1 | // $Id: Statistics.doxygen 1159 2008-02-26 15:12:30Z peter $ |
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2 | // |
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3 | // Copyright (C) 2005 Peter Johansson |
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4 | // Copyright (C) 2006 Jari Häkkinen, Markus Ringnér, Peter Johansson |
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5 | // Copyright (C) 2007, 2008 Peter Johansson |
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6 | // |
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7 | // This file is part of the yat library, http://trac.thep.lu.se/yat |
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8 | // |
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9 | // The yat library is free software; you can redistribute it and/or |
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10 | // modify it under the terms of the GNU General Public License as |
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11 | // published by the Free Software Foundation; either version 2 of the |
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12 | // License, or (at your option) any later version. |
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13 | // |
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14 | // The yat library is distributed in the hope that it will be useful, |
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15 | // but WITHOUT ANY WARRANTY; without even the implied warranty of |
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16 | // MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU |
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17 | // General Public License for more details. |
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18 | // |
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19 | // You should have received a copy of the GNU General Public License |
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20 | // along with this program; if not, write to the Free Software |
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21 | // Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA |
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22 | // 02111-1307, USA. |
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23 | |
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24 | |
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25 | /** |
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26 | \page weighted_statistics Weighted Statistics |
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27 | |
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28 | \section Introduction |
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29 | There are several different reasons why a statistical analysis needs |
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30 | to adjust for weighting. In literature reasons are mainly diveded in |
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31 | to groups. |
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32 | |
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33 | The first group is when some of the measurements are known to be more |
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34 | precise than others. The more precise a measurement is, the larger |
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35 | weight it is given. The simplest case is when the weight are given |
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36 | before the measurements and they can be treated as deterministic. It |
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37 | becomes more complicated when the weight can be determined not until |
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38 | afterwards, and even more complicated if the weight depends on the |
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39 | value of the observable. |
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40 | |
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41 | The second group of situations is when calculating averages over one |
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42 | distribution and sampling from another distribution. Compensating for |
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43 | this discrepency weights are introduced to the analysis. A simple |
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44 | example may be that we are interviewing people but for economical |
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45 | reasons we choose to interview more people from the city than from the |
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46 | countryside. When summarizing the statistics the answers from the city |
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47 | are given a smaller weight. In this example we are choosing the |
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48 | proportions of people from countryside and people from city being |
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49 | intervied. Hence, we can determine the weights before and consider |
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50 | them to be deterministic. In other situations the proportions are not |
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51 | deterministic, but rather a result from the sampling and the weights |
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52 | must be treated as stochastic and only in rare situations the weights |
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53 | can be treated as independent of the observable. |
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54 | |
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55 | Since there are various origins for a weight occuring in a statistical |
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56 | analysis, there are various ways to treat the weights and in general |
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57 | the analysis should be tailored to treat the weights correctly. We |
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58 | have not chosen one situation for our implementations, so see specific |
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59 | function documentation for what assumtions are made. Though, common |
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60 | for implementations are the following: |
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61 | |
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62 | - Setting all weights to unity yields the same result as the |
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63 | non-weighted version. |
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64 | - Rescaling the weights does not change any function. |
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65 | - Setting a weight to zero is equivalent to removing the data point. |
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66 | |
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67 | An important case is when weights are binary (either 1 or 0). Then we |
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68 | get the same result using the weighted version as using the data with |
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69 | weight not equal to zero and the non-weighted version. Hence, using |
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70 | binary weights and the weighted version missing values can be treated |
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71 | in a proper way. |
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72 | |
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73 | \section AveragerWeighted |
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74 | |
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75 | |
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76 | |
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77 | \subsection Mean |
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78 | |
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79 | For any situation the weight is always designed so the weighted mean |
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80 | is calculated as \f$ m=\frac{\sum w_ix_i}{\sum w_i} \f$, which obviously |
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81 | fulfills the conditions above. |
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82 | |
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83 | |
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84 | |
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85 | In the case of varying measurement error, it could be motivated that |
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86 | the weight shall be \f$ w_i = 1/\sigma_i^2 \f$. We assume measurement error |
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87 | to be Gaussian and the likelihood to get our measurements is |
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88 | \f$ L(m)=\prod |
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89 | (2\pi\sigma_i^2)^{-1/2}e^{-\frac{(x_i-m)^2}{2\sigma_i^2}} \f$. We |
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90 | maximize the likelihood by taking the derivity with respect to \f$ m \f$ on |
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91 | the logarithm of the likelihood \f$ \frac{d\ln L(m)}{dm}=\sum |
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92 | \frac{x_i-m}{\sigma_i^2} \f$. Hence, the Maximum Likelihood method yields |
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93 | the estimator \f$ m=\frac{\sum w_i/\sigma_i^2}{\sum 1/\sigma_i^2} \f$. |
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94 | |
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95 | |
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96 | \subsection Variance |
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97 | In case of varying variance, there is no point estimating a variance |
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98 | since it is different for each data point. |
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99 | |
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100 | Instead we look at the case when we want to estimate the variance over |
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101 | \f$f\f$ but are sampling from \f$ f' \f$. For the mean of an observable \f$ O \f$ we |
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102 | have \f$ \widehat O=\sum\frac{f}{f'}O_i=\frac{\sum w_iO_i}{\sum |
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103 | w_i} \f$. Hence, an estimator of the variance of \f$ X \f$ is |
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104 | |
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105 | \f$ |
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106 | s^2 = <X^2>-<X>^2= |
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107 | \f$ |
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108 | |
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109 | \f$ |
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110 | = \frac{\sum w_ix_i^2}{\sum w_i}-\frac{(\sum w_ix_i)^2}{(\sum w_i)^2}= |
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111 | \f$ |
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112 | |
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113 | \f$ |
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114 | = \frac{\sum w_i(x_i^2-m^2)}{\sum w_i}= |
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115 | \f$ |
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116 | |
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117 | \f$ |
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118 | = \frac{\sum w_i(x_i^2-2mx_i+m^2)}{\sum w_i}= |
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119 | \f$ |
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120 | |
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121 | \f$ |
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122 | = \frac{\sum w_i(x_i-m)^2}{\sum w_i} |
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123 | \f$ |
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124 | |
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125 | This estimator fulfills that it is invariant under a rescaling and |
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126 | having a weight equal to zero is equivalent to removing the data |
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127 | point. Having all weights equal to unity we get \f$ \sigma=\frac{\sum |
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128 | (x_i-m)^2}{N} \f$, which is the same as returned from Averager. Hence, |
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129 | this estimator is slightly biased, but still very efficient. |
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130 | |
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131 | \subsection standard_error Standard Error |
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132 | The standard error squared is equal to the expexted squared error of |
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133 | the estimation of \f$m\f$. The squared error consists of two parts, the |
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134 | variance of the estimator and the squared bias: |
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135 | |
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136 | \f$ |
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137 | <m-\mu>^2=<m-<m>+<m>-\mu>^2= |
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138 | \f$ |
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139 | \f$ |
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140 | <m-<m>>^2+(<m>-\mu)^2 |
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141 | \f$. |
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142 | |
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143 | In the case when weights are included in analysis due to varying |
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144 | measurement errors and the weights can be treated as deterministic, we |
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145 | have |
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146 | |
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147 | \f$ |
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148 | Var(m)=\frac{\sum w_i^2\sigma_i^2}{\left(\sum w_i\right)^2}= |
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149 | \f$ |
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150 | \f$ |
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151 | \frac{\sum w_i^2\frac{\sigma_0^2}{w_i}}{\left(\sum w_i\right)^2}= |
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152 | \f$ |
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153 | \f$ |
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154 | \frac{\sigma_0^2}{\sum w_i}, |
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155 | \f$ |
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156 | |
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157 | where we need to estimate \f$ \sigma_0^2 \f$. Again we have the likelihood |
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158 | |
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159 | \f$ |
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160 | 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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161 | \f$ |
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162 | and taking the derivity with respect to |
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163 | \f$\sigma_o^2\f$, |
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164 | |
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165 | \f$ |
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166 | \frac{d\ln L}{d\sigma_i^2}= |
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167 | \f$ |
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168 | \f$ |
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169 | \sum -\frac{1}{2\sigma_0^2}+\frac{w_i(x-m)^2}{2\sigma_0^2\sigma_o^2} |
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170 | \f$ |
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171 | |
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172 | which |
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173 | yields an estimator \f$ \sigma_0^2=\frac{1}{N}\sum w_i(x-m)^2 \f$. This |
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174 | estimator is not ignoring weights equal to zero, because deviation is |
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175 | most often smaller than the expected infinity. Therefore, we modify |
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176 | the expression as follows \f$\sigma_0^2=\frac{\sum w_i^2}{\left(\sum |
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177 | w_i\right)^2}\sum w_i(x-m)^2\f$ and we get the following estimator of |
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178 | the variance of the mean \f$\sigma_0^2=\frac{\sum w_i^2}{\left(\sum |
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179 | w_i\right)^3}\sum w_i(x-m)^2\f$. This estimator fulfills the conditions |
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180 | above: adding a weight zero does not change it: rescaling the weights |
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181 | does not change it, and setting all weights to unity yields the same |
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182 | expression as in the non-weighted case. |
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183 | |
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184 | In a case when it is not a good approximation to treat the weights as |
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185 | deterministic, there are two ways to get a better estimation. The |
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186 | first one is to linearize the expression \f$\left<\frac{\sum |
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187 | w_ix_i}{\sum w_i}\right>\f$. The second method when the situation is |
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188 | more complicated is to estimate the standard error using a |
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189 | bootstrapping method. |
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190 | |
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191 | \section AveragerPairWeighted |
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192 | Here data points come in pairs (x,y). We are sampling from \f$f'_{XY}\f$ |
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193 | but want to measure from \f$f_{XY}\f$. To compensate for this decrepency, |
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194 | averages of \f$g(x,y)\f$ are taken as \f$\sum \frac{f}{f'}g(x,y)\f$. Even |
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195 | though, \f$X\f$ and \f$Y\f$ are not independent \f$(f_{XY}\neq f_Xf_Y)\f$ we |
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196 | assume that we can factorize the ratio and get \f$\frac{\sum |
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197 | w_xw_yg(x,y)}{\sum w_xw_y}\f$ |
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198 | \subsection Covariance |
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199 | Following the variance calculations for AveragerWeighted we have |
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200 | \f$Cov=\frac{\sum w_xw_y(x-m_x)(y-m_y)}{\sum w_xw_y}\f$ where |
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201 | \f$m_x=\frac{\sum w_xw_yx}{\sum w_xw_y}\f$ |
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202 | |
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203 | \subsection Correlation |
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204 | |
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205 | As the mean is estimated as |
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206 | \f$ |
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207 | m_x=\frac{\sum w_xw_yx}{\sum w_xw_y} |
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208 | \f$, |
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209 | the variance is estimated as |
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210 | \f$ |
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211 | \sigma_x^2=\frac{\sum w_xw_y(x-m_x)^2}{\sum w_xw_y} |
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212 | \f$. |
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213 | As in the non-weighted case we define the correlation to be the ratio |
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214 | between the covariance and geometrical average of the variances |
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215 | |
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216 | \f$ |
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217 | \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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218 | w_xw_y(y-m_y)^2}} |
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219 | \f$. |
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220 | |
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221 | |
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222 | This expression fulfills the following |
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223 | - Having N equal weights the expression reduces to the non-weighted expression. |
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224 | - Adding a pair of data, in which one weight is zero is equivalent |
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225 | to ignoring the data pair. |
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226 | - Correlation is equal to unity if and only if \f$x\f$ is equal to |
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227 | \f$y\f$. Otherwise the correlation is between -1 and 1. |
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228 | |
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229 | \section Score |
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230 | |
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231 | \subsection Pearson |
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232 | |
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233 | \f$\frac{\sum w(x-m_x)(y-m_y)}{\sqrt{\sum w(x-m_x)^2\sum w(y-m_y)^2}}\f$. |
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234 | |
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235 | See AveragerPairWeighted correlation. |
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236 | |
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237 | \subsection ROC |
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238 | |
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239 | An interpretation of the ROC curve area is the probability that if we |
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240 | take one sample from class \f$+\f$ and one sample from class \f$-\f$, what is |
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241 | the probability that the sample from class \f$+\f$ has greater value. The |
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242 | ROC curve area calculates the ratio of pairs fulfilling this |
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243 | |
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244 | \f$ |
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245 | \frac{\sum_{\{i,j\}:x^-_i<x^+_j}1}{\sum_{i,j}1}. |
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246 | \f$ |
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247 | |
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248 | An geometrical interpretation is to have a number of squares where |
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249 | each square correspond to a pair of samples. The ROC curve follows the |
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250 | border between pairs in which the samples from class \f$+\f$ has a greater |
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251 | value and pairs in which this is not fulfilled. The ROC curve area is |
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252 | the area of those latter squares and a natural extension is to weight |
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253 | each pair with its two weights and consequently the weighted ROC curve |
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254 | area becomes |
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255 | |
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256 | \f$ |
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257 | \frac{\sum_{\{i,j\}:x^-_i<x^+_j}w^-_iw^+_j}{\sum_{i,j}w^-_iw^+_j} |
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258 | \f$ |
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259 | |
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260 | This expression is invariant under a rescaling of weight. Adding a |
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261 | data value with weight zero adds nothing to the exprssion, and having |
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262 | all weight equal to unity yields the non-weighted ROC curve area. |
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263 | |
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264 | \subsection tScore |
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265 | |
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266 | Assume that \f$x\f$ and \f$y\f$ originate from the same distribution |
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267 | \f$N(\mu,\sigma_i^2)\f$ where \f$\sigma_i^2=\frac{\sigma_0^2}{w_i}\f$. We then |
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268 | estimate \f$\sigma_0^2\f$ as |
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269 | \f$ |
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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 | \f$ |
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274 | The variance of difference of the means becomes |
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275 | \f$ |
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276 | Var(m_x)+Var(m_y)=\\\frac{\sum w_i^2Var(x_i)}{\left(\sum |
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277 | w_i\right)^2}+\frac{\sum w_i^2Var(y_i)}{\left(\sum w_i\right)^2}= |
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278 | \frac{\sigma_0^2}{\sum w_i}+\frac{\sigma_0^2}{\sum w_i}, |
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279 | \f$ |
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280 | and consequently the t-score becomes |
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281 | \f$ |
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282 | \frac{\sum w(x-m_x)^2+\sum w(y-m_y)^2} |
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283 | {\frac{\left(\sum w_x\right)^2}{\sum w_x^2}+ |
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284 | \frac{\left(\sum w_y\right)^2}{\sum w_y^2}-2} |
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285 | \left(\frac{1}{\sum w_i}+\frac{1}{\sum w_i}\right), |
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286 | \f$ |
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287 | |
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288 | For a \f$w_i=w\f$ we this expression get condensed down to |
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289 | \f$ |
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290 | \frac{w\sum (x-m_x)^2+w\sum (y-m_y)^2} |
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291 | {n_x+n_y-2} |
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292 | \left(\frac{1}{wn_x}+\frac{1}{wn_y}\right), |
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293 | \f$ |
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294 | in other words the good old expression as for non-weighted. |
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295 | |
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296 | \subsection FoldChange |
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297 | Fold-Change is simply the difference between the weighted mean of the |
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298 | two groups \f$\frac{\sum w_xx}{\sum w_x}-\frac{\sum w_yy}{\sum w_y}\f$ |
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299 | |
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300 | \subsection WilcoxonFoldChange |
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301 | Taking all pair samples (one from class \f$+\f$ and one from class \f$-\f$) |
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302 | and calculating the weighted median of the distances. |
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303 | |
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304 | \section Distance |
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305 | |
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306 | A Distance measures how far apart two ranges are. A Distance should |
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307 | preferably meet some criteria: |
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308 | |
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309 | - It is symmetric, \f$ d(x,y) = d(y,x) \f$, that is distance from \f$ |
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310 | x \f$ to \f$ y \f$ equals the distance from \f$ y \f$ to \f$ x \f$. |
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311 | - Zero self-distance: \f$ d(x,x) = 0 \f$ |
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312 | - Triangle inequality: \f$ d(x,z) \le d(x,y) + d(y,z) \f$ |
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313 | |
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314 | \subsection weighted_distance Weighted Distance |
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315 | |
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316 | Weighted Distance is an extension of usual unweighted distances, in |
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317 | which each data point is accompanied with a weight. A weighted |
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318 | distance should meet some criteria: |
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319 | |
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320 | - Having all unity weights should yield the unweighted case. |
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321 | - Rescaling invariant - \f$ w_i = Cw_i \f$ does not change the distance. |
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322 | - Having a \f$ w_x = 0 \f$ the distance should ignore corresponding |
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323 | \f$ x \f$, \f$ y \f$, and \f$ w_y \f$. |
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324 | - A zero weight should not result in a very different distance than a |
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325 | small weight, in other words, modifying a weight should change the |
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326 | distance in a continuous manner. |
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327 | - The duplicate property. If data is coming in duplicate such that |
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328 | \f$ x_{2i}=x_{2i+1} \f$, then the case when \f$ w_{2i}=w_{2i+1} \f$ |
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329 | should equal to if you set \f$ w_{2i}=0 \f$. |
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330 | |
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331 | For a weighted distance, meeting these criteria, it might be difficult |
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332 | to show that the triangle inequality is fulfilled. For most algorithms |
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333 | the triangle inequality is not essential for the distance to work |
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334 | properly, so if you need to choose between fulfilling triangle |
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335 | inequality and these latter criteria it is preferable to meet the |
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336 | latter criteria. Here follows some examples: |
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337 | |
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338 | \subsection EuclideanDistance |
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339 | |
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340 | \subsection PearsonDistance |
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341 | |
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342 | \section Kernel |
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343 | \subsection polynomial_kernel Polynomial Kernel |
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344 | The polynomial kernel of degree \f$N\f$ is defined as \f$(1+<x,y>)^N\f$, where |
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345 | \f$<x,y>\f$ is the linear kernel (usual scalar product). For the weighted |
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346 | case we define the linear kernel to be |
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347 | \f$<x,y>=\frac{\sum {w_xw_yxy}}{\sum{w_xw_y}}\f$ and the |
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348 | polynomial kernel can be calculated as before |
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349 | \f$(1+<x,y>)^N\f$. |
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350 | |
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351 | \subsection gaussian_kernel Gaussian Kernel |
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352 | We define the weighted Gaussian kernel as \f$\exp\left(-N\frac{\sum |
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353 | w_xw_y(x-y)^2}{\sum w_xw_y}\right)\f$. |
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354 | |
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355 | \section Regression |
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356 | \subsection Naive |
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357 | \subsection Linear |
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358 | We have the model |
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359 | |
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360 | \f$ |
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361 | y_i=\alpha+\beta (x-m_x)+\epsilon_i, |
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362 | \f$ |
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363 | |
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364 | where \f$\epsilon_i\f$ is the noise. The variance of the noise is |
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365 | inversely proportional to the weight, |
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366 | \f$Var(\epsilon_i)=\frac{\sigma^2}{w_i}\f$. In order to determine the |
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367 | model parameters, we minimimize the sum of quadratic errors. |
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368 | |
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369 | \f$ |
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370 | Q_0 = \sum \epsilon_i^2 |
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371 | \f$ |
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372 | |
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373 | Taking the derivity with respect to \f$\alpha\f$ and \f$\beta\f$ yields two conditions |
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374 | |
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375 | \f$ |
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376 | \frac{\partial Q_0}{\partial \alpha} = -2 \sum w_i(y_i - \alpha - |
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377 | \beta (x_i-m_x)=0 |
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378 | \f$ |
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379 | |
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380 | and |
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381 | |
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382 | \f$ \frac{\partial Q_0}{\partial \beta} = -2 \sum |
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383 | w_i(x_i-m_x)(y_i-\alpha-\beta(x_i-m_x)=0 |
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384 | \f$ |
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385 | |
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386 | or equivalently |
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387 | |
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388 | \f$ |
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389 | \alpha = \frac{\sum w_iy_i}{\sum w_i}=m_y |
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390 | \f$ |
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391 | |
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392 | and |
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393 | |
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394 | \f$ \beta=\frac{\sum w_i(x_i-m_x)(y-m_y)}{\sum |
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395 | w_i(x_i-m_x)^2}=\frac{Cov(x,y)}{Var(x)} |
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396 | \f$ |
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397 | |
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398 | Note, by having all weights equal we get back the unweighted |
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399 | case. Furthermore, we calculate the variance of the estimators of |
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400 | \f$\alpha\f$ and \f$\beta\f$. |
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401 | |
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402 | \f$ |
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403 | \textrm{Var}(\alpha )=\frac{w_i^2\frac{\sigma^2}{w_i}}{(\sum w_i)^2}= |
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404 | \frac{\sigma^2}{\sum w_i} |
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405 | \f$ |
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406 | |
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407 | and |
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408 | \f$ |
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409 | \textrm{Var}(\beta )= \frac{w_i^2(x_i-m_x)^2\frac{\sigma^2}{w_i}} |
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410 | {(\sum w_i(x_i-m_x)^2)^2}= |
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411 | \frac{\sigma^2}{\sum w_i(x_i-m_x)^2} |
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412 | \f$ |
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413 | |
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414 | Finally, we estimate the level of noise, \f$\sigma^2\f$. Inspired by the |
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415 | unweighted estimation |
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416 | |
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417 | \f$ |
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418 | s^2=\frac{\sum (y_i-\alpha-\beta (x_i-m_x))^2}{n-2} |
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419 | \f$ |
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420 | |
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421 | we suggest the following estimator |
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422 | |
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423 | \f$ s^2=\frac{\sum w_i(y_i-\alpha-\beta (x_i-m_x))^2}{\sum |
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424 | w_i-2\frac{\sum w_i^2}{\sum w_i}} \f$ |
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425 | |
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426 | */ |
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427 | |
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428 | |
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429 | |
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