# Market Impact: Going from 1/2 power to 3/2

So the standard market impact model (please correct me if I am wrong) says that for an order of size $$V_O$$ executed from time $$0$$ to time $$T$$: $$S_{T} = S_0\left(\alpha + \beta \sigma\sqrt{\frac{V_O}{V_T}} \right)$$where $$S_0$$ is the price at the start of the execution, $$V_T$$ is the total volume, $$\sigma$$ is some measure of daily volatility (in percentage terms), and $$\alpha$$, $$\beta$$ are coefficients, which are unknown but estimated from data.

When reading literature on optimization, I see that there is sometimes a penalty due to trading costs, of the form: $$\propto |w - w_0| ^ {3/2}$$ where $$w$$ is the new proposed set of weights and $$w_0$$ is the initial set of weights. The exponent $$3/2$$ seems to come from integrating $$x^{1/2}$$ (the square root above), suggesting that the square-root model is some "instantaneous" expression, and by integrating it, we get the full impact of the order. But that doesn't seem to be the case in what I've read and what I've written above, as it seems to describe the impact of a full order.

I ask about this distinction, because if we are doing classical mean-variance optimization, and want to add an expected cost term (let's drop the variance of the cost for now, as that's a little bit more challenging to calculate, although feasible w/ delta-method style approximations), then if we add a $$3/2$$ penalty to the problem, the objective remains convex as a function of $$w$$ (as power laws with exponent > $$1$$ are convex). However, if we use $$1/2$$ as the exponent, this is a concave function, and then adding it to the convex mean-variance problem no longer guarantees convexity nor concavity.

• Dec 11, 2022 at 19:45

So I just realized that the $$3/2$$ doesn't come from an integral, but from a "totaling".
If we have some quantity $$V_O$$ that we want to execute at price $$S_0$$, and we receive a price $$S_T$$ rather than $$S_0$$, our notional difference (in some currency) is: $$(S_T - S_0) V_O = S_0\left(\alpha + \frac{\beta \sigma}{\sqrt{V_T}}\sqrt{V_O}\right)V_O - S_0 V_O = S_0\left(c_1 V_O+ c_2 V_O^{\frac{3}{2}}\right)$$ Thus the dollar difference is a linear function of $$V_O$$, plus a $$3/2$$ power.