What does phi mean in this context?












1












$begingroup$


In an article on Selection Bias in A/B Testing, AirBnB proposes a solution to their own estimated bias. This solution is to subtract from the aggregated effect a bias estimate, captured by this equation:
$$
hat{beta} = sum_{i=1}^n W_i phi left( frac{W_i b_i - X_i}{W_i} right)
$$
where




$X_1, ldots, X_n$ are random variables defined on a same probability space, and each $X_i$ follows a distribution with finite mean $a_i$ and finite variance $sigma_i^2$ (the distributions are not necessarily identical.) We regard $a_i$ as the unknown true effect and usually estimate it by the unbiased estimate $X_i$;



$b_i$ is the cut-off from the reference distribution for significance level $alpha_i$, usually set at $0.05$; and



$W_i$ is the estimated standard deviation of $X_i$, to define the bias estimate




In this context, what does $phi$ stand for? My current understanding (by separating the two terms inside the parenthesis into
$$
b_i - X_i / W_i
$$
is that they're calculating the individual bias estimates as the difference between the cutoff and "how many" standard deviations fit into the estimate. Then they're adding these up, but I don't know what $W_iphi$ is doing inside the sum.






probability-theory probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    The phi coefficient is a correlation between two variables.
    $endgroup$
    – Phil H
    Oct 2 '18 at 0:38










  • $begingroup$
    So, the equation states that Wi is correlated to the expression in parenthesis, but it does not affect the actual sum?
    $endgroup$
    – Khashir
    Oct 3 '18 at 4:53
















1












$begingroup$


In an article on Selection Bias in A/B Testing, AirBnB proposes a solution to their own estimated bias. This solution is to subtract from the aggregated effect a bias estimate, captured by this equation:
$$
hat{beta} = sum_{i=1}^n W_i phi left( frac{W_i b_i - X_i}{W_i} right)
$$
where




$X_1, ldots, X_n$ are random variables defined on a same probability space, and each $X_i$ follows a distribution with finite mean $a_i$ and finite variance $sigma_i^2$ (the distributions are not necessarily identical.) We regard $a_i$ as the unknown true effect and usually estimate it by the unbiased estimate $X_i$;



$b_i$ is the cut-off from the reference distribution for significance level $alpha_i$, usually set at $0.05$; and



$W_i$ is the estimated standard deviation of $X_i$, to define the bias estimate




In this context, what does $phi$ stand for? My current understanding (by separating the two terms inside the parenthesis into
$$
b_i - X_i / W_i
$$
is that they're calculating the individual bias estimates as the difference between the cutoff and "how many" standard deviations fit into the estimate. Then they're adding these up, but I don't know what $W_iphi$ is doing inside the sum.






probability-theory probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    The phi coefficient is a correlation between two variables.
    $endgroup$
    – Phil H
    Oct 2 '18 at 0:38










  • $begingroup$
    So, the equation states that Wi is correlated to the expression in parenthesis, but it does not affect the actual sum?
    $endgroup$
    – Khashir
    Oct 3 '18 at 4:53














1












1








1





$begingroup$


In an article on Selection Bias in A/B Testing, AirBnB proposes a solution to their own estimated bias. This solution is to subtract from the aggregated effect a bias estimate, captured by this equation:
$$
hat{beta} = sum_{i=1}^n W_i phi left( frac{W_i b_i - X_i}{W_i} right)
$$
where




$X_1, ldots, X_n$ are random variables defined on a same probability space, and each $X_i$ follows a distribution with finite mean $a_i$ and finite variance $sigma_i^2$ (the distributions are not necessarily identical.) We regard $a_i$ as the unknown true effect and usually estimate it by the unbiased estimate $X_i$;



$b_i$ is the cut-off from the reference distribution for significance level $alpha_i$, usually set at $0.05$; and



$W_i$ is the estimated standard deviation of $X_i$, to define the bias estimate




In this context, what does $phi$ stand for? My current understanding (by separating the two terms inside the parenthesis into
$$
b_i - X_i / W_i
$$
is that they're calculating the individual bias estimates as the difference between the cutoff and "how many" standard deviations fit into the estimate. Then they're adding these up, but I don't know what $W_iphi$ is doing inside the sum.






probability-theory probability-distributions







share|cite|improve this question











$endgroup$




In an article on Selection Bias in A/B Testing, AirBnB proposes a solution to their own estimated bias. This solution is to subtract from the aggregated effect a bias estimate, captured by this equation:
$$
hat{beta} = sum_{i=1}^n W_i phi left( frac{W_i b_i - X_i}{W_i} right)
$$
where




$X_1, ldots, X_n$ are random variables defined on a same probability space, and each $X_i$ follows a distribution with finite mean $a_i$ and finite variance $sigma_i^2$ (the distributions are not necessarily identical.) We regard $a_i$ as the unknown true effect and usually estimate it by the unbiased estimate $X_i$;



$b_i$ is the cut-off from the reference distribution for significance level $alpha_i$, usually set at $0.05$; and



$W_i$ is the estimated standard deviation of $X_i$, to define the bias estimate




In this context, what does $phi$ stand for? My current understanding (by separating the two terms inside the parenthesis into
$$
b_i - X_i / W_i
$$
is that they're calculating the individual bias estimates as the difference between the cutoff and "how many" standard deviations fit into the estimate. Then they're adding these up, but I don't know what $W_iphi$ is doing inside the sum.






probability-theory probability-distributions




probability-theory probability-distributions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 2 '18 at 21:33









mwt

939416




939416










asked Oct 1 '18 at 23:49









KhashirKhashir

1062




1062












  • $begingroup$
    The phi coefficient is a correlation between two variables.
    $endgroup$
    – Phil H
    Oct 2 '18 at 0:38










  • $begingroup$
    So, the equation states that Wi is correlated to the expression in parenthesis, but it does not affect the actual sum?
    $endgroup$
    – Khashir
    Oct 3 '18 at 4:53


















  • $begingroup$
    The phi coefficient is a correlation between two variables.
    $endgroup$
    – Phil H
    Oct 2 '18 at 0:38










  • $begingroup$
    So, the equation states that Wi is correlated to the expression in parenthesis, but it does not affect the actual sum?
    $endgroup$
    – Khashir
    Oct 3 '18 at 4:53
















$begingroup$
The phi coefficient is a correlation between two variables.
$endgroup$
– Phil H
Oct 2 '18 at 0:38




$begingroup$
The phi coefficient is a correlation between two variables.
$endgroup$
– Phil H
Oct 2 '18 at 0:38












$begingroup$
So, the equation states that Wi is correlated to the expression in parenthesis, but it does not affect the actual sum?
$endgroup$
– Khashir
Oct 3 '18 at 4:53




$begingroup$
So, the equation states that Wi is correlated to the expression in parenthesis, but it does not affect the actual sum?
$endgroup$
– Khashir
Oct 3 '18 at 4:53










1 Answer
1






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oldest

votes


















0












$begingroup$

According to their paper, $phi$ refers to the density function of standard normal distribution.



Let me include some details:



They use the following result:




Suppose $Z$ follows standard normal distribution, then



$$f_{Z|Zge t}(s) = frac{phi(s)}{1-Phi(t)}$$
and hence
$$E[Z|Z ge t]=frac{int_t^infty sphi(s), ds}{1-Phi(t)}=frac{-int_t^infty phi'(s), ds}{1-Phi(t)}=frac{phi(t)}{1-Phi(t)}$$




Assuming $X_i$ follows normal distribution $N(a_i, sigma_i^2)$.
begin{align}
&beta = E[S_A-T_A] \&=sum_{i=1}^nE[I((X_i-a_i)>(b_isigma_i-a_i))(X_i-a_i)]\
&= sum_{i=1}^n sigma_i Eleft[Ileft(frac{X_i-a_i}{sigma_i}>frac{b_isigma_i-a_i}{sigma_i}right)frac{X_i-a_i}{sigma_i}right]\
&=sum_{i=1}^n sigma_iPleft(frac{X_i-a_i}{sigma_i}>frac{b_isigma_i-a_i}{sigma_i}right)Eleft[frac{X_i-a_i}{sigma_i}left|frac{X_i-a_i}{sigma}>frac{b_isigma_i-a_i}{sigma}right.right]\
&=sum_{i=1}^n sigma_i Pleft(frac{X_i-a_i}{sigma_i}>frac{b_isigma_i-a_i}{sigma_i}right)frac{phileft( frac{b_isigma_i-a_i}{sigma_i}right)}{Pleft(frac{X_i-a_i}{sigma_i}>frac{b_isigma_i-a_i}{sigma_i}right)} \
&=sum_{i=1}^n sigma_i phileft( frac{b_isigma_i-a_i}{sigma_i}right)\
end{align}



They then replace some parameters via estimation.






share|cite|improve this answer











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    $begingroup$

    According to their paper, $phi$ refers to the density function of standard normal distribution.



    Let me include some details:



    They use the following result:




    Suppose $Z$ follows standard normal distribution, then



    $$f_{Z|Zge t}(s) = frac{phi(s)}{1-Phi(t)}$$
    and hence
    $$E[Z|Z ge t]=frac{int_t^infty sphi(s), ds}{1-Phi(t)}=frac{-int_t^infty phi'(s), ds}{1-Phi(t)}=frac{phi(t)}{1-Phi(t)}$$




    Assuming $X_i$ follows normal distribution $N(a_i, sigma_i^2)$.
    begin{align}
    &beta = E[S_A-T_A] \&=sum_{i=1}^nE[I((X_i-a_i)>(b_isigma_i-a_i))(X_i-a_i)]\
    &= sum_{i=1}^n sigma_i Eleft[Ileft(frac{X_i-a_i}{sigma_i}>frac{b_isigma_i-a_i}{sigma_i}right)frac{X_i-a_i}{sigma_i}right]\
    &=sum_{i=1}^n sigma_iPleft(frac{X_i-a_i}{sigma_i}>frac{b_isigma_i-a_i}{sigma_i}right)Eleft[frac{X_i-a_i}{sigma_i}left|frac{X_i-a_i}{sigma}>frac{b_isigma_i-a_i}{sigma}right.right]\
    &=sum_{i=1}^n sigma_i Pleft(frac{X_i-a_i}{sigma_i}>frac{b_isigma_i-a_i}{sigma_i}right)frac{phileft( frac{b_isigma_i-a_i}{sigma_i}right)}{Pleft(frac{X_i-a_i}{sigma_i}>frac{b_isigma_i-a_i}{sigma_i}right)} \
    &=sum_{i=1}^n sigma_i phileft( frac{b_isigma_i-a_i}{sigma_i}right)\
    end{align}



    They then replace some parameters via estimation.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      According to their paper, $phi$ refers to the density function of standard normal distribution.



      Let me include some details:



      They use the following result:




      Suppose $Z$ follows standard normal distribution, then



      $$f_{Z|Zge t}(s) = frac{phi(s)}{1-Phi(t)}$$
      and hence
      $$E[Z|Z ge t]=frac{int_t^infty sphi(s), ds}{1-Phi(t)}=frac{-int_t^infty phi'(s), ds}{1-Phi(t)}=frac{phi(t)}{1-Phi(t)}$$




      Assuming $X_i$ follows normal distribution $N(a_i, sigma_i^2)$.
      begin{align}
      &beta = E[S_A-T_A] \&=sum_{i=1}^nE[I((X_i-a_i)>(b_isigma_i-a_i))(X_i-a_i)]\
      &= sum_{i=1}^n sigma_i Eleft[Ileft(frac{X_i-a_i}{sigma_i}>frac{b_isigma_i-a_i}{sigma_i}right)frac{X_i-a_i}{sigma_i}right]\
      &=sum_{i=1}^n sigma_iPleft(frac{X_i-a_i}{sigma_i}>frac{b_isigma_i-a_i}{sigma_i}right)Eleft[frac{X_i-a_i}{sigma_i}left|frac{X_i-a_i}{sigma}>frac{b_isigma_i-a_i}{sigma}right.right]\
      &=sum_{i=1}^n sigma_i Pleft(frac{X_i-a_i}{sigma_i}>frac{b_isigma_i-a_i}{sigma_i}right)frac{phileft( frac{b_isigma_i-a_i}{sigma_i}right)}{Pleft(frac{X_i-a_i}{sigma_i}>frac{b_isigma_i-a_i}{sigma_i}right)} \
      &=sum_{i=1}^n sigma_i phileft( frac{b_isigma_i-a_i}{sigma_i}right)\
      end{align}



      They then replace some parameters via estimation.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        According to their paper, $phi$ refers to the density function of standard normal distribution.



        Let me include some details:



        They use the following result:




        Suppose $Z$ follows standard normal distribution, then



        $$f_{Z|Zge t}(s) = frac{phi(s)}{1-Phi(t)}$$
        and hence
        $$E[Z|Z ge t]=frac{int_t^infty sphi(s), ds}{1-Phi(t)}=frac{-int_t^infty phi'(s), ds}{1-Phi(t)}=frac{phi(t)}{1-Phi(t)}$$




        Assuming $X_i$ follows normal distribution $N(a_i, sigma_i^2)$.
        begin{align}
        &beta = E[S_A-T_A] \&=sum_{i=1}^nE[I((X_i-a_i)>(b_isigma_i-a_i))(X_i-a_i)]\
        &= sum_{i=1}^n sigma_i Eleft[Ileft(frac{X_i-a_i}{sigma_i}>frac{b_isigma_i-a_i}{sigma_i}right)frac{X_i-a_i}{sigma_i}right]\
        &=sum_{i=1}^n sigma_iPleft(frac{X_i-a_i}{sigma_i}>frac{b_isigma_i-a_i}{sigma_i}right)Eleft[frac{X_i-a_i}{sigma_i}left|frac{X_i-a_i}{sigma}>frac{b_isigma_i-a_i}{sigma}right.right]\
        &=sum_{i=1}^n sigma_i Pleft(frac{X_i-a_i}{sigma_i}>frac{b_isigma_i-a_i}{sigma_i}right)frac{phileft( frac{b_isigma_i-a_i}{sigma_i}right)}{Pleft(frac{X_i-a_i}{sigma_i}>frac{b_isigma_i-a_i}{sigma_i}right)} \
        &=sum_{i=1}^n sigma_i phileft( frac{b_isigma_i-a_i}{sigma_i}right)\
        end{align}



        They then replace some parameters via estimation.






        share|cite|improve this answer











        $endgroup$



        According to their paper, $phi$ refers to the density function of standard normal distribution.



        Let me include some details:



        They use the following result:




        Suppose $Z$ follows standard normal distribution, then



        $$f_{Z|Zge t}(s) = frac{phi(s)}{1-Phi(t)}$$
        and hence
        $$E[Z|Z ge t]=frac{int_t^infty sphi(s), ds}{1-Phi(t)}=frac{-int_t^infty phi'(s), ds}{1-Phi(t)}=frac{phi(t)}{1-Phi(t)}$$




        Assuming $X_i$ follows normal distribution $N(a_i, sigma_i^2)$.
        begin{align}
        &beta = E[S_A-T_A] \&=sum_{i=1}^nE[I((X_i-a_i)>(b_isigma_i-a_i))(X_i-a_i)]\
        &= sum_{i=1}^n sigma_i Eleft[Ileft(frac{X_i-a_i}{sigma_i}>frac{b_isigma_i-a_i}{sigma_i}right)frac{X_i-a_i}{sigma_i}right]\
        &=sum_{i=1}^n sigma_iPleft(frac{X_i-a_i}{sigma_i}>frac{b_isigma_i-a_i}{sigma_i}right)Eleft[frac{X_i-a_i}{sigma_i}left|frac{X_i-a_i}{sigma}>frac{b_isigma_i-a_i}{sigma}right.right]\
        &=sum_{i=1}^n sigma_i Pleft(frac{X_i-a_i}{sigma_i}>frac{b_isigma_i-a_i}{sigma_i}right)frac{phileft( frac{b_isigma_i-a_i}{sigma_i}right)}{Pleft(frac{X_i-a_i}{sigma_i}>frac{b_isigma_i-a_i}{sigma_i}right)} \
        &=sum_{i=1}^n sigma_i phileft( frac{b_isigma_i-a_i}{sigma_i}right)\
        end{align}



        They then replace some parameters via estimation.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 30 '18 at 18:21

























        answered Nov 30 '18 at 18:11









        Siong Thye GohSiong Thye Goh

        100k1465117




        100k1465117






























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