### Serie 10 Problem 2.1 Retrieving delta*

Hi,

I have a question for the last exercise serie problem 2.1. I have been able to get the following equation for the adverserial problem like the solutionary,
$$min_w \frac{1}{n} \sum_{i=1}^n l(y \cdot w^t x - \epsilon ||w||_2)$$
however I am not sure how to get $$\delta^*$$ and $$\hat{x}^*$$ from this. Sorry if it is straightforward, however i just do not see it - and also do not understand in the solutionary which equality is referred to in order to obtain it!

Many thanks

Hi,

Thanks a lot for the question. I think this reference can be helpful:
https://math.berkeley.edu/~arash/54/notes/6_7.pdf
In particular Theorem 16 therein that discusses when do we have equality in the Cauchy-Schwartz inequality: when vector $$v$$ is a scalar multiple of $$w$$ or vice versa. Applying this to our notation from lab 10 it would mean that we need to select our adversarial perturbation $$\delta$$ to be collinear with the weight vector, i.e. $$\delta = \alpha w$$ where $$\alpha$$ is an appropriately chosen scalar to achieve the equality. In our case, by solving wrt $$\alpha$$ one gets $$\alpha = - \frac{y_i \varepsilon}{||w||_2}$$ which recovers the desired {\delta^\star).

I hope that helps.

Thank you very much! Have a great weekend

I still don't completely understand how to solve with respect to α. I'm guessing we could use the property in equation 1 (from the Berkeley notes) but I don't understand what <u,v> is. For this example to work, <u,v> would have to be ε. Is that correct? Could you explain in detail how to solve with respect to α? Thank you so much in advance :))

The notation $$\langle u, v\rangle$$ denotes the inner product between vectors $$u$$, $$v$$. For usual Euclidean spaces, it's just $$\langle u, v\rangle = u^T v = \sum_{i=1}^d u_i v_i$$.

For this example, it should hold that $$y_i \langle \delta, w \rangle = -\epsilon ||w||_2$$. But then we also know that $$\delta^* = \alpha w$$, thus we get that $$y_i \langle \alpha w, w \rangle = -\epsilon ||w||_2$$ or equivalently

$$\alpha = -\frac{\epsilon ||w||_2}{y_i \langle w, w \rangle} = -\frac{\epsilon ||w||_2}{y_i ||w||_2^2} = -\frac{\epsilon}{y_i ||w||_2} = -\frac{y_i \epsilon}{||w||_2},$$

where we used that $$\langle w, w \rangle = ||w||_2^2$$ and $$1/y_i = y_i$$ since $$y_i \in \{-1, 1\}$$.

I hope that helps.

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