Variance of average of $n$ correlated random variables












6












$begingroup$


Reading about deep leaning, I came across the following formula.



$$ mbox{var} left( frac{1}{n} sum_{i=1}^{n} X_i right) = rho sigma^2 + frac{1-rho}{n} sigma^2 $$



where $X_1, dots, X_n$ are identically distributed random variables with
pairwise correlation $rho > 0$ and variance $mbox{var}(X_i) = sigma^2$.




  1. How to derive this?

  2. How does bootstrap aggregating alleviate the effect of overfitting, according to this formula? What is the relationsip?






machine-learning deep-learning bootstrap regularization bagging







share|cite|improve this question











$endgroup$

















    6












    $begingroup$


    Reading about deep leaning, I came across the following formula.



    $$ mbox{var} left( frac{1}{n} sum_{i=1}^{n} X_i right) = rho sigma^2 + frac{1-rho}{n} sigma^2 $$



    where $X_1, dots, X_n$ are identically distributed random variables with
    pairwise correlation $rho > 0$ and variance $mbox{var}(X_i) = sigma^2$.




    1. How to derive this?

    2. How does bootstrap aggregating alleviate the effect of overfitting, according to this formula? What is the relationsip?






    machine-learning deep-learning bootstrap regularization bagging







    share|cite|improve this question











    $endgroup$















      6












      6








      6


      2



      $begingroup$


      Reading about deep leaning, I came across the following formula.



      $$ mbox{var} left( frac{1}{n} sum_{i=1}^{n} X_i right) = rho sigma^2 + frac{1-rho}{n} sigma^2 $$



      where $X_1, dots, X_n$ are identically distributed random variables with
      pairwise correlation $rho > 0$ and variance $mbox{var}(X_i) = sigma^2$.




      1. How to derive this?

      2. How does bootstrap aggregating alleviate the effect of overfitting, according to this formula? What is the relationsip?






      machine-learning deep-learning bootstrap regularization bagging







      share|cite|improve this question











      $endgroup$




      Reading about deep leaning, I came across the following formula.



      $$ mbox{var} left( frac{1}{n} sum_{i=1}^{n} X_i right) = rho sigma^2 + frac{1-rho}{n} sigma^2 $$



      where $X_1, dots, X_n$ are identically distributed random variables with
      pairwise correlation $rho > 0$ and variance $mbox{var}(X_i) = sigma^2$.




      1. How to derive this?

      2. How does bootstrap aggregating alleviate the effect of overfitting, according to this formula? What is the relationsip?






      machine-learning deep-learning bootstrap regularization bagging




      machine-learning deep-learning bootstrap regularization bagging






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 10 at 18:42









      Rodrigo de Azevedo

      730513




      730513










      asked Feb 10 at 14:58









      OmegaDOmegaD

      536




      536






















          1 Answer
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          8












          $begingroup$

          By definition, we have



          $$operatorname{var}left(sum_{i=1}^n{X_i}right)=operatorname{cov}left(sum_{i=1}^n{X_i},sum_{i=1}^n{X_i}right)=sum_{i=1}^n{operatorname{var}(X_i)}+sum_{ineq j}operatorname{cov}(X_i,X_j)$$



          which is $n operatorname{var}(X_i)+n(n-1)operatorname{cov}(X_i,X_j)=nsigma^2+n(n-1)rhosigma^2$, where $ineq j$. Substituting this into the original equation yields the following:



          $$operatorname{var}left(frac{1}{n}sum_{i=1}^nX_iright)=frac{1}{n^2}(nsigma^2+n(n-1)rhosigma^2)=rhosigma^2+frac{1-rho}{n}sigma^2$$



          Each $X_i$ can be thought of as a single decision mechanism, call it DM, (e.g. regressor). The variance of your decision was $sigma^2$. By using bootstrap samples and aggregating your DMs' outputs, you end up with a decision variance as above, which is strictly smaller than $sigma^2$ when $rho neq 1$ and $nneq 1$. DMs will have some degree of correlation of course, since they are trained over bootstrap samples obtained from the same base dataset, but the correlation between them most probably won't be equal to $1$. Overfitted mechanisms in general have large variance, so by aiming to decrease the variance of your DM, you actually address the problem of overfitting implicitly.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Fantastic! Thank you for so much for your answer. Quick question, in the term $n var(X_i) + n(n-1) cov(X_i,X_j)$ n and n-1 come from. Sorry if it is too obvious question.
            $endgroup$
            – OmegaD
            Feb 10 at 16:15






          • 1




            $begingroup$
            @OmegaD There are $n^2$ pairs of $i,j$, where $n$ of them have $i=j$, and $n^2-n=n(n-1)$ of them have $ineq j$.
            $endgroup$
            – gunes
            Feb 10 at 16:31













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          $begingroup$

          By definition, we have



          $$operatorname{var}left(sum_{i=1}^n{X_i}right)=operatorname{cov}left(sum_{i=1}^n{X_i},sum_{i=1}^n{X_i}right)=sum_{i=1}^n{operatorname{var}(X_i)}+sum_{ineq j}operatorname{cov}(X_i,X_j)$$



          which is $n operatorname{var}(X_i)+n(n-1)operatorname{cov}(X_i,X_j)=nsigma^2+n(n-1)rhosigma^2$, where $ineq j$. Substituting this into the original equation yields the following:



          $$operatorname{var}left(frac{1}{n}sum_{i=1}^nX_iright)=frac{1}{n^2}(nsigma^2+n(n-1)rhosigma^2)=rhosigma^2+frac{1-rho}{n}sigma^2$$



          Each $X_i$ can be thought of as a single decision mechanism, call it DM, (e.g. regressor). The variance of your decision was $sigma^2$. By using bootstrap samples and aggregating your DMs' outputs, you end up with a decision variance as above, which is strictly smaller than $sigma^2$ when $rho neq 1$ and $nneq 1$. DMs will have some degree of correlation of course, since they are trained over bootstrap samples obtained from the same base dataset, but the correlation between them most probably won't be equal to $1$. Overfitted mechanisms in general have large variance, so by aiming to decrease the variance of your DM, you actually address the problem of overfitting implicitly.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Fantastic! Thank you for so much for your answer. Quick question, in the term $n var(X_i) + n(n-1) cov(X_i,X_j)$ n and n-1 come from. Sorry if it is too obvious question.
            $endgroup$
            – OmegaD
            Feb 10 at 16:15






          • 1




            $begingroup$
            @OmegaD There are $n^2$ pairs of $i,j$, where $n$ of them have $i=j$, and $n^2-n=n(n-1)$ of them have $ineq j$.
            $endgroup$
            – gunes
            Feb 10 at 16:31


















          8












          $begingroup$

          By definition, we have



          $$operatorname{var}left(sum_{i=1}^n{X_i}right)=operatorname{cov}left(sum_{i=1}^n{X_i},sum_{i=1}^n{X_i}right)=sum_{i=1}^n{operatorname{var}(X_i)}+sum_{ineq j}operatorname{cov}(X_i,X_j)$$



          which is $n operatorname{var}(X_i)+n(n-1)operatorname{cov}(X_i,X_j)=nsigma^2+n(n-1)rhosigma^2$, where $ineq j$. Substituting this into the original equation yields the following:



          $$operatorname{var}left(frac{1}{n}sum_{i=1}^nX_iright)=frac{1}{n^2}(nsigma^2+n(n-1)rhosigma^2)=rhosigma^2+frac{1-rho}{n}sigma^2$$



          Each $X_i$ can be thought of as a single decision mechanism, call it DM, (e.g. regressor). The variance of your decision was $sigma^2$. By using bootstrap samples and aggregating your DMs' outputs, you end up with a decision variance as above, which is strictly smaller than $sigma^2$ when $rho neq 1$ and $nneq 1$. DMs will have some degree of correlation of course, since they are trained over bootstrap samples obtained from the same base dataset, but the correlation between them most probably won't be equal to $1$. Overfitted mechanisms in general have large variance, so by aiming to decrease the variance of your DM, you actually address the problem of overfitting implicitly.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Fantastic! Thank you for so much for your answer. Quick question, in the term $n var(X_i) + n(n-1) cov(X_i,X_j)$ n and n-1 come from. Sorry if it is too obvious question.
            $endgroup$
            – OmegaD
            Feb 10 at 16:15






          • 1




            $begingroup$
            @OmegaD There are $n^2$ pairs of $i,j$, where $n$ of them have $i=j$, and $n^2-n=n(n-1)$ of them have $ineq j$.
            $endgroup$
            – gunes
            Feb 10 at 16:31
















          8












          8








          8





          $begingroup$

          By definition, we have



          $$operatorname{var}left(sum_{i=1}^n{X_i}right)=operatorname{cov}left(sum_{i=1}^n{X_i},sum_{i=1}^n{X_i}right)=sum_{i=1}^n{operatorname{var}(X_i)}+sum_{ineq j}operatorname{cov}(X_i,X_j)$$



          which is $n operatorname{var}(X_i)+n(n-1)operatorname{cov}(X_i,X_j)=nsigma^2+n(n-1)rhosigma^2$, where $ineq j$. Substituting this into the original equation yields the following:



          $$operatorname{var}left(frac{1}{n}sum_{i=1}^nX_iright)=frac{1}{n^2}(nsigma^2+n(n-1)rhosigma^2)=rhosigma^2+frac{1-rho}{n}sigma^2$$



          Each $X_i$ can be thought of as a single decision mechanism, call it DM, (e.g. regressor). The variance of your decision was $sigma^2$. By using bootstrap samples and aggregating your DMs' outputs, you end up with a decision variance as above, which is strictly smaller than $sigma^2$ when $rho neq 1$ and $nneq 1$. DMs will have some degree of correlation of course, since they are trained over bootstrap samples obtained from the same base dataset, but the correlation between them most probably won't be equal to $1$. Overfitted mechanisms in general have large variance, so by aiming to decrease the variance of your DM, you actually address the problem of overfitting implicitly.






          share|cite|improve this answer











          $endgroup$



          By definition, we have



          $$operatorname{var}left(sum_{i=1}^n{X_i}right)=operatorname{cov}left(sum_{i=1}^n{X_i},sum_{i=1}^n{X_i}right)=sum_{i=1}^n{operatorname{var}(X_i)}+sum_{ineq j}operatorname{cov}(X_i,X_j)$$



          which is $n operatorname{var}(X_i)+n(n-1)operatorname{cov}(X_i,X_j)=nsigma^2+n(n-1)rhosigma^2$, where $ineq j$. Substituting this into the original equation yields the following:



          $$operatorname{var}left(frac{1}{n}sum_{i=1}^nX_iright)=frac{1}{n^2}(nsigma^2+n(n-1)rhosigma^2)=rhosigma^2+frac{1-rho}{n}sigma^2$$



          Each $X_i$ can be thought of as a single decision mechanism, call it DM, (e.g. regressor). The variance of your decision was $sigma^2$. By using bootstrap samples and aggregating your DMs' outputs, you end up with a decision variance as above, which is strictly smaller than $sigma^2$ when $rho neq 1$ and $nneq 1$. DMs will have some degree of correlation of course, since they are trained over bootstrap samples obtained from the same base dataset, but the correlation between them most probably won't be equal to $1$. Overfitted mechanisms in general have large variance, so by aiming to decrease the variance of your DM, you actually address the problem of overfitting implicitly.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 10 at 19:14









          StubbornAtom

          2,6451531




          2,6451531










          answered Feb 10 at 15:59









          gunesgunes

          5,7751115




          5,7751115












          • $begingroup$
            Fantastic! Thank you for so much for your answer. Quick question, in the term $n var(X_i) + n(n-1) cov(X_i,X_j)$ n and n-1 come from. Sorry if it is too obvious question.
            $endgroup$
            – OmegaD
            Feb 10 at 16:15






          • 1




            $begingroup$
            @OmegaD There are $n^2$ pairs of $i,j$, where $n$ of them have $i=j$, and $n^2-n=n(n-1)$ of them have $ineq j$.
            $endgroup$
            – gunes
            Feb 10 at 16:31




















          • $begingroup$
            Fantastic! Thank you for so much for your answer. Quick question, in the term $n var(X_i) + n(n-1) cov(X_i,X_j)$ n and n-1 come from. Sorry if it is too obvious question.
            $endgroup$
            – OmegaD
            Feb 10 at 16:15






          • 1




            $begingroup$
            @OmegaD There are $n^2$ pairs of $i,j$, where $n$ of them have $i=j$, and $n^2-n=n(n-1)$ of them have $ineq j$.
            $endgroup$
            – gunes
            Feb 10 at 16:31


















          $begingroup$
          Fantastic! Thank you for so much for your answer. Quick question, in the term $n var(X_i) + n(n-1) cov(X_i,X_j)$ n and n-1 come from. Sorry if it is too obvious question.
          $endgroup$
          – OmegaD
          Feb 10 at 16:15




          $begingroup$
          Fantastic! Thank you for so much for your answer. Quick question, in the term $n var(X_i) + n(n-1) cov(X_i,X_j)$ n and n-1 come from. Sorry if it is too obvious question.
          $endgroup$
          – OmegaD
          Feb 10 at 16:15




          1




          1




          $begingroup$
          @OmegaD There are $n^2$ pairs of $i,j$, where $n$ of them have $i=j$, and $n^2-n=n(n-1)$ of them have $ineq j$.
          $endgroup$
          – gunes
          Feb 10 at 16:31






          $begingroup$
          @OmegaD There are $n^2$ pairs of $i,j$, where $n$ of them have $i=j$, and $n^2-n=n(n-1)$ of them have $ineq j$.
          $endgroup$
          – gunes
          Feb 10 at 16:31




















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