Variance of parameter estimate using recursive least squares












1












$begingroup$


I am learning about recursive least squares estimation using a forgetting factor $lambda$ as a tool for treating time variations of model parameters and have become stuck on the following problem.



Question



Find an expression for $Vbig[hat{b}big]$ given
$$y_t = bu_t + e_t, quad t=1,...,N$$



Where $e_t$ is white Gaussian noise with variance $sigma^2_e$ and $u_t$ is a deterministic signal such that



$$lim_{Ntoinfty} frac{1}{N} sum_{t=1}^{N} u^2_t$$



is finite. The unknown parameter b is estimated as
$$hat{b}= operatorname*{argmin}_b sum_{t=1}^{N} lambda^{N-t}(y_t-bu_t)^2,$$ where $0<lambda leq 1$.



My attempt at a solution



I can be seen that the argument that minimises the above equation is $hat{b}= frac{y_t}{u_t}$. However when I try to calculate the variance I get



$$Vbig[hat{b}big]=Vbig[frac{y_t}{u_t}big].$$ But as $u_t$ is a deterministic signal and I am under the impression that the variance of a deterministic signal is zero this would give me a zero in the denominator?



Any help greatly appreciated.



Edit



After user617446 comment I went back and recalculated $hat{b}$ as follows-



$$frac{partial}{partial b}bigg[ sum_{t=1}^{N} lambda^{N-t}(y_t-bu_t)^2 bigg] = 2bsum_{t=1}^{N}lambda^{N-t}u_t^2-2sum_{t=1}^{N}lambda^{N-t}y_tu_t$$



setting this equal to zero then solving gave



$$hat{b}=frac{sum_{t=1}^{N}lambda^{N-t}y_tu_t}{sum_{t=1}^{N}lambda^{N-t}u_t^2}.$$



I believe this to be correct but I am now stuck once again on how to calculate the variance? Grateful for any and all help.






recursive-algorithms time-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $hat b$ is not time dependent and hence cannot be considered to be a ratio of $y_t / u_t$. Instead, ask yourself what single value of $b$ will minimize the expression, given $y_t, u_t$
    $endgroup$
    – user617446
    Jan 16 at 14:45










  • $begingroup$
    Thanks for the hint @user617446! I have updated the question with my new calculations for the b estimate but am still stuck on calculating the variance.
    $endgroup$
    – Eiraus
    Jan 19 at 12:30










  • $begingroup$
    Suppose you had to solve the variance of $b$ in the case $y=bu+e$ for a single $t$. Do you know what to do then?
    $endgroup$
    – user617446
    Jan 20 at 5:32










  • $begingroup$
    $Var[b]=Var[frac{1}{u}(y-e)]$ Is this what you mean? But for a single t would that not just be the variance of a constant? I.e zero? Thanks for sticking with me @user617446 !
    $endgroup$
    – Eiraus
    Jan 20 at 19:40










  • $begingroup$
    You are almost there. For a single equation, $hat{b}=y/u$ but $y=b u +e$ where $b$ is the "correct" value ( not the estimate). If we substitute we get $hat{b}=b+e/u$ and the variance is $sigma^2/u^2$ PS: I believe that if you read up on maximum likelihood estimation, you could get more insight into these kind of problems.
    $endgroup$
    – user617446
    Jan 21 at 5:53


















1












$begingroup$


I am learning about recursive least squares estimation using a forgetting factor $lambda$ as a tool for treating time variations of model parameters and have become stuck on the following problem.



Question



Find an expression for $Vbig[hat{b}big]$ given
$$y_t = bu_t + e_t, quad t=1,...,N$$



Where $e_t$ is white Gaussian noise with variance $sigma^2_e$ and $u_t$ is a deterministic signal such that



$$lim_{Ntoinfty} frac{1}{N} sum_{t=1}^{N} u^2_t$$



is finite. The unknown parameter b is estimated as
$$hat{b}= operatorname*{argmin}_b sum_{t=1}^{N} lambda^{N-t}(y_t-bu_t)^2,$$ where $0<lambda leq 1$.



My attempt at a solution



I can be seen that the argument that minimises the above equation is $hat{b}= frac{y_t}{u_t}$. However when I try to calculate the variance I get



$$Vbig[hat{b}big]=Vbig[frac{y_t}{u_t}big].$$ But as $u_t$ is a deterministic signal and I am under the impression that the variance of a deterministic signal is zero this would give me a zero in the denominator?



Any help greatly appreciated.



Edit



After user617446 comment I went back and recalculated $hat{b}$ as follows-



$$frac{partial}{partial b}bigg[ sum_{t=1}^{N} lambda^{N-t}(y_t-bu_t)^2 bigg] = 2bsum_{t=1}^{N}lambda^{N-t}u_t^2-2sum_{t=1}^{N}lambda^{N-t}y_tu_t$$



setting this equal to zero then solving gave



$$hat{b}=frac{sum_{t=1}^{N}lambda^{N-t}y_tu_t}{sum_{t=1}^{N}lambda^{N-t}u_t^2}.$$



I believe this to be correct but I am now stuck once again on how to calculate the variance? Grateful for any and all help.






recursive-algorithms time-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $hat b$ is not time dependent and hence cannot be considered to be a ratio of $y_t / u_t$. Instead, ask yourself what single value of $b$ will minimize the expression, given $y_t, u_t$
    $endgroup$
    – user617446
    Jan 16 at 14:45










  • $begingroup$
    Thanks for the hint @user617446! I have updated the question with my new calculations for the b estimate but am still stuck on calculating the variance.
    $endgroup$
    – Eiraus
    Jan 19 at 12:30










  • $begingroup$
    Suppose you had to solve the variance of $b$ in the case $y=bu+e$ for a single $t$. Do you know what to do then?
    $endgroup$
    – user617446
    Jan 20 at 5:32










  • $begingroup$
    $Var[b]=Var[frac{1}{u}(y-e)]$ Is this what you mean? But for a single t would that not just be the variance of a constant? I.e zero? Thanks for sticking with me @user617446 !
    $endgroup$
    – Eiraus
    Jan 20 at 19:40










  • $begingroup$
    You are almost there. For a single equation, $hat{b}=y/u$ but $y=b u +e$ where $b$ is the "correct" value ( not the estimate). If we substitute we get $hat{b}=b+e/u$ and the variance is $sigma^2/u^2$ PS: I believe that if you read up on maximum likelihood estimation, you could get more insight into these kind of problems.
    $endgroup$
    – user617446
    Jan 21 at 5:53
















1












1








1





$begingroup$


I am learning about recursive least squares estimation using a forgetting factor $lambda$ as a tool for treating time variations of model parameters and have become stuck on the following problem.



Question



Find an expression for $Vbig[hat{b}big]$ given
$$y_t = bu_t + e_t, quad t=1,...,N$$



Where $e_t$ is white Gaussian noise with variance $sigma^2_e$ and $u_t$ is a deterministic signal such that



$$lim_{Ntoinfty} frac{1}{N} sum_{t=1}^{N} u^2_t$$



is finite. The unknown parameter b is estimated as
$$hat{b}= operatorname*{argmin}_b sum_{t=1}^{N} lambda^{N-t}(y_t-bu_t)^2,$$ where $0<lambda leq 1$.



My attempt at a solution



I can be seen that the argument that minimises the above equation is $hat{b}= frac{y_t}{u_t}$. However when I try to calculate the variance I get



$$Vbig[hat{b}big]=Vbig[frac{y_t}{u_t}big].$$ But as $u_t$ is a deterministic signal and I am under the impression that the variance of a deterministic signal is zero this would give me a zero in the denominator?



Any help greatly appreciated.



Edit



After user617446 comment I went back and recalculated $hat{b}$ as follows-



$$frac{partial}{partial b}bigg[ sum_{t=1}^{N} lambda^{N-t}(y_t-bu_t)^2 bigg] = 2bsum_{t=1}^{N}lambda^{N-t}u_t^2-2sum_{t=1}^{N}lambda^{N-t}y_tu_t$$



setting this equal to zero then solving gave



$$hat{b}=frac{sum_{t=1}^{N}lambda^{N-t}y_tu_t}{sum_{t=1}^{N}lambda^{N-t}u_t^2}.$$



I believe this to be correct but I am now stuck once again on how to calculate the variance? Grateful for any and all help.






recursive-algorithms time-series







share|cite|improve this question











$endgroup$




I am learning about recursive least squares estimation using a forgetting factor $lambda$ as a tool for treating time variations of model parameters and have become stuck on the following problem.



Question



Find an expression for $Vbig[hat{b}big]$ given
$$y_t = bu_t + e_t, quad t=1,...,N$$



Where $e_t$ is white Gaussian noise with variance $sigma^2_e$ and $u_t$ is a deterministic signal such that



$$lim_{Ntoinfty} frac{1}{N} sum_{t=1}^{N} u^2_t$$



is finite. The unknown parameter b is estimated as
$$hat{b}= operatorname*{argmin}_b sum_{t=1}^{N} lambda^{N-t}(y_t-bu_t)^2,$$ where $0<lambda leq 1$.



My attempt at a solution



I can be seen that the argument that minimises the above equation is $hat{b}= frac{y_t}{u_t}$. However when I try to calculate the variance I get



$$Vbig[hat{b}big]=Vbig[frac{y_t}{u_t}big].$$ But as $u_t$ is a deterministic signal and I am under the impression that the variance of a deterministic signal is zero this would give me a zero in the denominator?



Any help greatly appreciated.



Edit



After user617446 comment I went back and recalculated $hat{b}$ as follows-



$$frac{partial}{partial b}bigg[ sum_{t=1}^{N} lambda^{N-t}(y_t-bu_t)^2 bigg] = 2bsum_{t=1}^{N}lambda^{N-t}u_t^2-2sum_{t=1}^{N}lambda^{N-t}y_tu_t$$



setting this equal to zero then solving gave



$$hat{b}=frac{sum_{t=1}^{N}lambda^{N-t}y_tu_t}{sum_{t=1}^{N}lambda^{N-t}u_t^2}.$$



I believe this to be correct but I am now stuck once again on how to calculate the variance? Grateful for any and all help.






recursive-algorithms time-series




recursive-algorithms time-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 20 at 3:20







Eiraus

















asked Jan 16 at 12:09









EirausEiraus

7910




7910








  • 1




    $begingroup$
    $hat b$ is not time dependent and hence cannot be considered to be a ratio of $y_t / u_t$. Instead, ask yourself what single value of $b$ will minimize the expression, given $y_t, u_t$
    $endgroup$
    – user617446
    Jan 16 at 14:45










  • $begingroup$
    Thanks for the hint @user617446! I have updated the question with my new calculations for the b estimate but am still stuck on calculating the variance.
    $endgroup$
    – Eiraus
    Jan 19 at 12:30










  • $begingroup$
    Suppose you had to solve the variance of $b$ in the case $y=bu+e$ for a single $t$. Do you know what to do then?
    $endgroup$
    – user617446
    Jan 20 at 5:32










  • $begingroup$
    $Var[b]=Var[frac{1}{u}(y-e)]$ Is this what you mean? But for a single t would that not just be the variance of a constant? I.e zero? Thanks for sticking with me @user617446 !
    $endgroup$
    – Eiraus
    Jan 20 at 19:40










  • $begingroup$
    You are almost there. For a single equation, $hat{b}=y/u$ but $y=b u +e$ where $b$ is the "correct" value ( not the estimate). If we substitute we get $hat{b}=b+e/u$ and the variance is $sigma^2/u^2$ PS: I believe that if you read up on maximum likelihood estimation, you could get more insight into these kind of problems.
    $endgroup$
    – user617446
    Jan 21 at 5:53
















  • 1




    $begingroup$
    $hat b$ is not time dependent and hence cannot be considered to be a ratio of $y_t / u_t$. Instead, ask yourself what single value of $b$ will minimize the expression, given $y_t, u_t$
    $endgroup$
    – user617446
    Jan 16 at 14:45










  • $begingroup$
    Thanks for the hint @user617446! I have updated the question with my new calculations for the b estimate but am still stuck on calculating the variance.
    $endgroup$
    – Eiraus
    Jan 19 at 12:30










  • $begingroup$
    Suppose you had to solve the variance of $b$ in the case $y=bu+e$ for a single $t$. Do you know what to do then?
    $endgroup$
    – user617446
    Jan 20 at 5:32










  • $begingroup$
    $Var[b]=Var[frac{1}{u}(y-e)]$ Is this what you mean? But for a single t would that not just be the variance of a constant? I.e zero? Thanks for sticking with me @user617446 !
    $endgroup$
    – Eiraus
    Jan 20 at 19:40










  • $begingroup$
    You are almost there. For a single equation, $hat{b}=y/u$ but $y=b u +e$ where $b$ is the "correct" value ( not the estimate). If we substitute we get $hat{b}=b+e/u$ and the variance is $sigma^2/u^2$ PS: I believe that if you read up on maximum likelihood estimation, you could get more insight into these kind of problems.
    $endgroup$
    – user617446
    Jan 21 at 5:53










1




1




$begingroup$
$hat b$ is not time dependent and hence cannot be considered to be a ratio of $y_t / u_t$. Instead, ask yourself what single value of $b$ will minimize the expression, given $y_t, u_t$
$endgroup$
– user617446
Jan 16 at 14:45




$begingroup$
$hat b$ is not time dependent and hence cannot be considered to be a ratio of $y_t / u_t$. Instead, ask yourself what single value of $b$ will minimize the expression, given $y_t, u_t$
$endgroup$
– user617446
Jan 16 at 14:45












$begingroup$
Thanks for the hint @user617446! I have updated the question with my new calculations for the b estimate but am still stuck on calculating the variance.
$endgroup$
– Eiraus
Jan 19 at 12:30




$begingroup$
Thanks for the hint @user617446! I have updated the question with my new calculations for the b estimate but am still stuck on calculating the variance.
$endgroup$
– Eiraus
Jan 19 at 12:30












$begingroup$
Suppose you had to solve the variance of $b$ in the case $y=bu+e$ for a single $t$. Do you know what to do then?
$endgroup$
– user617446
Jan 20 at 5:32




$begingroup$
Suppose you had to solve the variance of $b$ in the case $y=bu+e$ for a single $t$. Do you know what to do then?
$endgroup$
– user617446
Jan 20 at 5:32












$begingroup$
$Var[b]=Var[frac{1}{u}(y-e)]$ Is this what you mean? But for a single t would that not just be the variance of a constant? I.e zero? Thanks for sticking with me @user617446 !
$endgroup$
– Eiraus
Jan 20 at 19:40




$begingroup$
$Var[b]=Var[frac{1}{u}(y-e)]$ Is this what you mean? But for a single t would that not just be the variance of a constant? I.e zero? Thanks for sticking with me @user617446 !
$endgroup$
– Eiraus
Jan 20 at 19:40












$begingroup$
You are almost there. For a single equation, $hat{b}=y/u$ but $y=b u +e$ where $b$ is the "correct" value ( not the estimate). If we substitute we get $hat{b}=b+e/u$ and the variance is $sigma^2/u^2$ PS: I believe that if you read up on maximum likelihood estimation, you could get more insight into these kind of problems.
$endgroup$
– user617446
Jan 21 at 5:53






$begingroup$
You are almost there. For a single equation, $hat{b}=y/u$ but $y=b u +e$ where $b$ is the "correct" value ( not the estimate). If we substitute we get $hat{b}=b+e/u$ and the variance is $sigma^2/u^2$ PS: I believe that if you read up on maximum likelihood estimation, you could get more insight into these kind of problems.
$endgroup$
– user617446
Jan 21 at 5:53












1 Answer
1






active

oldest

votes


















0












$begingroup$

Firstly, the least square estmation can be found via differentiation of sum by the parameter $hat b,$ so the expression
$$hat b =dfrac{sumlimits_{t=1}^Nlambda^{N-t}y_t u_t}{sumlimits_{t=1}^Nlambda^{N-t}u_t^2}$$
is correct.



Parameter $hat b$ should be considered as random variable whose value depends on the specific white noise sample,
$$hat b =dfrac{sumlimits_{t=1^N}lambda^{N-t}(e_t+u_t b) u_t}{sumlimits_{t=1}^Nlambda^{N-t}u_t^2}
=b + dfrac{sumlimits_{t=1}^Nlambda^{N-t}e_t u_t}{sumlimits_{t=1}^Nlambda^{N-t}u_t^2}.$$
There are not reasons why the sums ratio can deviate the average mean of the random variable $hat b,$ so
$$M(hat b) = b.$$
Then the variance is
$$V(hat b) = M((hat b-b)^2) = Mleft(left(dfrac{sumlimits_{t=1}^Nlambda^{N-t}e_t u_t}{sumlimits_{t=1}^Nlambda^{N-t}u_t^2}right)^2right)\[4pt]
= dfrac{Mleft(sumlimits_{t=1}^N lambda^{2(N-t)}u_t^2e_t^2right)
+Mleft(sumlimits_{1leq t_1 < t_2leq N} lambda^{2N-t_1-t_2}u_{t_1}u_{t_2}e_{t_1} e_{t_2}right)}{left(sumlimits_{t=0}^Nlambda^{N-t}u_t^2right)^2}
= color{brown}{mathbf{dfrac{sumlimits_{t=1}^N lambda^{2(N-t)}u_t^2}{left(sumlimits_{t=1}^Nlambda^{N-t}u_t^2right)^2}cdotsigma_e^2}}.$$






share|cite|improve this answer









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    1 Answer
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    1 Answer
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    active

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    0












    $begingroup$

    Firstly, the least square estmation can be found via differentiation of sum by the parameter $hat b,$ so the expression
    $$hat b =dfrac{sumlimits_{t=1}^Nlambda^{N-t}y_t u_t}{sumlimits_{t=1}^Nlambda^{N-t}u_t^2}$$
    is correct.



    Parameter $hat b$ should be considered as random variable whose value depends on the specific white noise sample,
    $$hat b =dfrac{sumlimits_{t=1^N}lambda^{N-t}(e_t+u_t b) u_t}{sumlimits_{t=1}^Nlambda^{N-t}u_t^2}
    =b + dfrac{sumlimits_{t=1}^Nlambda^{N-t}e_t u_t}{sumlimits_{t=1}^Nlambda^{N-t}u_t^2}.$$
    There are not reasons why the sums ratio can deviate the average mean of the random variable $hat b,$ so
    $$M(hat b) = b.$$
    Then the variance is
    $$V(hat b) = M((hat b-b)^2) = Mleft(left(dfrac{sumlimits_{t=1}^Nlambda^{N-t}e_t u_t}{sumlimits_{t=1}^Nlambda^{N-t}u_t^2}right)^2right)\[4pt]
    = dfrac{Mleft(sumlimits_{t=1}^N lambda^{2(N-t)}u_t^2e_t^2right)
    +Mleft(sumlimits_{1leq t_1 < t_2leq N} lambda^{2N-t_1-t_2}u_{t_1}u_{t_2}e_{t_1} e_{t_2}right)}{left(sumlimits_{t=0}^Nlambda^{N-t}u_t^2right)^2}
    = color{brown}{mathbf{dfrac{sumlimits_{t=1}^N lambda^{2(N-t)}u_t^2}{left(sumlimits_{t=1}^Nlambda^{N-t}u_t^2right)^2}cdotsigma_e^2}}.$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Firstly, the least square estmation can be found via differentiation of sum by the parameter $hat b,$ so the expression
      $$hat b =dfrac{sumlimits_{t=1}^Nlambda^{N-t}y_t u_t}{sumlimits_{t=1}^Nlambda^{N-t}u_t^2}$$
      is correct.



      Parameter $hat b$ should be considered as random variable whose value depends on the specific white noise sample,
      $$hat b =dfrac{sumlimits_{t=1^N}lambda^{N-t}(e_t+u_t b) u_t}{sumlimits_{t=1}^Nlambda^{N-t}u_t^2}
      =b + dfrac{sumlimits_{t=1}^Nlambda^{N-t}e_t u_t}{sumlimits_{t=1}^Nlambda^{N-t}u_t^2}.$$
      There are not reasons why the sums ratio can deviate the average mean of the random variable $hat b,$ so
      $$M(hat b) = b.$$
      Then the variance is
      $$V(hat b) = M((hat b-b)^2) = Mleft(left(dfrac{sumlimits_{t=1}^Nlambda^{N-t}e_t u_t}{sumlimits_{t=1}^Nlambda^{N-t}u_t^2}right)^2right)\[4pt]
      = dfrac{Mleft(sumlimits_{t=1}^N lambda^{2(N-t)}u_t^2e_t^2right)
      +Mleft(sumlimits_{1leq t_1 < t_2leq N} lambda^{2N-t_1-t_2}u_{t_1}u_{t_2}e_{t_1} e_{t_2}right)}{left(sumlimits_{t=0}^Nlambda^{N-t}u_t^2right)^2}
      = color{brown}{mathbf{dfrac{sumlimits_{t=1}^N lambda^{2(N-t)}u_t^2}{left(sumlimits_{t=1}^Nlambda^{N-t}u_t^2right)^2}cdotsigma_e^2}}.$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Firstly, the least square estmation can be found via differentiation of sum by the parameter $hat b,$ so the expression
        $$hat b =dfrac{sumlimits_{t=1}^Nlambda^{N-t}y_t u_t}{sumlimits_{t=1}^Nlambda^{N-t}u_t^2}$$
        is correct.



        Parameter $hat b$ should be considered as random variable whose value depends on the specific white noise sample,
        $$hat b =dfrac{sumlimits_{t=1^N}lambda^{N-t}(e_t+u_t b) u_t}{sumlimits_{t=1}^Nlambda^{N-t}u_t^2}
        =b + dfrac{sumlimits_{t=1}^Nlambda^{N-t}e_t u_t}{sumlimits_{t=1}^Nlambda^{N-t}u_t^2}.$$
        There are not reasons why the sums ratio can deviate the average mean of the random variable $hat b,$ so
        $$M(hat b) = b.$$
        Then the variance is
        $$V(hat b) = M((hat b-b)^2) = Mleft(left(dfrac{sumlimits_{t=1}^Nlambda^{N-t}e_t u_t}{sumlimits_{t=1}^Nlambda^{N-t}u_t^2}right)^2right)\[4pt]
        = dfrac{Mleft(sumlimits_{t=1}^N lambda^{2(N-t)}u_t^2e_t^2right)
        +Mleft(sumlimits_{1leq t_1 < t_2leq N} lambda^{2N-t_1-t_2}u_{t_1}u_{t_2}e_{t_1} e_{t_2}right)}{left(sumlimits_{t=0}^Nlambda^{N-t}u_t^2right)^2}
        = color{brown}{mathbf{dfrac{sumlimits_{t=1}^N lambda^{2(N-t)}u_t^2}{left(sumlimits_{t=1}^Nlambda^{N-t}u_t^2right)^2}cdotsigma_e^2}}.$$






        share|cite|improve this answer









        $endgroup$



        Firstly, the least square estmation can be found via differentiation of sum by the parameter $hat b,$ so the expression
        $$hat b =dfrac{sumlimits_{t=1}^Nlambda^{N-t}y_t u_t}{sumlimits_{t=1}^Nlambda^{N-t}u_t^2}$$
        is correct.



        Parameter $hat b$ should be considered as random variable whose value depends on the specific white noise sample,
        $$hat b =dfrac{sumlimits_{t=1^N}lambda^{N-t}(e_t+u_t b) u_t}{sumlimits_{t=1}^Nlambda^{N-t}u_t^2}
        =b + dfrac{sumlimits_{t=1}^Nlambda^{N-t}e_t u_t}{sumlimits_{t=1}^Nlambda^{N-t}u_t^2}.$$
        There are not reasons why the sums ratio can deviate the average mean of the random variable $hat b,$ so
        $$M(hat b) = b.$$
        Then the variance is
        $$V(hat b) = M((hat b-b)^2) = Mleft(left(dfrac{sumlimits_{t=1}^Nlambda^{N-t}e_t u_t}{sumlimits_{t=1}^Nlambda^{N-t}u_t^2}right)^2right)\[4pt]
        = dfrac{Mleft(sumlimits_{t=1}^N lambda^{2(N-t)}u_t^2e_t^2right)
        +Mleft(sumlimits_{1leq t_1 < t_2leq N} lambda^{2N-t_1-t_2}u_{t_1}u_{t_2}e_{t_1} e_{t_2}right)}{left(sumlimits_{t=0}^Nlambda^{N-t}u_t^2right)^2}
        = color{brown}{mathbf{dfrac{sumlimits_{t=1}^N lambda^{2(N-t)}u_t^2}{left(sumlimits_{t=1}^Nlambda^{N-t}u_t^2right)^2}cdotsigma_e^2}}.$$







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        answered Jan 24 at 5:40









        Yuri NegometyanovYuri Negometyanov

        12.5k1729




        12.5k1729






























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