What does phi mean in this context?
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$endgroup$
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
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
$endgroup$
add a comment |
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1 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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited Nov 30 '18 at 18:21
answered Nov 30 '18 at 18:11
Siong Thye GohSiong Thye Goh
100k1465117
100k1465117
add a comment |
add a comment |
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$begingroup$
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