argmin with logic statements [solved]

Question

Could anyone explain me the equation below? Are we trying the find the T value where the $\theta_T < E_0[t]?$

$T^* = \underset{T}{Argmin} (\theta_T \geq E_0[T])$

I understand how argmin works for an equation like $\underset{x}{Argmin} f(x)$. 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!

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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=\begin{cases}b_{t-1}, & \text{if}\ \Delta p_t = 0 \\ \frac{|\Delta p_t|}{\Delta p_t} ,& \text{if} \Delta p_t \neq 0 \end{cases} $

• $p_t$ = the price associated with tick t
• $v_t$ = volume associated with tick t

$\theta_T = \sum_{t = 1}^{T}b_t$

T = tick index

$T^* = \underset{T}{Argmin} (\theta_T \geq E_0[T])$

$E_0[T]$ is calculated by exponentially weighted moving average of T

Answer 1

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.