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二项树收敛到正态分布（3D）

tag-icon • tag-icon tikz-pgf pgfplots 3d tikz-3dplot fpu
二项树收敛到正态分布（3D）

我尝试得到同样的想法布朗运动 3D 正态分布但用重组二项树代替布朗运动，并且每个节点都有相应的二项分布。这个想法是为了展示二项树如何收敛到正态分布。下图是 2D（不太清晰）版本。如何让二项树在一个平面上生长，并看到二项分布收敛到正态分布？

在此处输入图片描述

在二项分布我们得到 binom 函数

declare function={binom(\k,\n,\p)=\n!/(\k!*(\n-\k)!)*\p^\k*(1-\p)^(\n-\k);}

从重组二项式树我们得到了递归二叉树的思想。然而我甚至无法给出一个 MWE。

答案1

从@marmot 的回答开始，但我不太了解坐标和-0.5我需要在x坐标上进行的转变。

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{tikz-3dplot}
\usetikzlibrary{3d}
\usetikzlibrary{fpu}

\usepackage{xintexpr}
% 8 digits precision is enough
\xintDigits := 8;
% I don't like the order of variables but I picked it up from OP
\xintdeffloatfunc binom(k, n, p) := binomial(n, k)*p^k*(1-p)^(n-k);

\begin{document}

\tdplotsetmaincoords{110}{-30}

\begin{tikzpicture}[binom tree/.style={insert path={%
  foreach \X in {1,...,#1}
 {foreach \Y in {1,...,\X} {(\X,\Y-\X/2) -- (\X+1,\Y+1/2-\X/2)
 (\X,\Y-\X/2) -- (\X+1,\Y-1/2-\X/2)} }}},tdplot_main_coords]

 \begin{scope}[canvas is xy plane at z=0,scale=0.5]

  \draw[blue,binom tree=15];

 \end{scope}

 \tdplotsetrotatedcoords{0}{0}{0}

 \begin{scope}[tdplot_rotated_coords]

  \xintFor* #1 in {\xintSeq{2}{16}}:
  {
  \begin{scope}[canvas is yz plane at x=#1/2, scale=0.5]

   \draw[red,thick] plot[smooth] coordinates {\xintthecoords
     \xintfloatexpr seq((x-#1/2+0.5,8*binom(x, #1, 0.5)), x=0..#1)\relax};

  \end{scope}
  }

 \end{scope}
\end{tikzpicture}
\end{document}

在此处输入图片描述

我更了解坐标。我认为使用重新缩放的条形图并考虑到“ecart 类型”更接近原始问题。（但我保留了 marmot 答案中的 3D 设计）。

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{tikz-3dplot}
\usetikzlibrary{3d}
\usetikzlibrary{fpu}

\usepackage{xintexpr}
% 8 digits precision is enough
\xintDigits := 8;
% I don't like the order of variables but I picked it up from OP
\xintdeffloatfunc binom(k, n, p) := binomial(n, k)*p^k*(1-p)^(n-k);

\begin{document}

\tdplotsetmaincoords{110}{-30}

\begin{tikzpicture}[binom tree/.style={insert path={%
  foreach \X in {1,...,#1}
 {foreach \Y in {1,...,\X} {(\X,\Y-\X/2) -- (\X+1,\Y+1/2-\X/2)
 (\X,\Y-\X/2) -- (\X+1,\Y-1/2-\X/2)} }}},tdplot_main_coords]

 \begin{scope}[canvas is xy plane at z=0]

  \draw[blue,binom tree=48];

 \end{scope}

 \tdplotsetrotatedcoords{0}{0}{0}

 \begin{scope}[tdplot_rotated_coords]

  \xintFor* #1 in {\xintSeq{0}{48}}:
  {
  \begin{scope}[canvas is yz plane at x=#1+1]
  % ybar means in z direction here, as horizontal coordinate is y
   \draw[red,thick] plot[ybar] coordinates {\xintthecoords
     \xintfloatexpr subs(
           seq((y-#1/2+0.5,A*binom(y, #1, 0.5)), y=0..#1)
          ,A=sqrt(#1))\relax};

  \end{scope}
  }

 \end{scope}
\end{tikzpicture}
\end{document}

在此处输入图片描述

以下是

\begin{tikzpicture}[scale=0.5,binom tree/.style={insert path={%
  foreach \X in {1,...,#1}
 {foreach \Y in {1,...,\X} {(\X,\Y-\X/2) -- (\X+1,\Y+1/2-\X/2)
 (\X,\Y-\X/2) -- (\X+1,\Y-1/2-\X/2)} }}},tdplot_main_coords]

 \begin{scope}[canvas is xy plane at z=0]

  \draw[blue,binom tree=96];

 \end{scope}

 \tdplotsetrotatedcoords{0}{0}{0}

 \begin{scope}[tdplot_rotated_coords]

  \xintFor* #1 in {\xintSeq[6]{0}{96}}:
  {
  \begin{scope}[canvas is yz plane at x=#1+1]
  % ybar means in z direction here, as horizontal coordinate is y
   \fill[red,thick] plot[ybar] coordinates {\xintthecoords
     \xintfloatexpr subs(
           seq((y-#1/2+0.5,A*binom(y, #1, 0.5)), y=0..#1)
          ,A=4*sqrt(#1))\relax};

  \end{scope}
  }

 \end{scope}
\end{tikzpicture}

大部分 (超过 3/4) 的时间都花在蓝色画布上。

在此处输入图片描述

这里是p=0.25。

 \begin{scope}[tdplot_rotated_coords]

  \xintFor* #1 in {\xintSeq[6]{0}{96}}:
  {
  \begin{scope}[canvas is yz plane at x=#1+1]
  % ybar means in z direction here, as horizontal coordinate is y
   \fill[red,thick] plot[ybar] coordinates {\xintthecoords
     \xintfloatexpr subs(
           seq((y-#1/2+0.5,A*binom(y, #1, 0.25)), y=0..#1)
          ,A=4*sqrt(#1))\relax};

  \end{scope}
  }

 \end{scope}

在此处输入图片描述

完整代码和动画（p=0.25）

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{tikz-3dplot}
\usetikzlibrary{3d}
\usetikzlibrary{fpu}

\usepackage{xintexpr}
% 8 digits precision is enough
\xintDigits := 8;
% I don't like the order of variables but I picked it up from OP
\xintdeffloatfunc binom(k, n, p) := binomial(n, k)*p^k*(1-p)^(n-k);

\begin{document}

\tdplotsetmaincoords{110}{-30}

  \xintFor* #1 in {\xintSeq[3]{2}{50}}:
  {
\begin{tikzpicture}[scale=0.5,binom tree/.style={insert path={%
  foreach \X in {1,...,#1}
 {foreach \Y in {1,...,\X} {(\X,\Y-\X/2) -- (\X+1,\Y+1/2-\X/2)
 (\X,\Y-\X/2) -- (\X+1,\Y-1/2-\X/2)} }}},tdplot_main_coords]

% \xintifForLast {
 \begin{scope}[canvas is xy plane at z=0]

  \draw[very thin] (1,-25.5)--(51,-25.5) -- (51,25.5)-- (1, 25.5) -- cycle;
  %\draw[blue, very thick] (0,0)--(33,-16.5) -- (33,16.5)-- cycle;
  % \draw[blue,binom tree=32];
  % \draw[blue,binom tree=96];
  \draw[blue, binom tree=#1];

  \draw (#1+1, 25.5) node [right, scale=2] {\smash{\rlap{$#1$}}};
 \end{scope}
% }{}

 \tdplotsetrotatedcoords{0}{0}{0}

 \begin{scope}[tdplot_rotated_coords]

  \begin{scope}[canvas is yz plane at x=#1+1]
  % ybar means in z direction here, as horizontal coordinate is y
   \fill[red,thick] plot[ybar] coordinates {\xintthecoords
     \xintfloatexpr subs(
           seq((y-#1/2+0.5,A*binom(y, #1, 0.25)), y=0..#1)
          ,A=4*sqrt(#1))\relax};

  \end{scope}

 \end{scope}
\end{tikzpicture}
  }

\end{document}

执行pdflatex此操作然后

convert -verbose -delay 75 -loop 0 -density 100 binomialplotV.pdf binomialplotV.gif

在此处输入图片描述

注意p=0.15, 0.5, 0.85，中心情况更加分散，这是正常的，因为它的方差较大。

在此处输入图片描述

（放大时显示效果更佳）

答案2

这是一个快速写成的提案，其中我用高斯分布来近似二项分布。

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{tikz-3dplot}
\usetikzlibrary{3d}
\begin{document}
\tdplotsetmaincoords{110}{-30}
\begin{tikzpicture}[binom tree/.style={insert path={%
foreach \X in {1,...,#1}
 { foreach \Y in {1,...,\X} {(\X,\Y-\X/2) -- (\X+1,\Y+1/2-\X/2)
 (\X,\Y-\X/2) -- (\X+1,\Y-1/2-\X/2)} }}},tdplot_main_coords]
 \begin{scope}[canvas is xy plane at z=0,scale=0.5]
  \draw[blue,binom tree=15];
 \end{scope} 
 \tdplotsetrotatedcoords{0}{0}{0}
 \begin{scope}[tdplot_rotated_coords,declare function={gauss(\x,\y,\z)=1/(\y*sqrt(2*pi))*exp(-((\x-\z)^2)/(2*\y^2));}]
  \begin{scope}[canvas is yz plane at x=4,scale=0.5]
   \draw[red,thick] plot[variable=\x,domain=-4:4,samples=51,smooth] (\x+0.5,{6*gauss(\x,1,0)});
  \end{scope}
  \begin{scope}[canvas is yz plane at x=8,scale=0.5]
   \draw[red,thick] plot[variable=\x,domain=-8:8,samples=51,smooth] (\x+0.5,{6*gauss(\x,2,0)});
  \end{scope}
 \end{scope}
\end{tikzpicture}
\end{document}

在此处输入图片描述

利用该fpu库您可以绘制一些二项分布，但不幸的是，它并不能带给您很大的帮助。

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{tikz-3dplot}
\usetikzlibrary{3d}
\usetikzlibrary{fpu} 
\begin{document}
\tdplotsetmaincoords{110}{-30}
\begin{tikzpicture}[binom tree/.style={insert path={%
foreach \X in {1,...,#1}
 { foreach \Y in {1,...,\X} {(\X,\Y-\X/2) -- (\X+1,\Y+1/2-\X/2)
 (\X,\Y-\X/2) -- (\X+1,\Y-1/2-\X/2)} }}},tdplot_main_coords]
 \begin{scope}[canvas is xy plane at z=0,scale=0.5]
  \draw[blue,binom tree=15];
 \end{scope} 
 \tdplotsetrotatedcoords{0}{0}{0}
 \begin{scope}[tdplot_rotated_coords,declare
 function={gauss(\x,\y,\z)=1/(\y*sqrt(2*pi))*exp(-((\x-\z)^2)/(2*\y^2));},
 declare function={binom(\k,\n,\p)=\n!/(\k!*(\n-\k)!)*\p^\k*(1-\p)^(\n-\k);}]
  \begin{scope}[canvas is yz plane at x=4,scale=0.5]
   \pgfkeys{/pgf/fpu,/pgf/fpu/output format=fixed} 
   \pgfmathsetmacro{\Z}{4*binom(0,8,0.5)}
   \xdef\lstX{(-3.5,\Z)}
   \foreach \X [count=\Y] in {1,...,8}
   {
   \pgfmathsetmacro{\Z}{4*binom(\X,8,0.5)}
   \xdef\lstX{\lstX (\X-3.5,\Z)}
   }
   \draw[red,thick] plot[smooth] coordinates {\lstX};
  \end{scope}
  \begin{scope}[canvas is yz plane at x=6,scale=0.5]
%    \pgfmathsetmacro{\Z}{binom(1,12,0.5)}
%    \xdef\lstX{(-6,\Z)}
%    \foreach \X [count=\Y] in {1,...,12}
%    {
%    \pgfmathsetmacro{\Z}{binom(\X,12,0.5)}
%    \typeout{\Z}
%    \xdef\lstX{\lstX (\X-6,\Z)}
%    }
%    \draw[red,thick] plot[smooth] coordinates {\lstX};
  \end{scope}
 \end{scope}
\end{tikzpicture}
\end{document}

在此处输入图片描述

解决这个问题的最可能的方法是使用伽马函数。但我不确定这是否可以与高斯区分开来。

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