Upload 10 files

.gitattributes

@@ -33,3 +33,4 @@ saved_model/**/* filter=lfs diff=lfs merge=lfs -text
 *.zip filter=lfs diff=lfs merge=lfs -text
 *.zst filter=lfs diff=lfs merge=lfs -text
 *tfevents* filter=lfs diff=lfs merge=lfs -text
+Manually[[:space:]]Disecting[[:space:]]arXiv2512.15720/main.pdf filter=lfs diff=lfs merge=lfs -text

Manually Disecting arXiv2512.15720/main.pdf

@@ -0,0 +1,3 @@
+version https://git-lfs.github.com/spec/v1
+oid sha256:73fd211afee01f6d12dbeac2a58255eb7059ad0867b3a45510791f97d6ebdfb2
+size 305424

Manually Disecting arXiv2512.15720/main.tex

@@ -0,0 +1,519 @@
+
+\documentclass[11pt]{article}
+\usepackage[margin=1in]{geometry}
+
+% Core packages
+\usepackage{amsmath,amssymb}
+\usepackage{tikz-cd}
+\usepackage{multicol}
+\newcommand{\mathcolorbox}[2]{\begingroup\setlength{\fboxsep}{1pt}\colorbox{#1}{$\displaystyle #2$}\endgroup}
+
+% Paragraphs
+\setlength{\parindent}{0pt}
+\setlength{\parskip}{1\baselineskip}
+
+\title{Manually Disecting arXiv:2512.15720}
+\author{algorembrant}
+\date{\today}
+
+\begin{document}
+\maketitle
+
+The contents of this paper may or may not related directly to the paper 	arXiv:2512.15720, its freestyle. And also im using a simple 2-states with 20 sample data rather than 15-states with many data as what said on the paper, just to keep things simpler.
+
+\begin{align}
+\includegraphics[width=0.5\textwidth]
+{image.png} \\
+\includegraphics[width=0.5\textwidth]{image_2.png} \\
+\includegraphics[width=0.5\textwidth]{image_3.png}
+\end{align}
+
+
+\[
+VMA =\frac{1}{2} (\sum_{a=1}^{n} V_{t-a})/n
+\]
+
+\[
+\sum_{i \in{S}}
+\]
+
+\[
+s_t = sgn(P_t - P_{t-1})
+\]
+
+\[
+v_t = ceil(5 \cdot F_{V,t}(V_t) )
+\]
+
+\[
+S_t = (q_t,v_t)
+\]
+
+\[
+\hat{P} = 
+\]
+
+\begin{align*}
+\intertext{I think the starting probabilities doesnt matter since if we keep multiplying the transition matrix by itself (following the power rule), then the resulting probabilities will settle at some point and we get better proxy. True in finite markov chain.}
+\intertext{In the context of Markov chains and repeatedly multiplying the transition matrix by itself (i.e., raising it to a high power), the resulting probabilities that the system settles into are called the stationary distribution (or steady‑state distribution).}
+\end{align*}
+
+\[
+\pi \cdot \hat{P}_t = \pi    
+\]
+
+\begin{equation}
+H_t := -\frac{1}{\log K} \sum_{i \in S} \pi_i (\hat{P_t}) \sum_{j \in S} \hat{p}_{{ij},t} \log \hat{p}_{{ij},t}
+\end{equation}
+
+\begin{align*}
+    \intertext{
+        This is the entropy formula, the signal.
+    }
+    \intertext{
+        And let's break it down piece by piece.
+    }
+\end{align*}
+
+
+\begin{equation}
+H_t := -\frac{1}{\log K} \sum_{i \in S} \pi_i (\hat{P_t}) \boxed{\sum_{j \in S} \hat{p}_{{ij},t} \log \hat{p}_{{ij},t}}
+\end{equation}
+
+where:
+\[
+    RowEntropy(i) = - \sum_{j \in S} \hat{p}_{{ij},t} \log \hat{p}_{{ij},t}
+\]
+
+and the original Entropy formula is:
+\[
+    H = \sum_{i = 1}^c -P_i \log_2 (P_i)
+\]
+
+
+\begin{equation}
+H_t := -\frac{1}{\log K} \boxed{\sum_{i \in S} \pi_i (\hat{P_t})}\boxed{\sum_{j \in S} \hat{p}_{{ij},t} \log \hat{p}_{{ij},t}}
+\end{equation}
+
+\[
+\begin{pmatrix}
+a_{11} & a_{12} & \cdots & a_{1n} \\
+a_{21} & a_{22} & \cdots & a_{2n} \\
+\vdots & \vdots & \ddots & \vdots \\
+a_{m1} & a_{m2} & \cdots & a_{mn}
+\end{pmatrix}
+\]
+
+\[
+\begin{vmatrix}
+p & q \\
+r & s
+\end{vmatrix}
+\]
+
+
+\[
+P^2 = \begin{bmatrix}
+    a_1^2 & a_1^2 \\
+    a_2^2 & a_2^2)
+\end{bmatrix}    
+\]
+
+\newpage
+
+Lets try breaking down how to multiply the matrix by itself using power rule, given:
+\[
+    P = 
+    \begin{bmatrix}
+        a & b \\
+        c & d
+    \end{bmatrix}
+\]
+
+becomes $P$ into $P^2$, therefore:
+\[
+    P^2 = 
+    \begin{bmatrix}
+        a & b \\
+        c & d
+    \end{bmatrix}
+    \begin{bmatrix}
+        a & b \\
+        c & d
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+    aa+bc & ab+bd \\
+    ca+dc & cb + dd
+    \end{bmatrix}
+    \quad
+    \text{or} 
+    \quad
+    \left[\begin{matrix}
+    a^2+bc & ab+bd \\
+    ca+dc & cb + d^2
+    \end{matrix}\right]
+\]
+
+\[
+    P^2 = 
+    \begin{bmatrix}
+    \mathcolorbox{yellow!50}{a} & \mathcolorbox{yellow!75}{b} \\
+    c & d
+    \end{bmatrix}
+    \begin{bmatrix}
+        \mathcolorbox{yellow!50}{a} & b \\
+        \mathcolorbox{yellow!75}{c} & d
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+    \mathcolorbox{yellow!50}{aa+bc} & ab+bd \\
+    ca+dc & cb + dd
+    \end{bmatrix}
+    \quad
+    \text{or} 
+    \quad
+    \left[\begin{matrix}
+    a^2+bc & ab+bd \\
+    ca+dc & cb + d^2
+    \end{matrix}\right]
+\]
+
+\[
+    P^2 = 
+    \begin{bmatrix}
+    \mathcolorbox{yellow!50}{a} & \mathcolorbox{yellow!75}{b} \\
+    c & d
+    \end{bmatrix}
+    \begin{bmatrix}
+        a & \mathcolorbox{yellow!50}{b} \\
+        c & \mathcolorbox{yellow!75}{d}
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+    aa+bc & \mathcolorbox{yellow!50}{ab+bd} \\
+    ca+dc & cb + dd
+    \end{bmatrix}
+    \quad
+    \text{or} 
+    \quad
+    \left[\begin{matrix}
+    a^2+bc & ab+bd \\
+    ca+dc & cb + d^2
+    \end{matrix}\right]
+\]
+
+\[
+    P^2 = 
+    \begin{bmatrix}
+    a & b \\
+    \mathcolorbox{yellow!50}{c} & \mathcolorbox{yellow!75}{d}
+    \end{bmatrix}
+    \begin{bmatrix}
+        \mathcolorbox{yellow!50}{a} & b \\
+        \mathcolorbox{yellow!75}{c} & d
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+    aa+bc & ab+bd \\
+    \mathcolorbox{yellow!50}{ca+dc} & cb + dd
+    \end{bmatrix}
+    \quad
+    \text{or} 
+    \quad
+    \left[\begin{matrix}
+    a^2+bc & ab+bd \\
+    ca+dc & cb + d^2
+    \end{matrix}\right]
+\]
+
+\[
+    P^2 = 
+    \begin{bmatrix}
+    a & b \\
+    \mathcolorbox{yellow!50}{c} & \mathcolorbox{yellow!75}{d}
+    \end{bmatrix}
+    \begin{bmatrix}
+        a & \mathcolorbox{yellow!50}{b} \\
+        c & \mathcolorbox{yellow!75}{d}
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+    aa+bc & ab+bd \\
+    ca+dc & \mathcolorbox{yellow!50}{cb+dd}
+    \end{bmatrix}
+    \quad
+    \text{or} 
+    \quad
+    \left[\begin{matrix}
+    a^2+bc & ab+bd \\
+    ca+dc & cb + d^2
+    \end{matrix}\right]
+\]
+
+
+breakdown
+
+\[
+    \begin{bmatrix}
+        a & b
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        a \\ 
+        c
+    \end{bmatrix}
+    =
+    aa +bc
+\]
+
+\[
+    \begin{bmatrix}
+        a & b
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        b \\ 
+        d
+    \end{bmatrix}
+    =
+    ab +bd
+\]
+
+\[
+    \begin{bmatrix}
+        c & d
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        a \\ 
+        c
+    \end{bmatrix}
+    =
+    ca +dc
+\]
+
+\[
+    \begin{bmatrix}
+        c & d
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        b \\ 
+        d
+    \end{bmatrix}
+    =
+    cb +dd
+\]
+
+\[
+    \begin{bmatrix}
+        c & d
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        b \\ 
+        d
+    \end{bmatrix}
+    =
+    cb + d^2
+\]
+
+therefore:
+\[
+P^2 = \begin{bmatrix}
+    aa+bc & ab+bd \\
+    ca+dc & cb+dd
+    \end{bmatrix}
+\]
+
+or
+
+\[
+P^2 = \begin{bmatrix}
+    a^2+bc & ab+bd \\
+    ca+dc & cb + d^2
+    \end{bmatrix}   
+\]
+
+\newpage
+
+\begin{equation}
+H_t := -\frac{1}{\log K} \sum_{i \in S} \pi_i (\hat{P_t}) \sum_{j \in S} \hat{p}_{{ij},t} \log \hat{p}_{{ij},t}
+\end{equation}
+
+\begin{equation}
+H_t := -\frac{1}{\log K} \sum_{i \in S} \pi_i (\hat{P_t}) \textcolor{red}{\sum_{j \in S} \hat{p}_{{ij},t} \log \hat{p}_{{ij},t}}
+\end{equation}
+
+supposed we have a $\hat{P}_t$
+\[
+    \hat{P}_t = 
+    \begin{bmatrix}
+    (buy,buy) & (buy,sell) \\
+    (sell,buy) & (sell,sell)
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+    0.125 & 0.875 \\
+    0.5454545455 & 0.4545454545
+    \end{bmatrix}
+\]
+
+then the settle-probabilities is at $\hat{P}_t^7$
+
+\[
+    \hat{P}_t^7 = 
+    \begin{bmatrix}
+    0.384 & 0.616 \\
+    0.384 & 0.616 
+    \end{bmatrix}
+    \quad
+    \text{therefore}
+    \quad
+    \pi =
+    \begin{bmatrix}
+    0.384 & 0.616 
+    \end{bmatrix}
+\]
+
+then we calculate for the entropy of each row by following the general equation 
+
+\[
+RowEntopy(i) = \sum_{j \in S} \hat{p}_{{ij},t} \log \hat{p}_{{ij},t}
+\]
+
+If this equation is from a paper rooted in Physics (Statistical Mechanics), Mathematics, or Pure Statistics, the base is typically the natural logarithm, base $e$. Therefore,
+
+\[
+RowEntopy(i) = \sum_{j \in S} \hat{p}_{{ij},t} \textcolor{red}{\ln} \hat{p}_{{ij},t}
+\]
+
+\begin{align*}
+RowEntopy(\textcolor{red}{1}) &= \sum_{j \in S} \hat{p}_{{\textcolor{red}{1}j},t} \log \hat{p}_{{\textcolor{red}{1}j},t} \\
+&= [0.125 \cdot \log(0.125)] + [0.875 \cdot \log(0.875)] \\
+&= [-0.2599301927] + [-0.1168399685] \\
+&= \mathcolorbox{red!10}{-0.3767701613} \\
+\end{align*}
+
+\begin{align*}
+RowEntopy(\textcolor{red}{2}) &= \sum_{j \in S} \hat{p}_{{\textcolor{red}{2}j},t} \log \hat{p}_{{\textcolor{red}{2}j},t} \\
+&= [0.5454545455 \cdot \log(0.4545454545)] + [0.875 \cdot \log(0.875)] \\
+&= [-0.3306195292] + [-0.3583897093] \\
+&= \mathcolorbox{red!10}{-0.6890092385} \\
+\end{align*}
+
+
+
+\begin{equation}
+H_t := -\frac{1}{\log K} \sum_{i \in S} \textcolor{red}{\pi_i (\hat{P_t})} \sum_{j \in S} \hat{p}_{{ij},t} \log \hat{p}_{{ij},t}
+\end{equation}
+
+is the same as
+
+\begin{equation}
+H_t := -\frac{1}{\log K} \sum_{i \in S} \textcolor{red}{\pi_i} \sum_{j \in S} \hat{p}_{{ij},t} \log \hat{p}_{{ij},t}
+\end{equation}
+
+where:
+
+\begin{align*}
+\pi_1 &=
+\left[
+0.384 \quad 0.616
+\right], \\
+\pi_2 &=
+\left[
+0.384 \quad 0.616
+\right], \\
+\mathrm{RowEntropy}_{\textcolor{red}{1}}
+&= -0.3767701613, \\
+\mathrm{RowEntropy}_{\textcolor{red}{2}}
+&= -0.6890092385
+\end{align*}
+
+therefore:
+
+\begin{equation}
+H_t :=
+-\frac{1}{\log K}
+\textcolor{red}{
+\sum_{i \in S}
+\colorbox{orange!50}{$\pi_i$}
+\colorbox{yellow!50}{$\sum_{j \in S} \hat{p}_{ij,t} \log \hat{p}_{ij,t}$}
+}
+\end{equation}
+
+is
+\begin{align*}
+    &=[\mathcolorbox{orange!50}{0.384} \cdot \mathcolorbox{yellow!50}{(-0.3767701613)}] \textcolor{red}{+} [\mathcolorbox{orange!50}{0.616} \cdot \mathcolorbox{yellow!50}{(-0.6890092385)}] \\
+    &= -0.5691094329
+\end{align*}
+
+in here, we have $K = 2$ becuase we have buy or sell as states,$K$ refers to the total number of posssible states hence
+\begin{align*}
+    H_t &:= -\frac{1}{\log K}(\textcolor{red}{-0.5691094329}) \\
+    &:= -\frac{1}{\log \textcolor{red}{2}} (\textcolor{red}{-0.5691094329}) \\
+    &:= -(-0.8210513566) \\
+    &:= \mathcolorbox{gray!50}{0.8210513566}
+\end{align*}
+
+the paper says high entropy inficates unpredictable transitions, meanwhile low entropy indicates structure
+
+\begin{align*}
+    H_t \geq 0.95 \quad \text{means towards max unpredictable next state is} \\
+    H_t \geq 0.05 \quad \text{means towards min unpredictable next state is}
+\end{align*}
+
+this values depend on the creator, but the paper saus 0.05 amd 0.95 as treshold to trigger possitions.
+
+the paper has 3 criteria before entering a trade,
+one,
+
+\[
+    H_t < H_{5th percentile}
+    \]
+two,
+
+\[
+    V_t > V_{95th percentile}
+    \]
+
+three,
+
+\[
+    5 bps \leq |trailing^{5min}_{return}| \leq  20 bps
+\]
+
+and after all conditions is met, he enters a trade 
+
+\[
+    Direction = sgn(trailing_5min_return), \quad +1 = buy ; -1  = sell
+    \]
+
+and after entry, he places a stoploss
+\[
+    stoploss = entry^{price} \pm 5bps
+    \]
+
+then takeprofit
+\[
+    Take-Profit = Entry Price ± TP_threshold
+    \]
+
+and emergency exit
+
+\[
+    Max hold time = 300 seconds (5 minutes)
+    \]
+
+\section{failed attempt}
+at first, i actually tried manually recreating the paper using 15-state transition matrix markov chain model but in the long run i found out the formula is general and applicable to any given number of state hence i used simpler 2-state model
+
+\begin{align}
+\includegraphics[width=0.5\textwidth]{image_4.png} \\
+\includegraphics[width=0.5\textwidth]{image_5.png}
+\end{align}
+
+\end{document}
+

Manually Disecting arXiv2512.15720/matrix_power/main.tex

@@ -0,0 +1,245 @@
+
+\documentclass[11pt]{article}
+\usepackage[margin=1in]{geometry}
+
+% Core packages
+\usepackage{amsmath,amssymb}
+\usepackage{tikz-cd}
+\usepackage{multicol}
+
+% Paragraphs
+\setlength{\parindent}{0pt}
+\setlength{\parskip}{1\baselineskip}
+
+\title{multiplying matrix on power rule, through manual code}
+\author{algorembrant}
+\date{\today}
+
+\begin{document}
+\maketitle
+
+\section{Solve for $P^2}
+
+\text{Lets try breaking down how to multiply the matrix by itself using power rule, given:}
+\[
+    P = 
+    \begin{bmatrix}
+        a & b \\
+        c & d
+    \end{bmatrix}
+\]
+
+\text{becomes $P$ into $P^2$, therefore:}
+\[
+    P^2 = 
+    \begin{bmatrix}
+        a & b \\
+        c & d
+    \end{bmatrix}
+    \begin{bmatrix}
+        a & b \\
+        c & d
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+    aa+bc & ab+bd \\
+    ca+dc & cb + dd
+    \end{bmatrix}
+    \quad
+    \text{or} 
+    \quad
+    \boxed{\begin{bmatrix}
+    a^2+bc & ab+bd \\
+    ca+dc & cb + d^2
+    \end{bmatrix}}
+
+\]
+
+\[
+    P^2 = 
+    \begin{bmatrix}
+    \colorbox{yellow!50}{a} & \colorbox{yellow!75}{b} \\
+    c & d
+    \end{bmatrix}
+    \begin{bmatrix}
+        \colorbox{yellow!50}{a} & b \\
+        \colorbox{yellow!75}{c} & d
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+    \colorbox{yellow!50}{aa+bc} & ab+bd \\
+    ca+dc & cb + dd
+    \end{bmatrix}
+    \quad
+    \text{or} 
+    \quad
+    \boxed{\begin{bmatrix}
+    a^2+bc & ab+bd \\
+    ca+dc & cb + d^2
+    \end{bmatrix}}
+
+\]
+
+\[
+    P^2 = 
+    \begin{bmatrix}
+    \colorbox{yellow!50}{a} & \colorbox{yellow!75}{b} \\
+    c & d
+    \end{bmatrix}
+    \begin{bmatrix}
+        a & \colorbox{yellow!50}{b} \\
+        c & \colorbox{yellow!75}{d}
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+    aa+bc & \colorbox{yellow!50}{ab+bd} \\
+    ca+dc & cb + dd
+    \end{bmatrix}
+    \quad
+    \text{or} 
+    \quad
+    \boxed{\begin{bmatrix}
+    a^2+bc & ab+bd \\
+    ca+dc & cb + d^2
+    \end{bmatrix}}
+
+\]
+
+\[
+    P^2 = 
+    \begin{bmatrix}
+    a & b \\
+    \colorbox{yellow!50}{c} & \colorbox{yellow!75}{d}
+    \end{bmatrix}
+    \begin{bmatrix}
+        \colorbox{yellow!50}{a} & b \\
+        \colorbox{yellow!75}{c} & d
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+    aa+bc & ab+bd \\
+    \colorbox{yellow!50}{ca+dc} & cb + dd
+    \end{bmatrix}
+    \quad
+    \text{or} 
+    \quad
+    \boxed{\begin{bmatrix}
+    a^2+bc & ab+bd \\
+    ca+dc & cb + d^2
+    \end{bmatrix}}
+
+\]
+
+\[
+    P^2 = 
+    \begin{bmatrix}
+    a & b \\
+    \colorbox{yellow!50}{c} & \colorbox{yellow!75}{d}
+    \end{bmatrix}
+    \begin{bmatrix}
+        a & \colorbox{yellow!50}{b} \\
+        c & \colorbox{yellow!75}{d}
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+    aa+bc & ab+bd \\
+    ca+dc & \colorbox{yellow!50}{cb+dd}
+    \end{bmatrix}
+    \quad
+    \text{or} 
+    \quad
+    \boxed{\begin{bmatrix}
+    a^2+bc & ab+bd \\
+    ca+dc & cb + d^2
+    \end{bmatrix}}
+
+\]
+
+
+\text{breakdown}
+
+\[
+    \begin{bmatrix}
+        a & b
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        a \\ 
+        c
+    \end{bmatrix}
+    =
+    aa +bc
+\]
+
+\[
+    \begin{bmatrix}
+        a & b
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        b \\ 
+        d
+    \end{bmatrix}
+    =
+    ab +bd
+\]
+
+\[
+    \begin{bmatrix}
+        c & d
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        a \\ 
+        c
+    \end{bmatrix}
+    =
+    ca +dc
+\]
+
+\[
+    \begin{bmatrix}
+        c & d
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        b \\ 
+        d
+    \end{bmatrix}
+    =
+    cb +dd
+\]
+
+\[
+    \begin{bmatrix}
+        c & d
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        b \\ 
+        d
+    \end{bmatrix}
+    =
+    cb + d^2
+\]
+
+\text{therefore:}
+\[
+P^2 = \begin{bmatrix}
+    aa+bc & ab+bd \\
+    ca+dc & cb+dd
+    \end{bmatrix}
+\]
+
+\text{or}
+
+\[
+P^2 = \begin{bmatrix}
+    a^2+bc & ab+bd \\
+    ca+dc & cb + d^2
+    \end{bmatrix}   
+\]
+
+
+\end{document}
+