The solution of the model contains constant: $k = \alpha K$, it relates to: (i) probability of getting a fill ($\alpha$) and (ii) market impact ($K$).
Estimating (i). The author proposes that the market order sizes follow a power law distribution:
$$f(x)^Q \propto x^{-1-\alpha}$$
So no problem estimating this.
Estimating (ii). There are two equations of interest here:
$$\Delta p \propto \ln Q \tag{11}$$
$$\begin{align} \lambda (\delta) &= \Lambda \mathbb{P} [\Delta p > \delta] \\ &= \Lambda \mathbb{P}[\ln Q > K \delta] \\ &= \Lambda \mathbb{P} [Q > \exp \left( K \delta \right)] \\ &= \Lambda \int_{\exp (K \delta)}^{\infty} x^{-1-\alpha} dx \\ &= A \exp \left( -k \delta \right) \end{align} \tag{12}$$
According to the above two:
Line 1 in (12) says $\Delta p > \delta$, but we know that $\Delta p = c \ln Q$, therefore we can re-write:
$$\begin{align} \mathbb{P} [\Delta p > \delta] &= \mathbb{P} [c \ln Q > \delta] \\ &= \mathbb{P} [\ln Q > \frac{1}{c} \delta] \end{align}$$
Therefore, $K = \frac{1}{c}$, the inverse of the proportionality constant that we find in relation $\Delta p \propto \ln Q$.
Is this correct? (The estimation of $K$).
Note: I am aware of Sophie Laruelle's paper. But, unfortunately, I do not know French.
Super Note: My understanding of this answer relating to Sophie's paper:
Definitions:
$t_0$ - start time (this is the start time point in the interarrival time slots in a Poisson Process). Simply put, if you get hit, that is a new time: $t_1$, which becomes your new start time.
$\delta P$ - distance of your order from the mid price. If the mid price is \$100, and your bid is at \$85, then this quantity is \$15. Same for asks, obviously.
$P^m (t) $ - is your mid price at time $t$.
Procedure:
- Record start time: $t_0$ and corresponding mid price: $P^m(t_0)$ at this time.
- Record the time $t_1$ when the bid/ask is hit and the $\delta P$s. The reason I have added plural "s" ending in there is because you might get hit at a number of levels. An example is asked for here. This is your order book (OB):
|--- qty ---| --- size --- |
asks
$105 10
$104 5
bids
$100 6
$99 5
Imagine a market order comes in that eats 8 units on the bid size (so it was a market sell). You record the change in price, $\delta P = \$102 - \$100 = \$2$ at first level and the corresponding time of this trade, $t_1$. You also record $\delta P = \$102 - \$99 = \$3$ at the same time, $t_1$ (in this case). If there was a market order that only took portion of the best ask/bid, or full best ask/bid and did not go deeper, we would only collect a single $\delta P$ with its corresponding $t$. Note that you can define the point at which the order is hit differently, this is up to you. - You now have time lengths and corresponding sizes. Interarrival times in a Poisson Process are exponential:
$$\mathbb{P}[X_1 > t] = \mathbb{P}\left[ \texttt{no arrival in time (0, t]} \right]= e^{-\lambda t} $$
So now for each bucket of price changes like: $[\$1, \$1.5]$ i.e. all of the $\delta P$ that are between 1 and 1.5 bucks, you have a list of interarrival times: $[0.3, 0.2, 0.5, 0.7, 1.1]$ in seconds. Fit the above exponential distribution to each data bucket to obtain some empirical estimate of $\lambda$.
You now have figure 1 from Sophie's paper. Well done.
Problems ahead:
What now? So the question is, if you fit the regression to this, what is your $k$, what is your $A$?
Not so important. What is this equation referring to, what is $P_1$, what is $P_2$, what is $P$:
$$k = \mathbb{E}_{P1,P2}\left(\frac{\log\lambda(\Delta P1) - \log(\lambda(\Delta P2)}{\Delta P1 - \Delta P2}\right),\quad A=\mathbb{E}_{P}(\lambda(\Delta P) \exp k \Delta P)$$