1 | \documentclass[12pt]{article} |
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2 | |
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3 | % $Id: Statistics.tex 831 2007-03-27 13:22:09Z peter $ |
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4 | % |
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5 | % Copyright (C) 2005 Peter Johansson |
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6 | % Copyright (C) 2006 Jari Häkkinen, Markus Ringnér, Peter Johansson |
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7 | % Copyright (C) 2007 Peter Johansson |
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8 | % |
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9 | % This file is part of the yat library, http://lev.thep.lu.se/trac/yat |
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10 | % |
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11 | % The yat library is free software; you can redistribute it and/or |
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14 | % License, or (at your option) any later version. |
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15 | % |
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16 | % The yat library is distributed in the hope that it will be useful, |
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17 | % but WITHOUT ANY WARRANTY; without even the implied warranty of |
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19 | % General Public License for more details. |
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20 | % |
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21 | % You should have received a copy of the GNU General Public License |
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23 | % Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA |
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24 | % 02111-1307, USA. |
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25 | |
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26 | |
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27 | |
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28 | \flushbottom |
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29 | \footskip 54pt |
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35 | \textheight 230mm |
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36 | \textwidth 165mm |
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44 | |
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45 | \renewcommand{\d}{{\mathrm{d}}} |
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46 | \newcommand{\nd}{$^{\mathrm{nd}}$} |
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47 | \newcommand{\eg}{{\it {e.g.}}} |
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48 | \newcommand{\ie}{{\it {i.e., }}} |
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49 | \newcommand{\etal}{{\it {et al.}}} |
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50 | \newcommand{\eref}[1]{Eq.~(\ref{e:#1})} |
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51 | \newcommand{\fref}[1]{Fig.~\ref{f:#1}} |
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52 | \newcommand{\ovr}[2]{\left(\begin{array}{c} #1 \\ #2 \end{array}\right)} |
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53 | |
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54 | \begin{document} |
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55 | |
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56 | \large |
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57 | {\bf Weighted Statistics} |
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58 | \normalsize |
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59 | |
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60 | \tableofcontents |
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61 | \clearpage |
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62 | |
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63 | \section{Introduction} |
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64 | There are several different reasons why a statistical analysis needs |
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65 | to adjust for weighting. In literature reasons are mainly diveded in |
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66 | to groups. |
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67 | |
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68 | The first group is when some of the measurements are known to be more |
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69 | precise than others. The more precise a measurement is, the larger |
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70 | weight it is given. The simplest case is when the weight are given |
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71 | before the measurements and they can be treated as deterministic. It |
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72 | becomes more complicated when the weight can be determined not until |
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73 | afterwards, and even more complicated if the weight depends on the |
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74 | value of the observable. |
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75 | |
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76 | The second group of situations is when calculating averages over one |
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77 | distribution and sampling from another distribution. Compensating for |
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78 | this discrepency weights are introduced to the analysis. A simple |
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79 | example may be that we are interviewing people but for economical |
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80 | reasons we choose to interview more people from the city than from the |
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81 | countryside. When summarizing the statistics the answers from the city |
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82 | are given a smaller weight. In this example we are choosing the |
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83 | proportions of people from countryside and people from city being |
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84 | intervied. Hence, we can determine the weights before and consider |
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85 | them to be deterministic. In other situations the proportions are not |
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86 | deterministic, but rather a result from the sampling and the weights |
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87 | must be treated as stochastic and only in rare situations the weights |
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88 | can be treated as independent of the observable. |
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89 | |
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90 | Since there are various origins for a weight occuring in a statistical |
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91 | analysis, there are various ways to treat the weights and in general |
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92 | the analysis should be tailored to treat the weights correctly. We |
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93 | have not chosen one situation for our implementations, so see specific |
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94 | function documentation for what assumtions are made. Though, common |
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95 | for implementations are the following: |
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96 | \begin{itemize} |
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97 | \item Setting all weights to unity yields the same result as the |
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98 | non-weighted version. |
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99 | \item Rescaling the weights does not change any function. |
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100 | \item Setting a weight to zero is equivalent to removing the data point. |
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101 | \end{itemize} |
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102 | An important case is when weights are binary (either 1 or 0). Then we |
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103 | get the same result using the weighted version as using the data with |
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104 | weight not equal to zero and the non-weighted version. Hence, using |
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105 | binary weights and the weighted version missing values can be treated |
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106 | in a proper way. |
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107 | |
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108 | \section{AveragerWeighted} |
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109 | |
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110 | |
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111 | |
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112 | \subsection{Mean} |
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113 | |
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114 | For any situation the weight is always designed so the weighted mean |
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115 | is calculated as $m=\frac{\sum w_ix_i}{\sum w_i}$, which obviously |
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116 | fulfills the conditions above. |
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117 | |
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118 | In the case of varying measurement error, it could be motivated that |
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119 | the weight shall be $w_i = 1/\sigma_i^2$. We assume measurement error |
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120 | to be Gaussian and the likelihood to get our measurements is |
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121 | $L(m)=\prod |
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122 | (2\pi\sigma_i^2)^{-1/2}e^{-\frac{(x_i-m)^2}{2\sigma_i^2}}$. We |
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123 | maximize the likelihood by taking the derivity with respect to $m$ on |
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124 | the logarithm of the likelihood $\frac{d\ln L(m)}{dm}=\sum |
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125 | \frac{x_i-m}{\sigma_i^2}$. Hence, the Maximum Likelihood method yields |
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126 | the estimator $m=\frac{\sum w_i/\sigma_i^2}{\sum 1/\sigma_i^2}$. |
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127 | |
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128 | |
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129 | \subsection{Variance} |
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130 | In case of varying variance, there is no point estimating a variance |
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131 | since it is different for each data point. |
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132 | |
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133 | Instead we look at the case when we want to estimate the variance over |
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134 | $f$ but are sampling from $f'$. For the mean of an observable $O$ we |
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135 | have $\widehat O=\sum\frac{f}{f'}O_i=\frac{\sum w_iO_i}{\sum |
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136 | w_i}$. Hence, an estimator of the variance of $X$ is |
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137 | \begin{eqnarray} |
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138 | \sigma^2=<X^2>-<X>^2= |
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139 | \\\frac{\sum w_ix_i^2}{\sum w_i}-\frac{(\sum w_ix_i)^2}{(\sum w_i)^2}= |
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140 | \\\frac{\sum w_i(x_i^2-m^2)}{\sum w_i}= |
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141 | \\\frac{\sum w_i(x_i^2-2mx_i+m^2)}{\sum w_i}= |
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142 | \\\frac{\sum w_i(x_i-m)^2}{\sum w_i} |
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143 | \end{eqnarray} |
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144 | This estimator fulfills that it is invariant under a rescaling and |
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145 | having a weight equal to zero is equivalent to removing the data |
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146 | point. Having all weights equal to unity we get $\sigma=\frac{\sum |
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147 | (x_i-m)^2}{N}$, which is the same as returned from Averager. Hence, |
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148 | this estimator is slightly biased, but still very efficient. |
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149 | |
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150 | \subsection{Standard Error} |
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151 | The standard error squared is equal to the expexted squared error of |
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152 | the estimation of $m$. The squared error consists of two parts, the |
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153 | variance of the estimator and the squared |
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154 | bias. $<m-\mu>^2=<m-<m>+<m>-\mu>^2=<m-<m>>^2+(<m>-\mu)^2$. In the |
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155 | case when weights are included in analysis due to varying measurement |
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156 | errors and the weights can be treated as deterministic ,we have |
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157 | \begin{equation} |
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158 | Var(m)=\frac{\sum w_i^2\sigma_i^2}{\left(\sum w_i\right)^2}= |
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159 | \frac{\sum w_i^2\frac{\sigma_0^2}{w_i}}{\left(\sum w_i\right)^2}= |
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160 | \frac{\sigma_0^2}{\sum w_i}, |
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161 | \end{equation} |
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162 | where we need to estimate $\sigma_0^2$. Again we have the likelihood |
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163 | $L(\sigma_0^2)=\prod\frac{1}{\sqrt{2\pi\sigma_0^2/w_i}}\exp{-\frac{w_i(x-m)^2}{2\sigma_0^2}}$ |
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164 | and taking the derivity with respect to $\sigma_o^2$, $\frac{d\ln |
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165 | L}{d\sigma_i^2}=\sum |
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166 | -\frac{1}{2\sigma_0^2}+\frac{w_i(x-m)^2}{2\sigma_0^2\sigma_o^2}$ which |
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167 | yields an estimator $\sigma_0^2=\frac{1}{N}\sum w_i(x-m)^2$. This |
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168 | estimator is not ignoring weights equal to zero, because deviation is |
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169 | most often smaller than the expected infinity. Therefore, we modify |
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170 | the expression as follows $\sigma_0^2=\frac{\sum w_i^2}{\left(\sum |
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171 | w_i\right)^2}\sum w_i(x-m)^2$ and we get the following estimator of |
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172 | the variance of the mean $\sigma_0^2=\frac{\sum w_i^2}{\left(\sum |
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173 | w_i\right)^3}\sum w_i(x-m)^2$. This estimator fulfills the conditions |
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174 | above: adding a weight zero does not change it: rescaling the weights |
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175 | does not change it, and setting all weights to unity yields the same |
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176 | expression as in the non-weighted case. |
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177 | |
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178 | In a case when it is not a good approximation to treat the weights as |
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179 | deterministic, there are two ways to get a better estimation. The |
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180 | first one is to linearize the expression $\left<\frac{\sum |
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181 | w_ix_i}{\sum w_i}\right>$. The second method when the situation is |
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182 | more complicated is to estimate the standard error using a |
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183 | bootstrapping method. |
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184 | |
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185 | \section{AveragerPairWeighted} |
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186 | Here data points come in pairs (x,y). We are sampling from $f'_{XY}$ |
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187 | but want to measure from $f_{XY}$. To compensate for this decrepency, |
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188 | averages of $g(x,y)$ are taken as $\sum \frac{f}{f'}g(x,y)$. Even |
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189 | though, $X$ and $Y$ are not independent $(f_{XY}\neq f_Xf_Y)$ we |
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190 | assume that we can factorize the ratio and get $\frac{\sum |
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191 | w_xw_yg(x,y)}{\sum w_xw_y}$ |
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192 | \subsection{Covariance} |
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193 | Following the variance calculations for AveragerWeighted we have |
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194 | $Cov=\frac{\sum w_xw_y(x-m_x)(y-m_y)}{\sum w_xw_y}$ where |
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195 | $m_x=\frac{\sum w_xw_yx}{\sum w_xw_y}$ |
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196 | |
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197 | \subsection{correlation} |
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198 | |
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199 | As the mean is estimated as $m_x=\frac{\sum w_xw_yx}{\sum w_xw_y}$, |
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200 | the variance is estimated as $\sigma_x^2=\frac{\sum |
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201 | w_xw_y(x-m_x)^2}{\sum w_xw_y}$. As in the non-weighted case we define |
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202 | the correlation to be the ratio between the covariance and geometrical |
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203 | average of the variances |
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204 | |
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205 | $\frac{\sum w_xw_y(x-m_x)(y-m_y)}{\sqrt{\sum w_xw_y(x-m_x)^2\sum |
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206 | w_xw_y(y-m_y)^2}}$. |
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207 | |
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208 | This expression fulfills the following |
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209 | \begin{itemize} |
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210 | \item Having N weights the expression reduces to the non-weighted expression. |
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211 | \item Adding a pair of data, in which one weight is zero is equivalent |
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212 | to ignoring the data pair. |
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213 | \item Correlation is equal to unity if and only if $x$ is equal to |
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214 | $y$. Otherwise the correlation is between -1 and 1. |
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215 | \end{itemize} |
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216 | \section{Score} |
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217 | |
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218 | |
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219 | \subsection{Pearson} |
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220 | |
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221 | $\frac{\sum w(x-m_x)(y-m_y)}{\sqrt{\sum w(x-m_x)^2\sum w(y-m_y)^2}}$. |
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222 | |
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223 | See AveragerPairWeighted correlation. |
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224 | |
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225 | \subsection{ROC} |
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226 | |
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227 | An interpretation of the ROC curve area is the probability that if we |
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228 | take one sample from class $+$ and one sample from class $-$, what is |
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229 | the probability that the sample from class $+$ has greater value. The |
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230 | ROC curve area calculates the ratio of pairs fulfilling this |
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231 | |
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232 | \begin{equation} |
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233 | \frac{\sum_{\{i,j\}:x^-_i<x^+_j}1}{\sum_{i,j}1}. |
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234 | \end{equation} |
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235 | |
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236 | An geometrical interpretation is to have a number of squares where |
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237 | each square correspond to a pair of samples. The ROC curve follows the |
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238 | border between pairs in which the samples from class $+$ has a greater |
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239 | value and pairs in which this is not fulfilled. The ROC curve area is |
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240 | the area of those latter squares and a natural extension is to weight |
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241 | each pair with its two weights and consequently the weighted ROC curve |
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242 | area becomes |
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243 | |
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244 | \begin{equation} |
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245 | \frac{\sum_{\{i,j\}:x^-_i<x^+_j}w^-_iw^+_j}{\sum_{i,j}w^-_iw^+_j} |
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246 | \end{equation} |
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247 | |
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248 | This expression is invariant under a rescaling of weight. Adding a |
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249 | data value with weight zero adds nothing to the exprssion, and having |
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250 | all weight equal to unity yields the non-weighted ROC curve area. |
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251 | |
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252 | \subsection{tScore} |
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253 | |
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254 | Assume that $x$ and $y$ originate from the same distribution |
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255 | $N(\mu,\sigma_i^2)$ where $\sigma_i^2=\frac{\sigma_0^2}{w_i}$. We then |
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256 | estimate $\sigma_0^2$ as |
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257 | \begin{equation} |
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258 | \frac{\sum w(x-m_x)^2+\sum w(y-m_y)^2} |
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259 | {\frac{\left(\sum w_x\right)^2}{\sum w_x^2}+ |
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260 | \frac{\left(\sum w_y\right)^2}{\sum w_y^2}-2} |
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261 | \end{equation} |
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262 | The variance of difference of the means becomes |
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263 | \begin{eqnarray} |
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264 | Var(m_x)+Var(m_y)=\\\frac{\sum w_i^2Var(x_i)}{\left(\sum |
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265 | w_i\right)^2}+\frac{\sum w_i^2Var(y_i)}{\left(\sum w_i\right)^2}= |
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266 | \frac{\sigma_0^2}{\sum w_i}+\frac{\sigma_0^2}{\sum w_i}, |
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267 | \end{eqnarray} |
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268 | and consequently the t-score becomes |
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269 | \begin{equation} |
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270 | \frac{\sum w(x-m_x)^2+\sum w(y-m_y)^2} |
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271 | {\frac{\left(\sum w_x\right)^2}{\sum w_x^2}+ |
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272 | \frac{\left(\sum w_y\right)^2}{\sum w_y^2}-2} |
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273 | \left(\frac{1}{\sum w_i}+\frac{1}{\sum w_i}\right), |
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274 | \end{equation} |
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275 | |
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276 | For a $w_i=w$ we this expression get condensed down to |
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277 | \begin{equation} |
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278 | \frac{w\sum (x-m_x)^2+w\sum (y-m_y)^2} |
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279 | {n_x+n_y-2} |
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280 | \left(\frac{1}{wn_x}+\frac{1}{wn_y}\right), |
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281 | \end{equation} |
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282 | in other words the good old expression as for non-weighted. |
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283 | |
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284 | \subsection{FoldChange} |
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285 | Fold-Change is simply the difference between the weighted mean of the |
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286 | two groups //$\frac{\sum w_xx}{\sum w_x}-\frac{\sum w_yy}{\sum w_y}$ |
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287 | |
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288 | \subsection{WilcoxonFoldChange} |
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289 | Taking all pair samples (one from class $+$ and one from class $-$) |
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290 | and calculating the weighted median of the distances. |
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291 | |
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292 | \section{Kernel} |
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293 | \subsection{Polynomial Kernel} |
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294 | The polynomial kernel of degree $N$ is defined as $(1+<x,y>)^N$, where |
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295 | $<x,y>$ is the linear kernel (usual scalar product). For the weighted |
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296 | case we define the linear kernel to be $<x,y>=\sum {w_xw_yxy}$ and the |
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297 | polynomial kernel can be calculated as before |
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298 | $(1+<x,y>)^N$. Is this kernel a proper kernel (always being semi |
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299 | positive definite). Yes, because $<x,y>$ is obviously a proper kernel |
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300 | as it is a scalar product. Adding a positive constant to a kernel |
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301 | yields another kernel so $1+<x,y>$ is still a proper kernel. Then also |
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302 | $(1+<x,y>)^N$ is a proper kernel because taking a proper kernel to the |
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303 | $Nth$ power yields a new proper kernel (see any good book on SVM). |
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304 | \subsection{Gaussian Kernel} |
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305 | We define the weighted Gaussian kernel as $\exp\left(-\frac{\sum |
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306 | w_xw_y(x-y)^2}{\sum w_xw_y}\right)$, which fulfills the conditions |
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307 | listed in the introduction. |
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308 | |
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309 | Is this kernel a proper kernel? Yes, following the proof of the |
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310 | non-weighted kernel we see that $K=\exp\left(-\frac{\sum |
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311 | w_xw_yx^2}{\sum w_xw_y}\right)\exp\left(-\frac{\sum w_xw_yy^2}{\sum |
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312 | w_xw_y}\right)\exp\left(\frac{\sum w_xw_yxy}{\sum w_xw_y}\right)$, |
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313 | which is a product of two proper kernels. $\exp\left(-\frac{\sum |
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314 | w_xw_yx^2}{\sum w_xw_y}\right)\exp\left(-\frac{\sum w_xw_yy^2}{\sum |
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315 | w_xw_y}\right)$ is a proper kernel, because it is a scalar product and |
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316 | $\exp\left(\frac{\sum w_xw_yxy}{\sum w_xw_y}\right)$ is a proper |
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317 | kernel, because it a polynomial of the linear kernel with positive |
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318 | coefficients. As product of two kernel also is a kernel, the Gaussian |
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319 | kernel is a proper kernel. |
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320 | |
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321 | \section{Distance} |
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322 | |
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323 | \section{Regression} |
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324 | \subsection{Naive} |
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325 | \subsection{Linear} |
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326 | We have the model |
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327 | |
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328 | \begin{equation} |
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329 | y_i=\alpha+\beta (x-m_x)+\epsilon_i, |
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330 | \end{equation} |
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331 | |
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332 | where $\epsilon_i$ is the noise. The variance of the noise is |
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333 | inversely proportional to the weight, |
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334 | $Var(\epsilon_i)=\frac{\sigma^2}{w_i}$. In order to determine the |
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335 | model parameters, we minimimize the sum of quadratic errors. |
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336 | |
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337 | \begin{equation} |
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338 | Q_0 = \sum \epsilon_i^2 |
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339 | \end{equation} |
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340 | |
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341 | Taking the derivity with respect to $\alpha$ and $\beta$ yields two conditions |
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342 | |
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343 | \begin{equation} |
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344 | \frac{\partial Q_0}{\partial \alpha} = -2 \sum w_i(y_i - \alpha - |
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345 | \beta (x_i-m_x)=0 |
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346 | \end{equation} |
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347 | |
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348 | and |
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349 | |
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350 | \begin{equation} \frac{\partial Q_0}{\partial \beta} = -2 \sum |
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351 | w_i(x_i-m_x)(y_i-\alpha-\beta(x_i-m_x)=0 |
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352 | \end{equation} |
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353 | |
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354 | or equivalently |
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355 | |
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356 | \begin{equation} |
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357 | \alpha = \frac{\sum w_iy_i}{\sum w_i}=m_y |
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358 | \end{equation} |
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359 | |
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360 | and |
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361 | |
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362 | \begin{equation} \beta=\frac{\sum w_i(x_i-m_x)(y-m_y)}{\sum |
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363 | w_i(x_i-m_x)^2}=\frac{Cov(x,y)}{Var(x)} |
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364 | \end{equation} |
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365 | |
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366 | Note, by having all weights equal we get back the unweighted |
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367 | case. Furthermore, we calculate the variance of the estimators of |
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368 | $\alpha$ and $\beta$. |
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369 | |
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370 | \begin{equation} |
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371 | \textrm{Var}(\alpha )=\frac{w_i^2\frac{\sigma^2}{w_i}}{(\sum w_i)^2}= |
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372 | \frac{\sigma^2}{\sum w_i} |
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373 | \end{equation} |
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374 | |
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375 | and |
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376 | \begin{equation} |
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377 | \textrm{Var}(\beta )= \frac{w_i^2(x_i-m_x)^2\frac{\sigma^2}{w_i}} |
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378 | {(\sum w_i(x_i-m_x)^2)^2}= |
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379 | \frac{\sigma^2}{\sum w_i(x_i-m_x)^2} |
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380 | \end{equation} |
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381 | |
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382 | Finally, we estimate the level of noise, $\sigma^2$. Inspired by the |
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383 | unweighted estimation |
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384 | |
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385 | \begin{equation} |
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386 | s^2=\frac{\sum (y_i-\alpha-\beta (x_i-m_x))^2}{n-2} |
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387 | \end{equation} |
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388 | |
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389 | we suggest the following estimator |
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390 | |
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391 | \begin{equation} s^2=\frac{\sum w_i(y_i-\alpha-\beta (x_i-m_x))^2}{\sum |
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392 | w_i-2\frac{\sum w_i^2}{\sum w_i}} \end{equation} |
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393 | |
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394 | \section{Outlook} |
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395 | \subsection{Hierarchical clustering} |
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396 | A hierarchical clustering consists of two things: finding the two |
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397 | closest data points, merge these two data points two a new data point |
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398 | and calculate the new distances from this point to all other points. |
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399 | |
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400 | In the first item, we need a distance matrix, and if we use Euclidean |
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401 | distanses the natural modification of the expression would be |
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402 | |
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403 | \begin{equation} |
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404 | d(x,y)=\frac{\sum w_i^xw_j^y(x_i-y_i)^2}{\sum w_i^xw_j^y} |
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405 | \end{equation} |
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406 | |
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407 | For the second item, inspired by average linkage, we suggest |
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408 | |
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409 | \begin{equation} |
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410 | d(xy,z)=\frac{\sum w_i^xw_j^z(x_i-z_i)^2+\sum |
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411 | w_i^yw_j^z(y_i-z_i)^2}{\sum w_i^xw_j^z+\sum w_i^yw_j^z} |
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412 | \end{equation} |
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413 | |
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414 | to be the distance between the new merged point $xy$ and $z$, and we |
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415 | also calculate new weights for this point: $w^{xy}_i=w^x_i+w^y_i$ |
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416 | |
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417 | \end{document} |
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418 | |
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419 | |
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420 | |
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