argmin with logic statements [solved]Mathematical logic book with answers to exercisesCan this be solved?Logic behind combinations with repetition?Can we ever have E(argmin(f)) = argmin(E(f))?Proving MAE(mean absolute error) argmin is median using functionals?Properties of this argmin functionMonotonicity in argmin functionThe range of a argmin functionPermutation and combination(ALREADY SOLVED)Calculate average percentile and typical deviation with logic
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argmin with logic statements [solved]
Mathematical logic book with answers to exercisesCan this be solved?Logic behind combinations with repetition?Can we ever have E(argmin(f)) = argmin(E(f))?Proving MAE(mean absolute error) argmin is median using functionals?Properties of this argmin functionMonotonicity in argmin functionThe range of a argmin functionPermutation and combination(ALREADY SOLVED)Calculate average percentile and typical deviation with logic
$begingroup$
Could anyone explain me the equation below? Are we trying the find the T value where the $theta_T < E_0[t]?$
$T^* = undersetTArgmin (theta_T geq E_0[T])$
I understand how argmin works for an equation like $undersetxArgmin f(x)$. The question above is not a numeric solution (I think).
I would really appreciate if anyone can provide me guidance on this matter.
Thanks in advance!
Edit:
Here are the whole events leading upto the formula. This is taken from Advances in Financial Machine Learning by Marcos Lopez De Prado.
$b_t=begincasesb_t-1, & textif Delta p_t = 0 \ fracDelta p_t ,& textif Delta p_t neq 0 endcases $
$p_t$ = the price associated with tick t
$v_t$ = volume associated with tick t
$theta_T = sum_t = 1^Tb_t$
T = tick index
$T^* = undersetTArgmin (theta_T geq E_0[T])$
$E_0[T]$ is calculated by exponentially weighted moving average of T
statistics self-learning
asked yesterday
boniface316boniface316
1266
$endgroup$
|
show 1 more comment
$begingroup$
Could anyone explain me the equation below? Are we trying the find the T value where the $theta_T < E_0[t]?$
$T^* = undersetTArgmin (theta_T geq E_0[T])$
I understand how argmin works for an equation like $undersetxArgmin f(x)$. The question above is not a numeric solution (I think).
I would really appreciate if anyone can provide me guidance on this matter.
Thanks in advance!
Edit:
Here are the whole events leading upto the formula. This is taken from Advances in Financial Machine Learning by Marcos Lopez De Prado.
$b_t=begincasesb_t-1, & textif Delta p_t = 0 \ fracDelta p_t ,& textif Delta p_t neq 0 endcases $
$p_t$ = the price associated with tick t
$v_t$ = volume associated with tick t
$theta_T = sum_t = 1^Tb_t$
T = tick index
$T^* = undersetTArgmin (theta_T geq E_0[T])$
$E_0[T]$ is calculated by exponentially weighted moving average of T
statistics self-learning
asked yesterday
boniface316boniface316
1266
$endgroup$
$begingroup$
The statement as it stands is very weird; it may mean that $T^star$ is the smallest $T$ such that $theta_T geq E_0[T]$. However, the notation $E_0[T]$ seems to indicate that $T$ is a random variable; could you give more context?
$endgroup$
– Riccardo Sven Risuleo
yesterday
$begingroup$
You are correct about the T being a random variable. I couldnt believe that answer was obvious! Thanks @RiccardoSvenRisuleo!
$endgroup$
– boniface316
yesterday
$begingroup$
I'm not sure that that's the answer! As a matter of fact if T is a random variable, then $theta_T$ may also be a random variable; so the whole thing is a bit messy.
$endgroup$
– Riccardo Sven Risuleo
yesterday
$begingroup$
I can share the whole arguments if you would like....
$endgroup$
– boniface316
yesterday
$begingroup$
I'm just guessing, but if $theta_T$ is a random variable it may be that the expression is missing a probability? Something along the lines of $T^star = argmin_T P(theta_T geq E_0[T])$? Sadly, without reading the whole argument it is impossible to give a clear meaning to the statement.
$endgroup$
– Riccardo Sven Risuleo
yesterday
|
show 1 more comment
$begingroup$
Could anyone explain me the equation below? Are we trying the find the T value where the $theta_T < E_0[t]?$
$T^* = undersetTArgmin (theta_T geq E_0[T])$
I understand how argmin works for an equation like $undersetxArgmin f(x)$. The question above is not a numeric solution (I think).
I would really appreciate if anyone can provide me guidance on this matter.
Thanks in advance!
Edit:
Here are the whole events leading upto the formula. This is taken from Advances in Financial Machine Learning by Marcos Lopez De Prado.
$b_t=begincasesb_t-1, & textif Delta p_t = 0 \ fracDelta p_t ,& textif Delta p_t neq 0 endcases $
$p_t$ = the price associated with tick t
$v_t$ = volume associated with tick t
$theta_T = sum_t = 1^Tb_t$
T = tick index
$T^* = undersetTArgmin (theta_T geq E_0[T])$
$E_0[T]$ is calculated by exponentially weighted moving average of T
statistics self-learning
asked yesterday
boniface316boniface316
1266
$endgroup$
Could anyone explain me the equation below? Are we trying the find the T value where the $theta_T < E_0[t]?$
$T^* = undersetTArgmin (theta_T geq E_0[T])$
I understand how argmin works for an equation like $undersetxArgmin f(x)$. The question above is not a numeric solution (I think).
I would really appreciate if anyone can provide me guidance on this matter.
Thanks in advance!
Edit:
Here are the whole events leading upto the formula. This is taken from Advances in Financial Machine Learning by Marcos Lopez De Prado.
$b_t=begincasesb_t-1, & textif Delta p_t = 0 \ fracDelta p_t ,& textif Delta p_t neq 0 endcases $
$p_t$ = the price associated with tick t
$v_t$ = volume associated with tick t
$theta_T = sum_t = 1^Tb_t$
T = tick index
$T^* = undersetTArgmin (theta_T geq E_0[T])$
$E_0[T]$ is calculated by exponentially weighted moving average of T
statistics self-learning
statistics self-learning
asked yesterday
boniface316boniface316
1266
asked yesterday
boniface316boniface316
1266
edited 18 hours ago
boniface316
asked yesterday
boniface316boniface316
1266
asked yesterday
boniface316boniface316
1266
asked yesterday
boniface316boniface316
1266
1266
$begingroup$
The statement as it stands is very weird; it may mean that $T^star$ is the smallest $T$ such that $theta_T geq E_0[T]$. However, the notation $E_0[T]$ seems to indicate that $T$ is a random variable; could you give more context?
$endgroup$
– Riccardo Sven Risuleo
yesterday
$begingroup$
You are correct about the T being a random variable. I couldnt believe that answer was obvious! Thanks @RiccardoSvenRisuleo!
$endgroup$
– boniface316
yesterday
$begingroup$
I'm not sure that that's the answer! As a matter of fact if T is a random variable, then $theta_T$ may also be a random variable; so the whole thing is a bit messy.
$endgroup$
– Riccardo Sven Risuleo
yesterday
$begingroup$
I can share the whole arguments if you would like....
$endgroup$
– boniface316
yesterday
$begingroup$
I'm just guessing, but if $theta_T$ is a random variable it may be that the expression is missing a probability? Something along the lines of $T^star = argmin_T P(theta_T geq E_0[T])$? Sadly, without reading the whole argument it is impossible to give a clear meaning to the statement.
$endgroup$
– Riccardo Sven Risuleo
yesterday
|
show 1 more comment
$begingroup$
The statement as it stands is very weird; it may mean that $T^star$ is the smallest $T$ such that $theta_T geq E_0[T]$. However, the notation $E_0[T]$ seems to indicate that $T$ is a random variable; could you give more context?
$endgroup$
– Riccardo Sven Risuleo
yesterday
$begingroup$
You are correct about the T being a random variable. I couldnt believe that answer was obvious! Thanks @RiccardoSvenRisuleo!
$endgroup$
– boniface316
yesterday
$begingroup$
I'm not sure that that's the answer! As a matter of fact if T is a random variable, then $theta_T$ may also be a random variable; so the whole thing is a bit messy.
$endgroup$
– Riccardo Sven Risuleo
yesterday
$begingroup$
I can share the whole arguments if you would like....
$endgroup$
– boniface316
yesterday
$begingroup$
I'm just guessing, but if $theta_T$ is a random variable it may be that the expression is missing a probability? Something along the lines of $T^star = argmin_T P(theta_T geq E_0[T])$? Sadly, without reading the whole argument it is impossible to give a clear meaning to the statement.
$endgroup$
– Riccardo Sven Risuleo
yesterday
$begingroup$
The statement as it stands is very weird; it may mean that $T^star$ is the smallest $T$ such that $theta_T geq E_0[T]$. However, the notation $E_0[T]$ seems to indicate that $T$ is a random variable; could you give more context?
$endgroup$
– Riccardo Sven Risuleo
yesterday
$begingroup$
The statement as it stands is very weird; it may mean that $T^star$ is the smallest $T$ such that $theta_T geq E_0[T]$. However, the notation $E_0[T]$ seems to indicate that $T$ is a random variable; could you give more context?
$endgroup$
– Riccardo Sven Risuleo
yesterday
$begingroup$
You are correct about the T being a random variable. I couldnt believe that answer was obvious! Thanks @RiccardoSvenRisuleo!
$endgroup$
– boniface316
yesterday
$begingroup$
You are correct about the T being a random variable. I couldnt believe that answer was obvious! Thanks @RiccardoSvenRisuleo!
$endgroup$
– boniface316
yesterday
$begingroup$
I'm not sure that that's the answer! As a matter of fact if T is a random variable, then $theta_T$ may also be a random variable; so the whole thing is a bit messy.
$endgroup$
– Riccardo Sven Risuleo
yesterday
$begingroup$
I'm not sure that that's the answer! As a matter of fact if T is a random variable, then $theta_T$ may also be a random variable; so the whole thing is a bit messy.
$endgroup$
– Riccardo Sven Risuleo
yesterday
$begingroup$
I can share the whole arguments if you would like....
$endgroup$
– boniface316
yesterday
$begingroup$
I can share the whole arguments if you would like....
$endgroup$
– boniface316
yesterday
$begingroup$
I'm just guessing, but if $theta_T$ is a random variable it may be that the expression is missing a probability? Something along the lines of $T^star = argmin_T P(theta_T geq E_0[T])$? Sadly, without reading the whole argument it is impossible to give a clear meaning to the statement.
$endgroup$
– Riccardo Sven Risuleo
yesterday
$begingroup$
I'm just guessing, but if $theta_T$ is a random variable it may be that the expression is missing a probability? Something along the lines of $T^star = argmin_T P(theta_T geq E_0[T])$? Sadly, without reading the whole argument it is impossible to give a clear meaning to the statement.
$endgroup$
– Riccardo Sven Risuleo
yesterday
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Judging from the text around your expression, it seems plausible that $T^star$ is the smallest $T$ such that $theta_T geq E_0[T]$. These are not random variables because they are supposed to be computed at time $T$.
I think that the idea is that $theta_T$ collects the imbalance in the price; as soon as the imbalance exceeds what we were expecting (so the expected length of the bar, $E_0[T]$), we sample a new bar.
answered 15 hours ago
Riccardo Sven RisuleoRiccardo Sven Risuleo
87129
$endgroup$
$begingroup$
Thanks! This makes so much sense now!
$endgroup$
– boniface316
10 hours ago
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Judging from the text around your expression, it seems plausible that $T^star$ is the smallest $T$ such that $theta_T geq E_0[T]$. These are not random variables because they are supposed to be computed at time $T$.
I think that the idea is that $theta_T$ collects the imbalance in the price; as soon as the imbalance exceeds what we were expecting (so the expected length of the bar, $E_0[T]$), we sample a new bar.
answered 15 hours ago
Riccardo Sven RisuleoRiccardo Sven Risuleo
87129
$endgroup$
$begingroup$
Thanks! This makes so much sense now!
$endgroup$
– boniface316
10 hours ago
add a comment |
$begingroup$
Judging from the text around your expression, it seems plausible that $T^star$ is the smallest $T$ such that $theta_T geq E_0[T]$. These are not random variables because they are supposed to be computed at time $T$.
I think that the idea is that $theta_T$ collects the imbalance in the price; as soon as the imbalance exceeds what we were expecting (so the expected length of the bar, $E_0[T]$), we sample a new bar.
answered 15 hours ago
Riccardo Sven RisuleoRiccardo Sven Risuleo
87129
$endgroup$
$begingroup$
Thanks! This makes so much sense now!
$endgroup$
– boniface316
10 hours ago
add a comment |
$begingroup$
Judging from the text around your expression, it seems plausible that $T^star$ is the smallest $T$ such that $theta_T geq E_0[T]$. These are not random variables because they are supposed to be computed at time $T$.
I think that the idea is that $theta_T$ collects the imbalance in the price; as soon as the imbalance exceeds what we were expecting (so the expected length of the bar, $E_0[T]$), we sample a new bar.
answered 15 hours ago
Riccardo Sven RisuleoRiccardo Sven Risuleo
87129
$endgroup$
Judging from the text around your expression, it seems plausible that $T^star$ is the smallest $T$ such that $theta_T geq E_0[T]$. These are not random variables because they are supposed to be computed at time $T$.
I think that the idea is that $theta_T$ collects the imbalance in the price; as soon as the imbalance exceeds what we were expecting (so the expected length of the bar, $E_0[T]$), we sample a new bar.
answered 15 hours ago
Riccardo Sven RisuleoRiccardo Sven Risuleo
87129
answered 15 hours ago
Riccardo Sven RisuleoRiccardo Sven Risuleo
87129
answered 15 hours ago
Riccardo Sven RisuleoRiccardo Sven Risuleo
87129
answered 15 hours ago
Riccardo Sven RisuleoRiccardo Sven Risuleo
87129
87129
$begingroup$
Thanks! This makes so much sense now!
$endgroup$
– boniface316
10 hours ago
add a comment |
$begingroup$
Thanks! This makes so much sense now!
$endgroup$
– boniface316
10 hours ago
$begingroup$
Thanks! This makes so much sense now!
$endgroup$
– boniface316
10 hours ago
$begingroup$
Thanks! This makes so much sense now!
$endgroup$
– boniface316
10 hours ago
add a comment |
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$begingroup$
The statement as it stands is very weird; it may mean that $T^star$ is the smallest $T$ such that $theta_T geq E_0[T]$. However, the notation $E_0[T]$ seems to indicate that $T$ is a random variable; could you give more context?
$endgroup$
– Riccardo Sven Risuleo
yesterday
$begingroup$
You are correct about the T being a random variable. I couldnt believe that answer was obvious! Thanks @RiccardoSvenRisuleo!
$endgroup$
– boniface316
yesterday
$begingroup$
I'm not sure that that's the answer! As a matter of fact if T is a random variable, then $theta_T$ may also be a random variable; so the whole thing is a bit messy.
$endgroup$
– Riccardo Sven Risuleo
yesterday
$begingroup$
I can share the whole arguments if you would like....
$endgroup$
– boniface316
yesterday
$begingroup$
I'm just guessing, but if $theta_T$ is a random variable it may be that the expression is missing a probability? Something along the lines of $T^star = argmin_T P(theta_T geq E_0[T])$? Sadly, without reading the whole argument it is impossible to give a clear meaning to the statement.
$endgroup$
– Riccardo Sven Risuleo
yesterday