Timeline for Is there a way to simplify the following trace expression?

Current License: CC BY-SA 3.0

15 events

when toggle format	what		by	license	comment
Oct 7, 2015 at 22:44	comment	added	user34626		I don't think we need the "modulo" anymore: $\delta'_{ij}=1$ if $i=j\in\{0,1,2,\ldots M-1\}$ $\text{modulo}\,(N+M)$. I believe just $\delta'_{ij}=1$ if $i=j\in\{0,1,2,\ldots M-1\}$ will suffice.
Oct 7, 2015 at 22:00	vote	accept	user34626
Oct 7, 2015 at 22:00	comment	added	user34626		I edited the question to add $M<N$, also made a small correction to the conditions for $N_k$ to reflect that. Answer matches perfectly with simulation. Thank you.
S Oct 7, 2015 at 21:58	history	suggested	user34626	CC BY-SA 3.0	Corrected conditions for $N_k$ to reflect the fact that $M<N$. Also last interval for $k$ starts from $N+2$ rather than $N+1$.
Oct 7, 2015 at 21:54	review	Suggested edits
S Oct 7, 2015 at 21:58
Oct 7, 2015 at 8:51	history	edited	Carlo Beenakker	CC BY-SA 3.0	edited body
Oct 7, 2015 at 8:37	history	edited	Carlo Beenakker	CC BY-SA 3.0	added 165 characters in body
Oct 6, 2015 at 23:37	history	edited	Carlo Beenakker	CC BY-SA 3.0	deleted 417 characters in body
Oct 6, 2015 at 23:31	history	edited	Carlo Beenakker	CC BY-SA 3.0	deleted 417 characters in body
Oct 6, 2015 at 21:16	history	edited	Carlo Beenakker	CC BY-SA 3.0	added 236 characters in body
Oct 6, 2015 at 20:43	comment	added	user34626		$Y=\frac{1}{2}\sigma ^2N\text{tr}\mathbf{AA}^H$ does not match with simulation.
Oct 6, 2015 at 20:34	comment	added	user34626		Thank you for the answer. I'm trying to match the analytical solution you gave with simulation, currently they don't match. I'm trying to figure out why.
Oct 6, 2015 at 20:09	history	edited	Carlo Beenakker	CC BY-SA 3.0	added 421 characters in body
Oct 6, 2015 at 19:40	history	edited	Carlo Beenakker	CC BY-SA 3.0	deleted 2 characters in body
Oct 6, 2015 at 19:33	history	answered	Carlo Beenakker	CC BY-SA 3.0