Liquidity and the marginal value of information

Abstract

We revisit Kyle’s (Econometrica 53:1315–1335, 1985) model of price formation in the presence of private information. We begin by using Back’s (Rev Financ Stud 5(3):387–409, 1992) approach, demonstrating that if standard assumptions are imposed, the model has a unique equilibrium solution and that the insider’s trading strategy has a martingale property. That in turn implies that the insider’s strategies are linear in total order flow. We also show that for arbitrary prior distributions, the insider’s trading strategy is uniquely determined by a Doob $h$-transform that expresses the insider’s informational advantage. This allows us to reformulate the model so that Kyle’s liquidity parameter $\lambda$ is characterized by a Lagrange multiplier that is the marginal value or shadow price of information. Based on these findings, we can then interpret liquidity as the marginal value of information.

Appendices

Appendix A: Back’s (1992) approach: derivations and heuristic arguments

The value function is the sum of expected trading profits:

$$\begin{aligned} J(V,y,t)=E\left[ \int \limits _{t}^{T}q\left( t^{\prime }\right) \left( V-P(t^{\prime })\right) \text{ d }t^{\prime }\Big |{\fancyscript{F}}_{I}(t)\right] . \end{aligned}$$

Following Back (1992), we assume that the current price $P$ is obtained using a pricing rule $H\left( y,t\right)$ that may explicitly depend on time and on the state $y(t)$, but we make no other assumptions. Economically, the assumption is that the optimal pricing rule does not depend on a price path, but rather only on the current price.

Under these assumptions, the Bellman equation is

$$\begin{aligned} \max _{q\left( \cdot \right) } E \left[ \text{ d }J\left( V,y,t\right) \Big | {\fancyscript{F}}_I(t) \right] =-q\left( t\right) \left( V-H\left( y,t\right) \right) \text{ d }t, \end{aligned}$$

with

$$\begin{aligned} E \left[ \text{ d }J\left( V,y,t\right) \Big | {\fancyscript{F}}_I(t) \right] =\left( J_{t}+\frac{\sigma _{u}^{2}}{2}J_{yy}+q\left( t\right) J_{y}\right) \text{ d }t, \end{aligned}$$

which can be restated as

$$\begin{aligned} \max _{q\left( \cdot \right) }\left\{ J_{t}+\frac{\sigma _{u}^{2}}{2} J_{yy}+q\left( t\right) J_{y}+q\left( t\right) \left( V-H\left( y,t\right) \right) \right\} =0. \end{aligned}$$

(80)

It is evident from Eq. (80) that the optimization with respect to $q$ is linear, because the impact of $q(t)$ is not explicit in the pricing function $H$, and the optimal trading strategy therefore might not be unique. Consistent with the Markov strategy assumption that the pricing rule $H\left( y,t\right)$ only depends on the current value of the total order flow, that is, there is no path dependence, we assume that the optimal trading strategies $q^*$ are also path independent, i.e. $q^*=q^*\left( V,y,t\right)$. In what follows, we will sometimes use a short hand notation $q^*\left( t\right)$ assuming that the optimal trading strategy also depends on the realization of the fundamental and current value of the total order flow.

1.1 A.1: Solution of the insider’s problem

Our strategy is as follows. First, we use the first-order condition for the insider’s optimization problem (80) to show that the insider’s value function satisfies a second-order partial differential equation that is also a martingale condition. Rather than generating a boundary condition and solving for the value function, we show that the martingale condition for the value function implies that the pricing rule also satisfies that same martingale condition. It is this equation that we solve with boundary conditions. Those boundary conditions are driven by a no-arbitrage condition that must hold at the terminal time.

Returning to the recapitulation of Back’s analysis, we first establish that the insider’s value function is a martingale. To ensure internal solutions (with finite norm), the condition

$$\begin{aligned} J_{y}=-\left( V-H\left( y,t\right) \right) \end{aligned}$$

(81)

must be satisfied. Substituting this back into the Bellman Eq. (80), we obtain

$$\begin{aligned} J_{t}+\frac{\sigma _{u}^{2}}{2}J_{yy}=0 \end{aligned}$$

(82)

which is the inverse heat equation. The solution will therefore be a martingale.¹¹ The martingale property of $J$ immediately leads to a martingale property for the pricing function as well. Differentiating (82) w.r.t. $y$ and making use of (81) yields the same inverse heat equation for the pricing rule

$$\begin{aligned} H_{t}+\frac{\sigma _{u}^{2}}{2}H_{yy}=0, \end{aligned}$$

(83)

which implies that $H$ is also a martingale and can be solved using a Feynman–Kac theorem. We pursue this strategy in the main text.

Appendix B: Derivation of conditional the probability density of $V$ for the market makers

In this appendix, we demonstrate that even when the prior distribution of the fundamental value $V$ is not Gaussian, the market maker perceives the order flow as Brownian motion.

We make no assumptions about the structure of the insider’s order flow strategy $q(V,t)$ initially. The notation for the market maker’s conditional probability density function for the value $V$ is:

$$\begin{aligned} \rho =\rho (V,y,t)\equiv \rho (V,t). \end{aligned}$$

(84)

Using the shorthand notation

$$\begin{aligned} q(V,t)\equiv q(V,y,t) \end{aligned}$$

(85)

and Bayes’ rule (analogous to the linear case), the incremental density is

$$\begin{aligned} \rho (V,t+\text{ d }t)&\simeq \frac{\rho (V,t)\,e^{-\frac{({d} y-q(V,t)\mathrm{d}t)^{2}}{2\sigma _{u}^{2}{d} t}}}{\int \rho (V^{\prime },t)\,e^{-\frac{({d}y-q(V^{\prime },t){d}t)^{2}}{2\sigma _{u}^{2}{d}t}}\,{d}V^{\prime }}\end{aligned}$$

(86)

$$\begin{aligned}&\simeq \frac{\rho (V,t)\,e^{-\frac{({d}y)^{2}}{2\sigma _{u}^{2}{d}t}}e^{-\frac{ q{d}y}{\sigma _{u}^{2}}-\frac{q^{2}{d}t}{2\sigma _{u}^{2}}}}{\int \rho (V^{\prime },t)\,e^{-\frac{({d}y)^{2}}{2\sigma _{u}^{2}{d}t}}e^{-\frac{q{d}y}{ \sigma _{u}^{2}}-\frac{q^{2}{d}t}{2\sigma _{u}^{2}{d}t}}\,{d}V^{\prime }}\end{aligned}$$

(87)

$$\begin{aligned}&\simeq \frac{\rho (V,t)\,e^{-\frac{q{d}y}{\sigma _{u}^{2}}-\frac{q^{2}{d}t}{ 2\sigma _{u}^{2}}}}{\int \rho (V^{\prime },t)\,e^{-\frac{q{d}y}{\sigma _{u}^{2} }-\frac{q^{2}{d}t}{2\sigma _{u}^{2}}}\,{d}V^{\prime }}\end{aligned}$$

(88)

$$\begin{aligned}&\simeq \frac{\rho (V,t)\left( 1-\frac{q^{2}{d}t}{2\sigma _{u}^{2}}+\frac{q{d}y}{ \sigma _{u}^{2}}+\frac{1}{2}\frac{q^{2}\sigma _{u}^{2}{d}t}{\sigma _{u}^{4}} \right) }{\int \rho (V^{\prime },t)\,\left( 1-\frac{q^{2}}{2\sigma _{u}^{2}}+ \frac{1}{2}\frac{q^{2}\sigma _{u}^{2}{d}t}{\sigma _{u}^{4}}\right) \,\mathrm{d}V^{\prime }}\end{aligned}$$

(89)

$$\begin{aligned}&\simeq \frac{\rho (V,t)\left( 1+\frac{q{d}y}{\sigma _{u}^{2}}\right) }{\int \rho (V^{\prime },t)\,\left( 1+\frac{q{d}y}{\sigma _{u}^{2}}\right) \,{d}V^{\prime }} \end{aligned}$$

(90)

Therefore,

$$\begin{aligned} \rho (V,t+{d}t)\simeq \rho (V,t)\frac{\left( 1+q\frac{{d}y}{\sigma _{u}^{2}} \right) }{\left( 1+\bar{q}\frac{{d}y}{\sigma _{u}^{2}}\right) } \end{aligned}$$

(91)

where the expected insider trading strategy is defined as

$$\begin{aligned} \bar{q}\equiv \int \rho (V^{\prime },t)q(V^{\prime },t)\,\text{ d }V^{\prime }. \end{aligned}$$

(92)

Our demonstration that the order flow is Gaussian in the insider’s information will rest on showing that this quantity is zero.

Expanding the denominator of (91) yields

$$\begin{aligned} \rho (V,t+\text{ d }t)&\simeq \rho (V,t)\left( 1+q\frac{\text{ d }y}{\sigma _{u}^{2}}\right) \left( 1-\bar{q}\frac{\text{ d }y}{\sigma _{u}^{2}}+\bar{q}^{2}\frac{\text{ d }y^{2}}{\sigma _{u}^{4}}+\dots \right) \end{aligned}$$

(93)

$$\begin{aligned}&\simeq \rho (V,t)\left( 1+(q-\bar{q})\frac{\text{ d }y}{\sigma _{u}^{2}}-\bar{q}\frac{ (q-\bar{q})}{\sigma _{u}^{2}}\text{ d }t\right) \end{aligned}$$

(94)

which reduces to

$$\begin{aligned} \frac{\text{ d }\rho }{\rho }=\frac{(q-\bar{q})}{\sigma _{u}^{2}}\text{ d }y-\bar{q}\frac{(q- \bar{q})}{\sigma _{u}^{2}}\text{ d }t. \end{aligned}$$

(95)

Therefore, we have

$$\begin{aligned} \rho (V,t+\text{ d }t)\simeq \rho (V,t)\left( 1+\frac{(q-\bar{q})}{\sigma _{u}^{2}}\text{ d }y- \bar{q}\frac{(q-\bar{q})}{\sigma _{u}^{2}}\text{ d }t\right) . \end{aligned}$$

(96)

Because $P(y,t)=E_{V}\left[ V\Big |{\fancyscript{F}}_{M}(t)\right]$, where ${\fancyscript{F}}_{M}(t)$ is the market maker’s information at $t$, if we calculate the conditional expected value

$$\begin{aligned} E_{V}\left[ V\Big |{\fancyscript{F}}_{M}(t)\right] =\int V\text{ d }\rho (V,t)=\int \rho (V,t)V\frac{(q-\bar{q})}{\sigma _{u}^{2}}\left( \text{ d }y-\bar{q}\text{ d }t\right) dV \end{aligned}$$

(97)

we have

$$\begin{aligned} P(t+\text{ d }t)\simeq P(t)+\lambda \text{ d }y-\bar{q}\lambda \text{ d }t, \end{aligned}$$

(98)

where

$$\begin{aligned} \lambda (t)\equiv E_{V}\left[ \frac{(q-\bar{q})}{\sigma _{u}^{2}}V\Big |{ \fancyscript{F}}_{M}(t)\right] =\text{ cov }_{V}^{t}[q,V]\frac{1}{\sigma _{u}^{2}} \end{aligned}$$

(99)

which is the generalized analogue of Kyle’s $\lambda$.

1.1 B.1: Dynamics of price $P$

From (98), it follows that

$$\begin{aligned} \text{ d }P=\lambda (t)\text{ d }y-\bar{q}\lambda (t)\text{ d }t, \end{aligned}$$

(100)

and because $\text{ d }y=q\text{ d }t+\sigma _{u}\text{ d }z_{u}$,

$$\begin{aligned} \text{ d }P=\lambda (t)\left( (q-\bar{q})\text{ d }t+\sigma _{u}\text{ d }z_{u}\right) \end{aligned}$$

(101)

Equation (101) says that the price reacts only to the “surprise” or forecast error component of order flow, which consists of (i) the liquidity shock $\sigma _{u}\mathrm{d}z_{u}(t)$ and (ii) the unexpected component of the insider’s order flow, $(q-\bar{q})\mathrm{d}t$. In other words, the price does not react to the predictable component of order flow, $\bar{q} \mathrm{d}t$, because it does not transmit any additional information.

1.2 B.2: The insider’s problem

Proposition 6

In equilibrium the expected insider’s strategy in the market makers’ information set vanishes, i.e. $\bar{q}=0$.

We remark that we have not assumed Gaussianity of the prior distribution of the fundamental value $V$.

Proof

Consider the insider’s optimization problem. Under the assumption that the price is described by the path-independent pricing function

$$\begin{aligned} P=H(y,t), \end{aligned}$$

(102)

Back (1992) shows that the necessary condition for the existence of internal solutions for the insider’s problem is that the pricing function satisfies an inverse heat equation, which we have also derived in Eq. (3):

$$\begin{aligned} \left( \frac{\partial }{\partial t}+\frac{\sigma _{u}^{2}}{2}\frac{\partial ^{2}}{\partial y^{2}}\right) H\left( y,t\right) =0, \end{aligned}$$

(103)

Combining this result with Ito’s lemma yields

$$\begin{aligned} \text{ d }H\left( y,t\right) =\frac{\partial H\left( y,t\right) }{\partial y}\text{ d }y=\frac{ \partial H\left( y,t\right) }{\partial y}\left( q\text{ d }t+\sigma _{u}\text{ d }z_{u}\right) . \end{aligned}$$

(104)

In what follows, we assume that the price is indeed path independent. In equilibrium, the pricing function dynamics obtained from the insider’s optimization (104) has to match the dynamics from the market makers’ perspective given by

$$\begin{aligned} \text{ d }H\left( y,t\right) =\lambda (y,t)\left( (q-\bar{q})\text{ d }t+\sigma _{u}\text{ d }z_{u}\right) . \end{aligned}$$

(105)

Comparing (104) and (105), we immediately conclude that in equilibrium

$$\begin{aligned} \frac{\partial H\left( y,t\right) }{\partial y}&= \lambda (y,t), \nonumber \\ \bar{q}&= 0. \end{aligned}$$

(106)

In other words, the insider does not choose a strategy with nonzero expected component $\bar{q}$ at equilibrium and profit from anything that depends on $\bar{q}$, i.e., anything that is predictable by the market makers. $\square$

Appendix C: Dynamics of the Doob $h$-transform

Proof

(of Corollary 1) The dynamics of the total order flow $y\left( t\right)$ is given by (2). In the market makers’ information set ${\fancyscript{F}}_{M}$, the total order flow is a martingale and is represented by a Brownian motion, and is therefore driftless. As such, the probability density $g$ of the process satisfies the forward Kolmogorov equation,

$$\begin{aligned} \widehat{L}g(y,t)=0, \end{aligned}$$

(107)

with infinitesimal generator

$$\begin{aligned} \widehat{L}=\frac{\partial }{\partial t}-\frac{\sigma _{u}^{2}}{2}\frac{ \partial ^{2}}{\partial y^{2}}. \end{aligned}$$

(108)

In the insider’s information however, there is a requirement that the final price must equal the value, and thus satisfies a terminal condition

$$\begin{aligned} H(y(T),T) = h(y_{T}) =V, \end{aligned}$$

(109)

or

$$\begin{aligned} y_{T}=h^{-1}(V). \end{aligned}$$

(110)

The terminal condition means that the backward Kolmogorov equation applies, and the transition probability $f(y,t|y_{T},T)$ is a diffusion equation¹²

$$\begin{aligned} \widehat{L}_{f}f(y,t)=0, \end{aligned}$$

(111)

with infinitesimal generator

$$\begin{aligned} \widehat{L}_{f}=\frac{\partial }{\partial t}+\frac{\sigma _{u}^{2}}{2}\frac{ \partial ^{2}}{\partial y^{2}}. \end{aligned}$$

(112)

In the insider’s information set, there is a finite drift given by the insider’s strategy, which is generally nonzero. Consider a change of measure (that is, a Girsanov transformation) that transforms the market makers’ information ${\fancyscript{F}}_{M}$ into the insider’s information ${\fancyscript{F}} _{I}$, while maintaining the terminal condition. This transformation is described by a so-called Doob $h$-transform, which is a generalization of a Radon–Nikodym derivative.¹³ Denoting this measure by $m(y,t)$, it must be a martingale in the market makers’ information set (due to the normalization condition, that is, that it is a probability density) and therefore also satisfies the backward Kolmogorov equation,

$$\begin{aligned} \left( \frac{\partial }{\partial t}+\frac{\sigma _{u}^{2}}{2}\frac{\partial ^{2}}{\partial y^{2}}\right) m(y,t)=0. \end{aligned}$$

(113)

This equation, along with the normalization requirement that $m(y,t)$ is a probability density and the requirement that the terminal value be a Dirac $\delta$ function, defines a Cauchy problem for the fundamental solution of the inverse heat problem, which has a unique solution.

The transformed measure is given by $f(y,t)=g(y,t)m(y,t)$, and in this measure, the generator of the original diffusion is given by¹⁴

$$\begin{aligned} \widehat{L}_{h}g(y,t)=\frac{1}{m(y,t)}\widehat{L}\left( g(y,t)m(y,t)\right) . \end{aligned}$$

(114)

Because the Doob $h$-transform $m$ satisfies (113), the infinitesimal generator in the new measure takes the form

$$\begin{aligned} \widehat{L}_{h}=\widehat{L}+\mu \left( y,t\right) \frac{\partial }{\partial y }, \end{aligned}$$

(115)

with

$$\begin{aligned} \mu \left( y,t\right) =\sigma _{u}^{2}\frac{\frac{\partial }{\partial y} m(y,t)}{m(y,t)}, \end{aligned}$$

(116)

which means that the change of measure leads to an additional drift term, consistent with Girsanov’s theorem.¹⁵

In our case, (116) implies that the total order flow $y\left( t\right)$ satisfies the following stochastic differential equation in the transformed measure:

$$\begin{aligned} \text{ d }y(t)=\mu (y,t)\text{ d }t+\sigma _{u}\text{ d }z_{u}(t). \end{aligned}$$

(117)

The next task is to characterize the drift $\mu (y,t)$. We establish that it is a martingale by using the martingale properties of $m$ and applying Ito’s lemma.

Recalling that the Doob $h$-transform $m$ satisfies the diffusion Eq. (113) and directly differentiating (116) w.r.t. $t$, we obtain

$$\begin{aligned} \frac{1}{\sigma _{u}^{2}}\frac{\partial \mu \left( y,t\right) }{\partial t}&= -\frac{1}{m^{2}}\frac{\partial m}{\partial t}\frac{\partial m}{\partial y} +\frac{1}{m}\frac{\partial ^{2}m}{\partial t\partial y} \nonumber \\&= \frac{\sigma _{u}^{2}}{2}\left( \frac{1}{m^{2}}\frac{\partial m}{\partial y}\frac{\partial ^{2}m}{\partial y^{2}}-\frac{1}{m}\frac{\partial ^{3}m}{ \partial y^{3}}\right) . \end{aligned}$$

(118)

Differentiating (116) w.r.t. $y$, we obtain

$$\begin{aligned} \frac{1}{\sigma _{u}^{2}}\frac{\partial \mu \left( y,t\right) }{\partial y}=- \frac{1}{m^{2}}\left( \frac{\partial m}{\partial y}\right) ^{2}+\frac{1}{m} \frac{\partial ^{2}m}{\partial y^{2}}, \end{aligned}$$

(119)

and

$$\begin{aligned} \frac{1}{\sigma _{u}^{2}}\frac{\partial ^{2}\mu \left( y,t\right) }{\partial y^{2}}&= \frac{2}{m^{3}}\left( \frac{\partial m}{\partial y}\right) ^{3}- \frac{2}{m^{2}}\frac{\partial m}{\partial y}\frac{\partial ^{2}m}{\partial y^{2}}\nonumber \\&-\frac{1}{m^{2}}\frac{\partial m}{\partial y}\frac{\partial ^{2}m}{ \partial y^{2}}+\frac{1}{m}\frac{\partial ^{3}m}{\partial y^{3}}. \end{aligned}$$

(120)

Combining (118), (119) and (120) yields

$$\begin{aligned} \frac{1}{\sigma _{u}^{2}}\left( \frac{\partial }{\partial t}+\frac{\sigma _{u}^{2}}{2}\frac{\partial ^{2}}{\partial y^{2}}\right) \mu \left( y,t\right) =\frac{\sigma _{u}^{2}}{2}R\left( y,t\right) , \end{aligned}$$

(121)

with

$$\begin{aligned} R\left( y,t\right)&= \frac{1}{m^{2}}\frac{\partial m}{\partial y}\frac{ \partial ^{2}m}{\partial y^{2}}-\frac{1}{m}\frac{\partial ^{3}m}{\partial y^{3}}\nonumber \\&+\frac{1}{m}\frac{\partial ^{3}m}{\partial y^{3}}-\frac{1}{m^{2}}\frac{ \partial m}{\partial y}\frac{\partial ^{2}m}{\partial y^{2}}\nonumber \\&+\frac{2}{m^{3}}\left( \frac{\partial m}{\partial y}\right) ^{3}-\frac{2}{ m^{2}}\frac{\partial m}{\partial y}\frac{\partial ^{2}m}{\partial y^{2}}\nonumber \\&= \frac{2}{m^{3}}\left( \frac{\partial m}{\partial y}\right) ^{3}-\frac{2}{ m^{2}}\frac{\partial m}{\partial y}\frac{\partial ^{2}m}{\partial y^{2}}, \end{aligned}$$

(122)

and therefore

$$\begin{aligned} \left( \frac{\partial }{\partial t}+\frac{\sigma _{u}^{2}}{2}\frac{\partial ^{2}}{\partial y^{2}}\right) \mu \left( y,t\right) =\sigma _{u}^{2}\frac{1}{m }\frac{\partial m}{\partial y}\left( \frac{1}{m^{2}}\left( \frac{\partial m}{ \partial y}\right) ^{2}-\frac{1}{m}\frac{\partial ^{2}m}{\partial y^{2}} \right) . \end{aligned}$$

(123)

Substituting (119) into the right-hand side of (123), we immediately obtain

$$\begin{aligned} \left( \frac{\partial }{\partial t}+\frac{\sigma _{u}^{2}}{2}\frac{\partial ^{2}}{\partial y^{2}}\right) \mu \left( y,t\right) =-\sigma _{u}^{2}\frac{1}{ m}\frac{\partial m}{\partial y}\frac{\partial \mu \left( y,t\right) }{ \partial y}, \end{aligned}$$

(124)

or, equivalently

$$\begin{aligned} \left( \frac{\partial }{\partial t}+\mu \left( y,t\right) \frac{\partial }{ \partial y}+\frac{\sigma _{u}^{2}}{2}\frac{\partial ^{2}}{\partial y^{2}} \right) \mu (y,t)=0. \end{aligned}$$

(125)

Applying Ito’s lemma to $\mu$ and using Eq. (125), we obtain

$$\begin{aligned} \text{ d }\mu (y,t)=\frac{\partial \mu \left( y,t\right) }{\partial y}\sigma _{u}\text{ d }z_{u}\left( t\right) , \end{aligned}$$

(126)

i.e. the drift itself is a martingale in the insider’s information set.

Note that in our particular case, the trading strategy $q$ plays the role of a drift, and therefore, the insider’s trading strategy $q=\mu (y,t)$ is a martingale in the insider’s information set. $\square$

Appendix D: Proof of equation (33)

In this appendix, we demonstrate the validity of Eq. (33). The linearity of the insider’s strategies means that the information sets are normal conditioned on the market makers’ information, with p.d.f.

$$\begin{aligned} \rho \left( V,t\right) =\frac{1}{\sqrt{2\pi \varSigma _M(t) }}\exp \left( -\frac{ \left( V-P\left( t\right) \right) ^{2}}{2\varSigma _M(t) }\right) . \end{aligned}$$

(127)

When time increases from $t$ to $t+\text{ d }t$, the market makers observe the incremental order flow $\text{ d }y\left( t\right)$ and update their information sets. For the p.d.f. $\rho \left( V,t+\text{ d }t\right)$, we have

$$\begin{aligned} \rho \left( V,t+\text{ d }t\right)&= C\rho \left( V,t\right) \exp \left( -\frac{ \left( \text{ d }y-\beta \left( t\right) \left( V-P\left( t\right) \right) \text{ d }t\right) ^{2}}{2\sigma _{u}^{2}\text{ d }t}\right) \nonumber \\&= \frac{1}{\sqrt{2\pi \varSigma _M( t+\mathrm{d}t) }}\exp \left( -\frac{\left( V-P\left( t+\text{ d }t\right) \right) ^{2}}{2\varSigma _M( t+\text{ d }t) }\right) , \end{aligned}$$

(128)

where $C$ is a normalization constant. The first line in (128) notes, as did Kyle, that the stochastic evolution of order flow in the interval $\text{ d }t$ is driven by the noise trade. Comparing the exponential terms in the two lines in (128), matching the coefficients of $V^{1}$ and $V^{2}$ and keeping only the terms linear in $\text{ d }t$, we obtain

$$\begin{aligned} \frac{1}{\varSigma _M( t) }+\frac{\beta \left( t\right) ^{2} }{\sigma _{u}^{2}}\text{ d }t= \frac{1}{\varSigma _M( t+\text{ d }t) }, \nonumber \end{aligned}$$

and

$$\begin{aligned} \frac{P\left( t+\mathrm{d}t\right) -P\left( t\right) }{\varSigma _M(t) }=\frac{\beta \left( t\right) }{\sigma _{u}^{2}}\text{ d }y. \end{aligned}$$

(129)

In the limit,¹⁶

$$\begin{aligned} \text{ d }\left( \frac{1}{\varSigma _M( t) }\right) =\text{ d }\left( \frac{1}{\text{ Var }\left[ \text{ d }y \right] }\right) =\frac{\beta \left( t\right) ^{2} }{\sigma _{u}^{2}}\text{ d }t, \end{aligned}$$

(130)

Further manipulation yields (33). Note that the derivation is entirely probabilistic and does not rest on optimizing a quadratic loss function.