output of the ReLU network is $$v = \sum_{ij} X_i A_{ij} w^{(1)}_{ij} \cdots w^{(L)}_{ij} $$ where $i$ is the input $j$ is the path and $A_{ij}$ is $1$ if the path is open and $0$ otherwise. Now if you take the expectation and use independence of the noise and also assume that the noise has mean $0$ we get that asymptotically when $m$ is large we have $$\mathbb{E}(\lVert v - v^* \rVert_2) \leq \sqrt{\mathbb{E}(\Vert v - v^* \rVert_2^2)} \leq 1/10^{L/2}$$$$\mathbb{E}(\lVert v - v^* \rVert_2) \leq \sqrt{\mathbb{E}(\lVert v - v^* \rVert_2^2)} \leq 1/10^{L}$$ since $A_{ij}$ and the product of parameters are asymptotically independent. Note that the expectation of the output$v^*$ squared is athe sum of the expectation of the original network squared and the expectation of the network that contains only the noise squared.

Output of the ReLU network is $$v = \sum_{ij} X_i A_{ij} w^{(1)}_{ij} \cdots w^{(L)}_{ij} $$ where $i$ is the input $j$ is the path and $A_{ij}$ is $1$ if the path is open and $0$ otherwise. Now if you take the expectation and use independence of the noise and also assume that the noise has mean $0$ we get that asymptotically when $m$ is large we have $$\mathbb{E}(\lVert v - v^* \rVert_2) \leq \sqrt{\mathbb{E}(\Vert v - v^* \rVert_2^2)} \leq 1/10^{L/2}$$ since $A_{ij}$ and the product of parameters are asymptotically independent. Note that the expectation of the output is a sum of the expectation of the original network and the expectation of the network that contains only the noise.

Output of the ReLU network is $$v = \sum_{ij} X_i A_{ij} w^{(1)}_{ij} \cdots w^{(L)}_{ij} $$ where $i$ is the input $j$ is the path and $A_{ij}$ is $1$ if the path is open and $0$ otherwise. Now if you take the expectation and use independence of the noise and also assume that the noise has mean $0$ we get that asymptotically when $m$ is large we have $$\mathbb{E}(\lVert v - v^* \rVert_2) \leq \sqrt{\mathbb{E}(\lVert v - v^* \rVert_2^2)} \leq 1/10^{L}$$ since $A_{ij}$ and the product of parameters are asymptotically independent. Note that the expectation of $v^*$ squared is the sum of the expectation of the original network squared and the expectation of the network that contains only the noise squared.

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Output of the ReLU network is $$v = \sum_{ij} X_i A_{ij} w^{(1)}_{ij} \cdots w^{(L)}_{ij} $$ where $i$ is the input $j$ is the path and $A_{ij}$ is $1$ if the path is open and $0$ otherwise. Now if you take the expectation and use independence of the noise and also assume that the noise has mean $0$ we get that asymptotically when $m$ is large we have $$\mathbb{E}(\lVert v - v^* \rVert_2) \leq \sqrt{\mathbb{E}(\Vert v - v^* \rVert_2^2)} \leq 1/10^{L/2}$$ since $A_{ij}$ and the product of parameters are asymptotically independent. Note that the the expectation of the output is a sum of the expectation of the original network and the expectation of the network that contains only the noise.

Output of the ReLU network is $$v = \sum_{ij} X_i A_{ij} w^{(1)}_{ij} \cdots w^{(L)}_{ij} $$ where $i$ is the input $j$ is the path and $A_{ij}$ is $1$ if the path is open and $0$ otherwise. Now if you take the expectation and use independence of the noise and also assume that the noise has mean $0$ we get that asymptotically when $m$ is large we have $$\mathbb{E}(\lVert v - v^* \rVert_2) \leq \sqrt{\mathbb{E}(\Vert v - v^* \rVert_2^2)} \leq 1/10^{L/2}$$ since $A_{ij}$ and the product of parameters are asymptotically independent. Note that the the expectation of the output is a sum of the expectation of the original network and the expectation of the network that contains only the noise.

Output of the ReLU network is $$v = \sum_{ij} X_i A_{ij} w^{(1)}_{ij} \cdots w^{(L)}_{ij} $$ where $i$ is the input $j$ is the path and $A_{ij}$ is $1$ if the path is open and $0$ otherwise. Now if you take the expectation and use independence of the noise and also assume that the noise has mean $0$ we get that asymptotically when $m$ is large we have $$\mathbb{E}(\lVert v - v^* \rVert_2) \leq \sqrt{\mathbb{E}(\Vert v - v^* \rVert_2^2)} \leq 1/10^{L/2}$$ since $A_{ij}$ and the product of parameters are asymptotically independent. Note that the expectation of the output is a sum of the expectation of the original network and the expectation of the network that contains only the noise.

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Output of the ReLU network is $$v = \sum_{ij} X_i A_{ij} w^{(1)}_{ij} \cdots w^{(L)}_{ij} $$ where $i$ is the input $j$ is the path and $A_{ij}$ is $1$ if the path is open and $0$ otherwise. Now if you take the expectation and use independence of the noise and also assume that the noise has mean $0$ we get that asymptotically when $m$ is large we have $\mathbb{E}(\lVert v - v^* \rVert_2) \leq 1/10^L$ since $$\mathbb{E}(\lVert v - v^* \rVert_2) \leq \sqrt{\mathbb{E}(\Vert v - v^* \rVert_2^2)} \leq 1/10^{L/2}$$ since $A_{ij}$ and the product of parameters are asymptotically independent. Note that the the expectation of the output is a sum of the expectation of the original network and the expectation of the network that contains only the noise.

Output of the ReLU network is $$v = \sum_{ij} X_i A_{ij} w^{(1)}_{ij} \cdots w^{(L)}_{ij} $$ where $i$ is the input $j$ is the path and $A_{ij}$ is $1$ if the path is open and $0$ otherwise. Now if you take the expectation and use independence of the noise and also assume that the noise has mean $0$ we get that asymptotically when $m$ is large we have $\mathbb{E}(\lVert v - v^* \rVert_2) \leq 1/10^L$ since $A_{ij}$ and the product of parameters are asymptotically independent. Note that the the expectation of the output is a sum of the expectation of the original network and the expectation of the network that contains only the noise.

Output of the ReLU network is $$v = \sum_{ij} X_i A_{ij} w^{(1)}_{ij} \cdots w^{(L)}_{ij} $$ where $i$ is the input $j$ is the path and $A_{ij}$ is $1$ if the path is open and $0$ otherwise. Now if you take the expectation and use independence of the noise and also assume that the noise has mean $0$ we get that asymptotically when $m$ is large we have $$\mathbb{E}(\lVert v - v^* \rVert_2) \leq \sqrt{\mathbb{E}(\Vert v - v^* \rVert_2^2)} \leq 1/10^{L/2}$$ since $A_{ij}$ and the product of parameters are asymptotically independent. Note that the the expectation of the output is a sum of the expectation of the original network and the expectation of the network that contains only the noise.

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