Mean of L2 norm of random vector Ax+n

Click For Summary

Homework Help Overview

The discussion revolves around finding the expected value of the L2 norm of the expression ||Ax+n||, where x and n are uncorrelated random variables and the expected value of n is zero. The participants are examining the implications of this setup in the context of linear algebra and probability.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to derive the expected value by expanding the norm and considering the properties of uncorrelated variables. They are questioning whether certain terms can be simplified or rearranged, particularly in relation to the trace and scalar properties of the expressions involved.

Discussion Status

The discussion is active, with participants clarifying each other's expressions and exploring the simplifications of the expected value. There is a focus on ensuring the mathematical validity of their manipulations, but no consensus has been reached on further simplification.

Contextual Notes

Participants are working under the assumption that x and n are uncorrelated and that E[n] = 0, which influences their reasoning about the terms in the expected value calculation.

cutesteph
Messages
62
Reaction score
0

Homework Statement



What is the expected value of ||Ax+n|| where || || is the L2 norm and x and n are uncorrelated and E[n] = 0


The Attempt at a Solution


E[ norm of Y] = E[(Ax+n)' (Ax+n)] = E[(x'A'+n')(Ax+n)] = E[x'A'xA +x'T'n +n'Ax +n'n]
the three last terms = 0 due to uncorrelatedness so = E[x'A'xA] = E[tr(x'A'xA)] = E[tr(x'xA'A)] = A'AE[x'x] does this reduce any further?
 
Physics news on Phys.org
cutesteph said:
E[ norm of Y] = E[(Ax+n)' (Ax+n)] = E[(x'A'+n')(Ax+n)] = E[x'A'xA +x'T'n +n'Ax +n'n]
You mean E[x'A'Ax +x'A'n +n'Ax +n'n] , right?
 
Yes. But the result is still the same since it is a scalar and we can take the trace of it to rearrange. So does A'AE[x'x] simplify any further?
 
cutesteph said:
Yes. But the result is still the same since it is a scalar
x'A'Ax is a scalar; x'A'x is a scalar; (x'A'x)A is an n x n matrix; x'(A'xA) is meaningless (since xA is meaningless).
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
Replies
7
Views
7K
  • · Replies 1 ·
Replies
1
Views
4K