User Junaid ur Rehman | chat.stackexchange.com

Brown Ninja

Jan 2, 2019 08:45

Let $\hat{X} = \left( x^0,x^1,\cdots , x^{N-1}\right)$ be the sequence of $N$ i.i.d realizations of a random variable $X$ with pmf $P_1, P_2, \cdots , P_{d}$, where $P_i$ is the probability of appearance of symbol $i$. Can we show that
\begin{align}
\hat{P}_{i} &= \frac{N_i}{N}\\
&= \frac{1}{N}\sum_{k=0}^{N-1} I\left( x^k = i \right)
\end{align}
is a consistent estimator of $P_i$, where $I\left( x^k = i \right)$ is the indicator function (=1 when $x^k = i$, and 0 otherwise)?

Brown Ninja

Jan 2, 2019 08:44

Happy new year everyone

Brown Ninja

Oct 14, 2018 09:22

1

Q: Summation of roots of unity equal to zero

Let $u$ be a integer which has an odd prime $m$ as a divisor. So we let $u=u_1m$ where $u_1=\frac{u}{m}$. Consider the set of $u$th roots of unity, i.e., $A=\{e^{j\frac{2\pi}{u}n} | 0 \leq n \leq u-1\}=\{e^{j\frac{2\pi}{u_1m}n} | 0 \leq n \leq u_1m-1\}$. For convenience, let $e^{j\frac{2\pi}...

Brown Ninja

Oct 14, 2018 09:21

Any comments on this?

While working on Heisenberg-Weyl operators (specifically working on their eigenvalues), I have come across the following problem:
Given an $n$, $n$th roots of some complex number $p$,$|p|=1$ sum to zero. Let $A$ be the set that contains all these $n$ roots. My intuition was if we form a subset of $A$ whose elements also sum to zero, then this subset is necessarily a complete set of $m$th roots of some $p′$, where $m$ is a factor of $n$. However, this intuition turned out to be wrong

Brown Ninja

Oct 14, 2018 05:47

For example, for a prime $n$, can we show that there does not exist a subset of $A$ whose elements sum to zero?

Brown Ninja

Oct 14, 2018 05:43

While working on Heisenberg-Weyl operators (specifically working on their eigenvalues), I have come across the following problem:
Given an $n$, $n$th roots of some complex number $p, |p| = 1$ sum to zero. Let $A$ be the set that contains all these $n$ roots. My intuition was if we form a subset of $A$ whose elements also sum to zero, then this subset is necessarily a complete set of $m$th roots of some $p'$, where $m$ is a factor of $n$. However, this intuition turned out to be wrong
see: https://math.stackexchange.com/questions/490115/summation-of-roots-of-unity-equal-to-zero?rq=1

Brown Ninja

Aug 17, 2018 09:27

This chatroom has never disappointed me and I have always regarded this place as filled with people of great talents :)

Brown Ninja

Aug 17, 2018 09:26

This is what I was looking for. Thanks a lot @Holo :)

Brown Ninja

Aug 17, 2018 09:24

Definitely as this will help. I am just trying to verify it.

Brown Ninja

Aug 17, 2018 09:21

Ahan, it looks like it.

Brown Ninja

Aug 17, 2018 09:01

The product terms (inside the square brackets) increase recursively. I can post the series for $N=15$, if someone is interested in having a go at it. I have given the above example for $N=7$ for clarity.

Brown Ninja

Aug 17, 2018 08:57

This is for $N = 7$, and N is always of the form $N = 2^n - 1$ for some $n$.

Brown Ninja

Aug 17, 2018 08:56

\begin{align*}
&g\left(0\right)+\\
&f\left(\{0\},\{1\}\right)g\left(1\right)+\\
&f\left(\{0,1\},\{2,3\}\right)\Big[g\left(2\right)+f\left(\{2\},\{3\}\right)g\left(3\right)\Big]+\\&f\left(\{0,1,2,3\},\{4,5,6,7\}\right)\bigg[g\left(4\right)+f\left(\{4\},\{5\}\right)g\left(5\right)+f\left(\{4,5\},\{6,7\}\right)\left[g\left(6\right)+f\left(\{6\},\{7\}\right)g\left(7\right)\right]\bigg]
\end{align*}

Brown Ninja

Aug 17, 2018 08:56

I have a series problem where I am unable to identify the sequence. I need to display the sequence in a compact form using $\sum$ or/and $\prod$ form.

Brown Ninja

Jan 22, 2018 12:15

Maybe $X = Y^\dagger Z$ solves the problem :D

Brown Ninja

Jan 22, 2018 12:11

if there are some conditions, can you please tell me what they might be?

Brown Ninja

Jan 22, 2018 12:10

Hi,
$XY = Z$ where all $X$, $Y$, and $Z$ are unitaries. Is it possible to find an $X$ for any $Y$ and $Z$?

Brown Ninja

Jul 4, 2017 14:41

Thank you very much @Secret . You were of great help. Now if anyone knows how to save this chat session for later viewing?

Brown Ninja

Jul 4, 2017 14:40

exactly.

Brown Ninja

Jul 4, 2017 14:39

I also suspect that. Intuition being any rearrangement of $x^{\downarrow}$ to form $\vec{x}$ takes $x_{n,k}$ closer to a vector of uniform elements.

Brown Ninja

Jul 4, 2017 14:37

yes

Brown Ninja

Jul 4, 2017 14:36

$(3 \quad 3 \quad 8 \quad 8 \quad 18\quad 18)$

Brown Ninja

Jul 4, 2017 14:34

$k$ repitions of sum of $k$ elements for every element?

Brown Ninja

Jul 4, 2017 14:33

repeated elements?

Brown Ninja

Jul 4, 2017 14:31

Yes, $S^{T}S = k I_{\ell}$

Brown Ninja

Jul 4, 2017 14:31

Yes

Brown Ninja

Jul 4, 2017 14:21

I am sorry, I do not think $SS^{T} = k I$. For example
$$
S = \begin{bmatrix}
1 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\\ 0 & 0 & 1
\end{bmatrix}
$$

Brown Ninja

Jul 4, 2017 14:16

Or we can simply write $\vec{x} S = x^{\downarrow} S P$. Since $S$ is not orthogonal, so $S^{T} \neq S^{-1}$? Also since $S$ is not square, inverse does not exist, we need to look at the pseudo-inverse?

Brown Ninja

Jul 4, 2017 14:11

So, $\vec{x}_{n,k} = x^{\downarrow} T S$, and $x^{\downarrow}_{n,k} = x^{\downarrow} S$. So the condition for majorization is
$x^{\downarrow} T S = x^{\downarrow} S P$

Brown Ninja

Jul 4, 2017 14:05

Yes. Additionally, a necessary and sufficient condition for a vector $\vec{x}$ to majorize other vector $\vec{y}$ is if $\vec{y} = \vec{x} P$ where $P$ is a doubly stochastic matrix

Brown Ninja

Jul 4, 2017 14:03

We also have a relation between $\vec{x}$ and $x^{\downarrow}$. We can write $\vec{x} = x^{\downarrow} T$ where $T$ is a permutation matrix.

Brown Ninja

Jul 4, 2017 13:59

@Semiclassical true

Brown Ninja

Jul 4, 2017 13:59

@Semiclassical nice to see you :)
You helped me a lot when I asked a question here a few days back.

Brown Ninja

Jul 4, 2017 13:58

@Secret Right. So I was using the matrix $S$ in the context of $k$ element summation of an $n$ element vector. $\vec{x}_{n,k} = \vec{x} S$. Right?

Brown Ninja

Jul 4, 2017 13:51

Right. I will read it carefully.

Brown Ninja

Jul 4, 2017 13:49

with the definition of 'Orthogonal ( :p )' $S$ I provided?

Brown Ninja

Jul 4, 2017 13:45

@Secret I have written my question in a better way here. I am sorry earlier it was poorly defined.
https://math.stackexchange.com/questions/2346189/does-this-vector-majorizes-all-other-similar-vectors

Brown Ninja

Jul 4, 2017 13:29

@Secret Thank you very much. I was away, so did not see your message earlier.

Brown Ninja

Jul 4, 2017 08:39

$S$ is not $n \times k$, it is $n \times \ell$

Brown Ninja

Jul 4, 2017 08:36

I hope I made some sense.

Brown Ninja

Jul 4, 2017 08:36

So let $\vec{x}' = \vec{x}S$, and $\vec{y}' = \vec{y}S$ Does the majorization relation still hold between $\vec{x}'$ and $\vec{y}'$?

Brown Ninja

Jul 4, 2017 08:34

I have another vector $\vec{y}$ which majorizes $\vec{x}$

Brown Ninja

Jul 4, 2017 08:33

where the first column has $k$ ones at the first $k$ locations

Brown Ninja

Jul 4, 2017 08:31

this can be written as $\vec{x} S$ where $S$ is of the form
$$
\begin{bmatrix}
1 & 0 & 0 & \cdots & 0\\
1 & 0 & 0 & \cdots & 0\\
\vdots & \vdots & \vdots & \cdots & \vdots\\
0 & 1 & 0 & \cdots & 0\\
\end{bmatrix}
$$

Brown Ninja

Jul 4, 2017 08:30

to obtain an $\ell$ element vector

Brown Ninja

Jul 4, 2017 08:30

I have an $n$ element vector on which I am applying summation over every $k$ elements

Brown Ninja

Jul 4, 2017 08:29

No they should not be orthonormal. Let me explain the bigger problem I am dealing with.

Brown Ninja

Jul 4, 2017 08:27

to each other

Brown Ninja

Jul 4, 2017 08:27

orthogonal matrix -> all columns are orthogonal

Brown Ninja

Jul 4, 2017 08:27

Doubly stochastic -> all rows sum to 1 and all columns sum to 1

Junaid ur Rehman

Mathematics

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