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 Manually[[:space:]]Disecting[[:space:]]arXiv2512.15720/main.pdf filter=lfs diff=lfs merge=lfs -text
+simple_trend_filter[[:space:]]dominant_poc_displacement/main.pdf filter=lfs diff=lfs merge=lfs -text

simple_trend_filter dominant_poc_displacement/main.pdf

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+version https://git-lfs.github.com/spec/v1
+oid sha256:ec1738dbb5595b4b54fa6100b9f6689b50e758d6d3f5278fa1773d2ddde46015
+size 140624

simple_trend_filter dominant_poc_displacement/main.tex

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+
+\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{Simple Trend Filter: Dominant POC Displacement}
+\author{algorembrant}
+\date{\today}
+
+\begin{document}
+\maketitle
+
+I will formalize the simple trend filter as
+
+\begin{align*}
+    vali(buy) = \left(\frac{\sum_{\substack{a = 0 \\ \delta_{t-a} \textcolor{red}{>} 0}}^{n} |\delta_{t-a}|}{\sum_{\substack{a = 0 \\ \delta_{t-a} \gtrless 0}}^{n} |\delta_{t-a}|} \times 100 \right) > 90\%, \quad valid(sell) = \left(\frac{\sum_{\substack{a = 0 \\ \delta_{t-a} \textcolor{red}{<} 0}}^{n} |\delta_{t-a}|}{\sum_{\substack{a = 0 \\ \delta_{t-a} \gtrless 0}}^{n} |\delta_{t-a}|} \times 100 \right) > 90\%
+\end{align*}
+
+This formula can be used in any typed of dataset becuase the main component is a simple change $\delta$
+
+\section{origin}
+
+Consider the dataset below, let's say that from A to E is our setup (you may call this our data within a window size), and our developing point-of-control are as follows:
+
+\begin{tabular}{|l|c|r|}
+  \hline
+  index (t) & POC & $\delta$ \\
+  \hline
+  1 & \$4490.175 & \colorbox{gray!20}{\$0.000} \\
+  2 & \$490.179 & \colorbox{blue!20}{+\$0.004} \\
+  3 & \$490.008 & \colorbox{red!20}{- \$0.171} \\
+  4 & \$491.029 & \colorbox{blue!20}{+\$1.021} \\
+  5 & \$496.029 & \colorbox{blue!20}{+\$5.000} \\
+  \hline
+\end{tabular}
+
+And we are curently at timestep index = 5. Assume we are looking for somesort of confirmation. To solve for valid(buy,sell), the procedding calculation are as follows.
+
+where a simple change in POC is defined as:
+
+\begin{align*}
+\delta = POC_t - POC_{t-1}
+\end{align*}
+
+\begin{align*}
+    buy &= \sum_{\substack{a = 0 \\ \delta_{t-a} > 0}}^{4} \delta_{t-a} \\
+    buy &= 0.004_5 + 1.021_4 + 5.000_2 \\
+    buy &= 6.025 \\
+\end{align*}
+
+\begin{align*}
+    sell &= \sum_{\substack{a = 0 \\ \delta_{t-a} < 0}}^{4} | \delta_{t-a}| \\
+    sell &= |-0.171_3| \\
+    sell &= 0.171 \\
+\end{align*}
+
+therefore
+
+\begin{align*}
+    total &= buy + sell \\
+    total &= 6.025 + 0.171 \\
+    total &= 6.196
+\end{align*}
+
+and we calculate what side is dominant
+
+\begin{align*}
+buyside &= \frac{buy}{total} \quad sellside = \frac{sell}{total}\\
+buyside &= \frac{6.025}{6.196} \quad sellside = \frac{0.171}{6.196} \\
+buyside &= 0.97 \quad sellside = 0.03
+\end{align*}
+
+Valid creteria: if the dominance is above 90\% then its trending. Therefore in this example we consider the situation as buy trend.
+
+\newpage
+\section{formalization}
+
+\begin{align*}
+\delta &= POC_t - POC_{t-1}, \quad \text{change in POC} \\
+t &= \text{current timestep, \quad index} \\
+a,n &= \text{lookback parameters}
+\end{align*}
+
+\begin{align*}
+    dominance(buy) = \frac{\sum_{\substack{a = 0 \\ \delta_{t-a} \textcolor{red}{>} 0}}^{n} |\delta_{t-a}|}{\sum_{\substack{a = 0 \\ \delta_{t-a} \gtrless 0}}^{n} |\delta_{t-a}|}
+\end{align*}
+
+\begin{align*}
+    dominance(sell) = \frac{\sum_{\substack{a = 0 \\ \delta_{t-a} \textcolor{red}{<} 0}}^{n} |\delta_{t-a}|}{\sum_{\substack{a = 0 \\ \delta_{t-a} \gtrless 0}}^{n} |\delta_{t-a}|}
+\end{align*}
+
+\begin{align*}
+    valid(buy) = \left(\frac{\sum_{\substack{a = 0 \\ \delta_{t-a} \textcolor{red}{>} 0}}^{n} |\delta_{t-a}|}{\sum_{\substack{a = 0 \\ \delta_{t-a} \gtrless 0}}^{n} |\delta_{t-a}|} \times 100 \right) > 90\%
+\end{align*}
+
+\begin{align*}
+    valid(sell) = \left(\frac{\sum_{\substack{a = 0 \\ \delta_{t-a} \textcolor{red}{<} 0}}^{n} |\delta_{t-a}|}{\sum_{\substack{a = 0 \\ \delta_{t-a} \gtrless 0}}^{n} |\delta_{t-a}|} \times 100 \right) > 90\%
+\end{align*}
+
+\end{document}
+