Answers.Body: <p>I believe adversarial samples violate i.i.d. by definition.</p>
  <p>They are constructed by an adversary to undermine our model. Thus, they are not independently chosen, and are not randomly distributed.</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2022-12-16T16:52:29.640Z
Answers.LastActivityDate: 2022-12-16T16:52:29.640Z
Answers.UserDisplayName: chessprogrammer
Body: <p>I am trying to understand why adversarial attacks work in theory.</p>
  <p>I have read, that the image is perturbed by a special perturbation <span class="math-container">$X_{adv}=X_1+p$</span>, but i can't find any reference on that perturbation being an i.i.d. violation by itself.</p>
  <p>If the i.i.d. assumption is not violated, then the only meaningful explanation in my mind would be, that the current date observed is out of distribution. But wouldn't this be an i.i.d. violation also?</p>
  
Tags: <machine-learning><adversarial-ml><iid><adversarial-attacks>
Title: Do adversarial samples violate the i.i.d. assumption?
UserDisplayName: Jan
ThreadText: # Do adversarial samples violate the i.i.d. assumption?
  
  (Asked by: username_0 on 2022-12-16)
  
  I am trying to understand why adversarial attacks work in theory.
  
  I have read, that the image is perturbed by a special perturbation $X\_{adv}=X\_1+p$, but i can't find any reference on that perturbation being an i.i.d. violation by itself.
  
  If the i.i.d. assumption is not violated, then the only meaningful explanation in my mind would be, that the current date observed is out of distribution. But wouldn't this be an i.i.d. violation also?
  
  **Tags:** machine-learning, adversarial-ml, iid, adversarial-attacks
  
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  ## 1 answers
  
  ### Answer by username_1 on 2022-12-16. Score: 2 | Accepted answer
  
  I believe adversarial samples violate i.i.d. by definition.
  
  They are constructed by an adversary to undermine our model. Thus, they are not independently chosen, and are not randomly distributed.
  
  
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