Insider trading with different risk attitudes

Abstract

This paper examines the impact of different risk attitudes on the financial decisions of two insiders trading in the stock market. We consider a static version of the Kyle (Econometrica 53:1315–1335, 1985) model with two insiders. Insider 1 is risk neutral while insider 2 is risk averse with negative exponential utility. First, we analytically prove the existence of a unique linear equilibrium. Second, we carry out a comparative static analysis with respect to the duopoly case of risk-neutral insiders (Tighe in Three essays on insider trading. Unpublished Ph.D dissertation, University of Illinois at Champaign-Urbana, 1989) and with respect to the duopoly case of risk-averse insiders (Holden and Subrahmanyam in Econ Lett 44:181–190, 1994) models. Our findings reveal that the market depth and the information revelation are higher in Tighe (1989) than in our model. However, compared to Holden and Subrahmanyam (1994), we find that the market depth depends crucially on the degree of risk aversion. Finally, we show that regardless of the degree of risk aversion, the stock price reveals more information in our model than the stock price in Holden and Subrahmanyam (1994).

Appendices

Appendices

1.1 Appendix A: Proof of Proposition 1

We begin by the maximization problem of the risk neutral insider, i.e. insider 1. The decision rule of insider 1 is the function $x_{1}( {\tilde{z}})$. The expected profits after plugging the linear pricing function, become,

$$\begin{aligned} E[({\tilde{z}}-p(x_{1}({\tilde{z}})+x_{2}({\tilde{z}})+{\tilde{u}}))x_{1}({\tilde{z}} )] = E[({\tilde{z}}-\mu -\lambda (x_{1}({\tilde{z}})+x_{2}({\tilde{z}})+{\tilde{u}} ))x_{1}({\tilde{z}})] \end{aligned}$$

The first and second order conditions are

$$\begin{aligned} x_1({\tilde{z}})= \frac{{\tilde{z}}-\mu -\lambda x_{2}({\tilde{z}})}{2\lambda } \quad {\mathrm {and}} \quad \lambda > 0. \end{aligned}$$

(8)

We move now to solve the maximization of the risk averse insider, i.e. insider 2. The decision rule of insider 2 is the function $x_{2}( {\tilde{z}}))$. The expected profits after plugging the linear pricing function, become,

$$\begin{aligned} \displaystyle E[-e^{-A({\tilde{z}}-p(x_{1}({\tilde{z}})+x_{2}({\tilde{z}})+\tilde{u }))x_{2}({\tilde{z}})}|{\tilde{z}}] = \displaystyle E[-e^{-A({\tilde{z}}-\mu -\lambda (x_{1}({\tilde{z}})+x_{2}({\tilde{z}})+{\tilde{u}}))x_{2}({\tilde{z}})}| {\tilde{z}}] \end{aligned}$$

Using the normality and the independency of the noise traders orders ${\tilde{u}}$, the first and the second order conditions are

$$\begin{aligned} x_2({\tilde{z}})= \frac{{\tilde{z}}-\mu -\lambda x_{1}({\tilde{z}})}{\lambda (2+A\lambda \sigma _{u}^2) } \quad {\mathrm{and}} \quad \lambda (2+A\lambda \sigma _{u}^2)> 0. \end{aligned}$$

(9)

Combining Eqs. 8 and 9, we obtain

$$\begin{aligned} x_1({\tilde{z}})= \frac{({\tilde{z}}-\mu )(1+A \lambda \sigma _{u}^2) }{\lambda (3+ 2 A\lambda \sigma _{u}^2) } \quad {\mathrm{and}} \quad x_2({\tilde{z}} )= \frac{({\tilde{z}}-\mu )}{\lambda (3+ 2 A\lambda \sigma _{u}^2) } \end{aligned}$$

(10)

Regarding the price function coefficients, $\mu$ and $\lambda$, first note that the semi-strong market efficiency together with linear price function assumption lead to,

$$\begin{aligned} \mu + \lambda {\tilde{r}} = E[{\tilde{z}}|{\tilde{r}}] \end{aligned}$$

(11)

Evaluating the expectation on both sides of Eq. 11 and then applying the law of iterated expectations, we obtain

$$\begin{aligned} \mu + \lambda {\bar{r}} = {\bar{z}} \end{aligned}$$

(12)

where ${\bar{r}}= {\bar{x}}_1+{\bar{x}}_2+{\bar{u}}= {\bar{x}}_1+{\bar{x}}_2$. Using Eq. 10 to find the expression of ${\bar{r}}$ and plugging the result in Eq. 12, we obtain

$$\begin{aligned} \mu = {\bar{z}} \end{aligned}$$

(13)

To complete the proof, it remains to find a unique value of the price function slope $\lambda$. Indeed, note that the linear expressions of the insiders strategies decisions, ${\tilde{x}}_1$ and ${\tilde{x}}_2$, induce the normality distribution of the total order flow ${\tilde{r}}$. Thus, by applying the projection theorem to Eq. 11, we have

$$\begin{aligned} \lambda = \displaystyle \frac{cov({\tilde{z}}, {\tilde{r}})}{var({\tilde{r}})} \end{aligned}$$

(14)

Evaluating the right-hand side of Eq. 14 and after certain arrangement we find that $\lambda$ is a root of the following quadric equation

$$\begin{aligned} 4A^2\sigma _{u}^4 \mathbf {\lambda ^4} + 12A \sigma _{u}^2 \mathbf {\lambda ^3} +(9- A^2 \sigma _{u}^2 \sigma _{z}^2) \mathbf {\lambda ^2} -3A\sigma _{z}^2 \mathbf {\lambda } -2 \frac{\sigma _{z}^2}{\sigma _{u}^2} = 0. \end{aligned}$$

(15)

By Descartes’ rule of signs,¹⁴ there is only one positive root satisfying the second order condition which ends the proof. $\square$

1.2 Appendix B: Proof of Lemma 1

We first prove the results of Lemma 3 which will be used in many of the subsequent proofs. It should be pointed out that Lemma 3 provides a quantitative characterization for $\lambda$ as function of A.

Lemma 3

The following statements hold

• (i) $\displaystyle \lambda =\frac{1}{2}\frac{\sigma _{z}}{\sigma _{u}}$ is an asymptote in the $(A,\lambda )$ plane. Moreover, the curve defined by Eq.  5 , will not cross its asymptote.

• (ii) $\lambda$is increasing in A.

Proof of i. To find the asymptote, we view the quartic (Eq. 5)

$$\begin{aligned} f\left( \lambda ,A\right) =4A^{2}\sigma _{u}^{4}\lambda ^{4}+12A\sigma _{u}^{2}\lambda ^{3}+\left( 9-A^{2}\sigma _{u}^{2}\sigma _{z}^{2}\right) \lambda ^{2}-3A\sigma _{z}^{2}\lambda -2\frac{\sigma _{z}^{2}}{\sigma _{u}^{2}}=0, \end{aligned}$$

(16)

as a plane algebraic curve.

Note that most of the curves represented by $f\left( \lambda ,A;\sigma _{z},\sigma _{u}\right)$ are irreducible. We work under this assumption. The projective curve corresponding to the affine curve $f=0$ is

$$\begin{aligned} F\left( \lambda ,A,Z\right) =4A^{2}\sigma _{u}^{4}\lambda ^{4}+12A\sigma _{u}^{2}\lambda ^{3}Z^{2}+\left( 9Z^{4}-A^{2}\sigma _{u}^{2}\sigma _{z}^{2}Z^{2}\right) \lambda ^{2}-3A\sigma _{z}^{2}\lambda Z^{4}-2\frac{ \sigma _{z}^{2}}{\sigma _{u}^{2}}Z^{6}=0. \end{aligned}$$

(17)

It is not difficult to see that this projective curve has the singular points $\left( 0:1:0\right)$ and $\left( 1:0:0\right)$. Now, we consider the affine view $Z=1.$ Put $Z=0$ into the Eq. (17) to get $A^{2}\lambda ^{4}=0.$ So the points $\left( 1:0:0\right)$ and $\left( 0:1:0\right)$ are at infinity. The second of these points is on the A-axis, and as noted above it is singular. Set $A=1$ in the equation of the projective curve (17), we thus obtain the affine curve

$$\begin{aligned} 4\sigma _{u}^{4}\lambda ^{4}+12\sigma _{u}^{2}\lambda ^{3}Z^{2}+\left( 9Z^{4}-\sigma _{u}^{2}\sigma _{z}^{2}Z^{2}\right) \lambda ^{2}-3\sigma _{z}^{2}\lambda Z^{4}-2\frac{\sigma _{z}^{2}}{\sigma _{u}^{2}}Z^{6}=0. \end{aligned}$$

The lower order terms $4\sigma _{u}^{4}\lambda ^{4}-\sigma _{u}^{2}\sigma _{z}^{2}Z^{2}\lambda ^{2}$ give us the distinct tangents $\lambda =0,\lambda =\frac{1}{2}\sigma _{z}\frac{Z}{\sigma _{u}},\lambda =-\frac{1}{2}\sigma _{z} \frac{Z}{\sigma _{u}}.$ Now dehomogenize to obtain, for $\lambda >0,$ the affine asymptote $\lambda =\frac{\sigma _{z}}{2\sigma _{u}}.$

Next we show that the curve will not cross its asymptote. The equation of the affine curve above (16) can be written as

$$\begin{aligned} \left( 4\sigma _{u}^{2}\lambda ^{2}-\sigma _{z}^{2}\right) \left( \left( \sigma _{u}^{2}\lambda A+3\right) \lambda A+\frac{2}{\sigma _{u}^{2}}\right) =-\lambda ^{2}<0. \end{aligned}$$

This then implies that $4\sigma _{u}^{2}\lambda ^{2}-\sigma _{z}^{2}<0,$ and so $\lambda <\frac{\sigma _{z}}{2\sigma _{u}}$ for $\lambda >0.$$\square$

Proof of ii In this part of the proof, we show that $\lambda$ is increasing with respect to A where $\sigma _{u}$ and $\sigma _{z}$ are treated as parameters. By implicit differentiation, we obtain

$$\begin{aligned} \frac{d\lambda }{dA}=-\frac{\frac{\partial f}{\partial A}}{\frac{\partial f}{ \partial \lambda }}, \end{aligned}$$

where $\frac{\partial f}{\partial \lambda }$ must not be zero. It is then simple to find that

$$\begin{aligned} \frac{d\lambda }{dA}=\frac{-8A\sigma _{u}^{4}\lambda ^{4}-12\sigma _{u}^{2}\lambda ^{3}+2A\sigma _{u}^{2}\sigma _{z}^{2}\lambda ^{2}+3\sigma _{z}^{2}\lambda }{16A^{2}\sigma _{u}^{4}\lambda ^{3}+36A\sigma _{u}^{2}\lambda ^{2}+2\left( 9-A^{2}\sigma _{u}^{2}\sigma _{z}^{2}\right) \lambda -3A\sigma _{z}^{2}}. \end{aligned}$$

Note that the numerator has one positive root - by Descartes’ rule (since we are interested in $A\ge 0$). Now factor the expression, and so we have

$$\begin{aligned} \frac{d\lambda }{dA}= & {} \frac{-\lambda \left( 2\sigma _{u}\lambda -\sigma _{z}\right) \left( 2\sigma _{u}\lambda +\sigma _{z}\right) \left( 2\lambda A\sigma _{u}^{2}+3\right) }{ \left( 2\lambda A\sigma _{u}^{2}+3\right) \left( 8A\sigma _{u}^{2}\lambda ^{2}+6\lambda -A\sigma _{z}^{2}\right) } \\= & {} -\lambda \left( 2\sigma _{u}\lambda -\sigma _{z}\right) \frac{2\sigma _{u}\lambda +\sigma _{z}}{8A\sigma _{u}^{2}\lambda ^{2}+6\lambda -A\sigma _{z}^{2}} \end{aligned}$$

Clearly, $\frac{d\lambda }{dA}>0$ if $2\sigma _{u}\lambda -\sigma _{z}<0$, or $\lambda <\frac{\sigma _{z}}{2\sigma _{u}}$ and provided that $8A\sigma _{u}^{2}\lambda ^{2}+6\lambda -A\sigma _{z}^{2}>0.$

Consider the upper bound (from part i) to see what is the value of $f\left( \lambda ,A\right)$ for this value of $\lambda .$ The evaluation gives that,

$$\begin{aligned} 4A^{2}\sigma _{u}^{4}\left( \frac{\sigma _{z}}{2\sigma _{u}}\right) ^{4}+12A\sigma _{u}^{2}\left( \frac{\sigma _{z}}{2\sigma _{u}}\right) ^{3}+\left( 9-A^{2}\sigma _{u}^{2}\sigma _{z}^{2}\right) \left( \frac{\sigma _{z}}{2\sigma _{u}}\right) ^{2}-3A\sigma _{z}^{2}\left( \frac{\sigma _{z}}{ 2\sigma _{u}}\right) -2\frac{\sigma _{z}^{2}}{\sigma _{u}^{2}}= \frac{1}{4}\frac{\sigma _{z}^{2}}{\sigma _{u}^{2}}>0. \end{aligned}$$

On the other hand, for $\displaystyle \lambda =\frac{\sqrt{2}\sigma _{z}}{3\sigma _{u}},$ we have

$$\begin{aligned}&4A^{2}\sigma _{u}^{4}\left( \frac{\sqrt{2}\sigma _{z}}{3\sigma _{u}} \right) ^{4}+12A\sigma _{u}^{2}\left( \frac{\sqrt{2}\sigma _{z}}{3\sigma _{u} }\right) ^{3}+\left( 9-A^{2}\sigma _{u}^{2}\sigma _{z}^{2}\right) \left( \frac{\sqrt{2}\sigma _{z}}{3\sigma _{u}}\right) ^{2}-3A\sigma _{z}^{2}\left( \frac{\sqrt{2}\sigma _{z}}{3\sigma _{u}}\right) -2\frac{\sigma _{z}^{2}}{ \sigma _{u}^{2}} \\&\quad = -\frac{1}{81}\sigma _{z}^{3}\frac{A}{\sigma _{u}}\left( 2A\sigma _{z}\sigma _{u}+9\sqrt{2}\right) <0. \end{aligned}$$

This shows that the unique root $\lambda ^*$ is bracketed between $\frac{\sqrt{2 }\sigma _{z}}{3\sigma _{u}}$ and $\frac{\sigma _{z}}{2\sigma _{u}}$ (by the Intermediate Value Theorem).

It remains to show that $8A\sigma _{u}^{2}\lambda ^{2}+6\lambda -A\sigma _{z}^{2}>0.$ Solve

$$\begin{aligned} 8A\sigma _{u}^{2}\lambda ^{2}+6\lambda -A\sigma _{z}^{2}=0 \end{aligned}$$

for A, we get $A=-6\frac{\lambda }{8\sigma _{u}^{2}\lambda ^{2}-\sigma _{z}^{2}}.$ As we are interested in the part of the domain where $\lambda >0$ and $A>0,$ then we should have $8\sigma _{u}^{2}\lambda ^{2}-\sigma _{z}^{2}<0.$ That is $\lambda ^{2}<\frac{\sigma _{z}^{2}}{8\sigma _{u}^{2}}= \frac{1}{2}\left( \frac{\sigma _{z}}{2\sigma _{u}}\right) ^{2}$. But this is lower than the square of the lower bound as $\left( \frac{\sqrt{2}\sigma _{z} }{3\sigma _{u}}\right) ^{2}-\frac{1}{2}\left( \frac{\sigma _{z}}{2\sigma _{u} }\right) ^{2}=$$\frac{7}{72}\frac{\sigma _{z}^{2}}{\sigma _{u}^{2}}.$ Now as $\lambda >0,$ it is clear that

$$\begin{aligned} 8A\sigma _{u}^{2}\lambda ^{2}+6\lambda -A\sigma _{z}^{2}>\left( 8\sigma _{u}^{2}\lambda ^{2}-\sigma _{z}^{2}\right) A. \end{aligned}$$

As the lower bound on $\lambda ^*$ is $\frac{\sqrt{2}\sigma _{z}}{3\sigma _{u}} ,$ then $8\sigma _{u}^{2}\left( \frac{\sqrt{2}\sigma _{z}}{3\sigma _{u}} \right) ^{2}-\sigma _{z}^{2}$$=$$\frac{7}{9}\sigma _{z}^{2}>0.$ This means $r\left( \lambda ,A\right) =8A\sigma _{u}^{2}\lambda ^{2}+6\lambda -A\sigma _{z}^{2}$ is positive on the domain of interest. $\square$

We turn now to prove the results in Lemma 1. Lemma 3 shows that $\lambda ^*$ is increasing with to respect to A and converges to $\frac{\sigma _{z}}{2\sigma _{u}}$ as A getting large. Note that $\frac{\sigma _{z}}{2\sigma _{u}}$ is independent from A and thus it can be treated as constant in our analysis. Hence, one can see that $x_2({\tilde{z}})= \frac{({\tilde{z}}-\mu )}{\lambda ^*(3+ 2 A\lambda ^*\sigma _{u}^2) }$ converges to zero as A getting large. We show now that the trading order of the risk neutral insider will converge to the one of Kyle (1985) model. Recall the risk neutral insider order in Kyle model and in our model are respectively given by

$$\begin{aligned} x({\tilde{z}})= \frac{({\tilde{z}}-\mu )\sigma _{u}}{ \sigma _{z}} \quad {\mathrm {and}}\quad x_1({\tilde{z}})= \frac{({\tilde{z}}-\mu )(1+A \lambda ^*\sigma _{u}^2) }{\lambda ^*(3+ 2A\lambda ^*\sigma _{u}^2) } \end{aligned}$$

When A is getting large and taking into account that $\lambda ^*$ is treated as constant (by Lemma 3), one can notice that $x_1({\tilde{z}})$ converges to $\frac{({\tilde{z}}-\mu )}{2\lambda ^*}$. Since $\lambda ^*$ will converge to $\frac{\sigma _{z}}{2\sigma _{u}}$, replacing its expression in $x_1({\tilde{z}})$, we obtain the same expression as in Kyle (1985).

1.3 Appendix C: Proof of Lemma 2

Recall that the market depth parameter in Tighe (1989) is given by $\lambda ^T= \frac{\sqrt{2}}{3}\frac{\sigma _z}{\sigma _u}$ which corresponds to our market depth parameter when A converges to 0. Since we showed in Lemma 3 that $\lambda ^*$ is increasing in A , the relation in part (a) holds.

We turn now to prove part (b) of Lemma 2. Recall the one-shot game equilibrium of the Holden and Subrahmanyam (1994) model.

Proposition 3

(Holden and Subrahmanyam 1994) In the presence of two risk averse informed traders, a linear equilibrium exists and it is unique. It is characterized by,

(i):

$$\begin{aligned} x_{1}({\tilde{z}})=x_{2}({\tilde{z}})=\displaystyle \frac{({\tilde{z}}-\mu )}{ \lambda ^{HS}(3+A\lambda ^{HS}\sigma _{u}^{2})} \end{aligned}$$

(18)

(ii):

$p({\tilde{y}})=\mu +\lambda ^{HS}{\tilde{y}},$where $\mu ={\bar{z}}$, and $\lambda ^{HS}$is the unique strictly positive root of the following quadric equation:

$$\begin{aligned} A^{2}\sigma _{u}^{4}\mathbf {\lambda ^{4}}+6A\sigma _{u}^{2}\mathbf {\lambda ^{3}}+9\mathbf {\lambda ^{2}}-2A\sigma _{z}^{2}\mathbf {\lambda }-2\frac{ \sigma _{z}^{2}}{\sigma _{u}^{2}}=0. \end{aligned}$$

(19)

We move now to prove the relation between $\lambda ^{*}$ and $\lambda ^{HS}$. Note that (19) can also be considered as an algebraic curve $g\left( A,\lambda ;\sigma _{u},\sigma _{z}\right) =0,$ where $\sigma _{u}$ and $\sigma _{z}$ as viewed as parameters. By implicit differentiation, we obtain

$$\begin{aligned} \frac{d\lambda }{dA}=-\frac{\frac{\partial g}{\partial A}}{\frac{\partial g}{ \partial \lambda }}, \end{aligned}$$

where $\frac{\partial g}{\partial \lambda }$ must not be zero. Thus, we have

$$\begin{aligned} \frac{d\lambda }{dA}=\frac{-\left( 2A\sigma _{u}^{4}\lambda ^{4}+6\sigma _{u}^{2}\lambda ^{3}-2\sigma _{z}^{2}\lambda \right) }{4A^{2}\sigma _{u}^{4}\lambda ^{3}+18A\sigma _{u}^{2}\lambda ^{2}+18\lambda -2A\sigma _{z}^{2}}. \end{aligned}$$

Note that the numerator has one positive root - by Descartes’ rule (since we are interested in $A\ge 0$); and we can write this equation as

$$\begin{aligned} \frac{d\lambda }{dA}=\frac{-\lambda \left( \lambda ^{2}\sigma _{u}^{2}\left( \lambda A\sigma _{u}^{2}+3\right) -\sigma _{z}^{2}\right) }{\lambda \left( \lambda A\sigma _{u}^{2}+3\right) \left( 2\lambda A\sigma _{u}^{2}+3\right) -A\sigma _{z}^{2}}. \end{aligned}$$

Observe that $\frac{d\lambda }{dA}=0,$ implies that

$$\begin{aligned} \lambda \left( \lambda ^{2}\sigma _{u}^{2}\left( \lambda A\sigma _{u}^{2}+3\right) -\sigma _{z}^{2}\right) =0. \end{aligned}$$

Solving this equation for A,  gives $A=-\frac{1}{\lambda ^{3}} \frac{3\lambda ^{2}\sigma _{u}^{2}-\sigma _{z}^{2}}{\sigma _{u}^{4}}.$ Put this expression for A into the quartic to find $\lambda ,$ we then obtain

$\lambda =\frac{1}{2}\frac{\sigma _{z}}{\sigma _{u}}$ or $\lambda =-\frac{1}{ 2}\frac{\sigma _{z}}{\sigma _{u}}.$ Since $\lambda >0,$ we see that $g\left( A,\lambda \right) =0,$ for

$\left( A,\lambda \right) =\left( -\frac{1}{\lambda ^{3}}\frac{3\lambda ^{2}\sigma _{u}^{2}-\sigma _{z}^{2}}{\sigma _{u}^{4}},\frac{1}{2}\frac{ \sigma _{z}}{\sigma _{u}}\right) =\left( -\frac{1}{\left( \frac{1}{2}\frac{ \sigma _{z}}{\sigma _{u}}\right) ^{3}}\frac{3\left( \frac{1}{2}\frac{\sigma _{z}}{\sigma _{u}}\right) ^{2}\sigma _{u}^{2}-\sigma _{z}^{2}}{\sigma _{u}^{4}},\frac{1}{2}\frac{\sigma _{z}}{\sigma _{u}}\right)$ $=\left( \frac{ 2}{\sigma _{z}\sigma _{u}},\frac{1}{2}\frac{\sigma _{z}}{\sigma _{u}}\right) .$

On the other hand, solve (19) for A to obtain $A=\frac{1}{ 2\lambda ^{3}\sigma _{u}^{4}}\left( -6\lambda ^{2}\sigma _{u}^{2}+2\sigma _{z}^{2}\pm 2\sqrt{\left( -4\lambda ^{2}\sigma _{u}^{2}\sigma _{z}^{2}+\sigma _{z}^{4}\right) }\right)$.

In order to have real roots, we need $-4\lambda ^{2}\sigma _{u}^{2}\sigma _{z}^{2}+\sigma _{z}^{4}\ge 0,$ that is $-4\lambda ^{2}\sigma _{u}^{2}\sigma _{z}^{2}+\sigma _{z}^{4}$$=$$-\sigma _{z}^{2}\left( 2\lambda \sigma _{u}-\sigma _{z}\right) \left( 2\lambda \sigma _{u}+\sigma _{z}\right) \ge 0,$ or equivalently, $\left( 2\lambda \sigma _{u}-\sigma _{z}\right) \left( 2\lambda \sigma _{u}+\sigma _{z}\right) \le 0.$ This means that $\left( 2\lambda \sigma _{u}-\sigma _{z}\right) \le 0$, that is $\lambda \le \frac{1}{2}\frac{\sigma _{z}}{ \sigma _{u}}.$

So this bound is the maximal value of $\lambda .$ By the analysis above, this happens when $A=\frac{2}{\sigma _{z}\sigma _{u}}.$ We now study the sign of $\frac{d\lambda }{dA}.$ If we evaluate $\frac{d\lambda }{dA}$ at $A=0$, we get

$$\begin{aligned} \frac{d\lambda }{dA}|_{A=0}=\frac{-\left( 6\sigma _{u}^{2}\lambda ^{3}-2\sigma _{z}^{2}\lambda \right) }{18\lambda }=-\frac{1}{3}\lambda ^{2}\sigma _{u}^{2}+\frac{1}{9}\sigma _{z}^{2}. \end{aligned}$$

Now when $A=0$ , $g\left( 0,\lambda \right) =9\lambda ^{2}-2\frac{ \sigma _{z}^{2}}{\sigma _{u}^{2}}=0$, which has the solutions $\lambda =\pm \frac{1}{3}\sqrt{2}\frac{\sigma _{z}}{\sigma _{u}}.$ Since $\lambda >0,$ we get that

$$\begin{aligned} \frac{d\lambda }{dA}|_{A=0}=\frac{-\left( 6\sigma _{u}^{2}\lambda ^{3}-2\sigma _{z}^{2}\lambda \right) }{18\lambda }=-\frac{1}{3}\left( \frac{1 }{3}\sqrt{2}\frac{\sigma _{z}}{\sigma _{u}}\right) ^{2}\sigma _{u}^{2}+\frac{ 1}{9}\sigma _{z}^{2}=\frac{1}{27}\sigma _{z}^{2}>0. \end{aligned}$$

On the other hand, one can show that $\frac{d\lambda }{dA}|_{A= \frac{3}{\sigma _{z}\sigma _{u}}}<0$. Thus, the derivative at $A=0$ is positive, and to the right of A at $A=\frac{3}{\sigma _{z}\sigma _{u}}$ is negative. This shows that the plane curve described by $g\left( A,\lambda \right) =0,$ has a max given by $\left( A,\lambda \right) =\left( \frac{2}{ \sigma _{z}\sigma _{u}},\frac{1}{2}\frac{\sigma _{z}}{\sigma _{u}}\right) .$ So we can conclude that the algebraic curve $g=0$ is unimodal. In fact, this can be verified by computing the second derivative of lambda with respect to A, which evaluates at $\left( \frac{2}{\sigma _{u}\sigma _{z}},\frac{\sigma _{z}}{2\sigma _{u}}\right)$, is equal to $-\frac{1}{128} (\sigma _z)^3 (\sigma _u)<0.$ This shows that it is concave down on the region containing the maximum point.

Now, computing the first derivative of $\lambda ^*$ with respect to A when $A=0$, we obtain

$$\begin{aligned} \frac{d\lambda ^* }{dA}|_{A=0}= & {} -\frac{1}{6}\left( 2\lambda \sigma _{u}-\sigma _{z}\right) \left( 2\lambda \sigma _{u}+\sigma _{z}\right) \\= & {} -\frac{1}{6}\left( 2\left( \frac{1}{3}\sqrt{2}\frac{\sigma _{z}}{\sigma _{u}}\right) \sigma _{u}-\sigma _{z}\right) \left( 2\left( \frac{1}{3}\sqrt{2 }\frac{\sigma _{z}}{\sigma _{u}}\right) \sigma _{u}+\sigma _{z}\right) \\= & {} -\frac{1}{54}\sigma _{z}^{2}\left( 2\sqrt{2}-3\right) \left( 2\sqrt{2} +3\right) =\frac{\sigma _{z}^{2}}{54}>0 \end{aligned}$$

This shows that the initial slope in Holden and Subrahmanyam (1994) case is exactly twice that of obtained in our case. Consider now the difference $h=g-f$, we get

$$\begin{aligned} h\left( A,\lambda \right) =\lambda A\left( -3A\lambda ^{3}\sigma _{u}^{4}-6\lambda ^{2}\sigma _{u}^{2}+\sigma _{z}^{2}+A\lambda \sigma _{u}^{2}\sigma _{z}^{2}\right) \end{aligned}$$

So to find the intersection of f and g,  h must vanish. Then either $\lambda A=0$ or $-3A\lambda ^{3}\sigma _{u}^{4}-6\lambda ^{2}\sigma _{u}^{2}+\sigma _{z}^{2}+A\lambda \sigma _{u}^{2}\sigma _{z}^{2}=0.$ The cubic polynomial has one sign change, and therefore, it has one positive root by Descarte’s sign rule.

We show in the table below that at $A=\frac{3}{\sigma _{z}\sigma _{u}},$$f<g,$ but that at $A=\frac{4}{\sigma _{z}\sigma _{u}},$$f>g.$ It also shows the intersection point .

$$\begin{aligned} \begin{array}{llll} \lambda \backslash A &{} \frac{3}{\sigma _{z}\sigma _{u}} &{} \frac{4}{\sigma _{z}\sigma _{u}} &{} \frac{3.5}{\sigma _{z}\sigma _{u}} \\ \lambda {\text { from }}f=0 &{} .\, 492\,91\frac{\sigma _{z}}{\sigma _{u}} &{} .\, 494\,81\frac{\sigma _{z}}{\sigma _{u}} &{} .\, 493\,97\frac{\sigma _{z}}{\sigma _{u}} \\ \lambda {\text { from }}g=0 &{} .\, 497\frac{\sigma _{z}}{\sigma _{u}} &{} .\, 490\,51\frac{\sigma _{z}}{\sigma _{u}} &{} .\, 494\,02\frac{\sigma _{z}}{\sigma _{u}} \end{array} \end{aligned}$$

Note that the intersection point is roughly $\left( {\tilde{A}},{\tilde{\lambda }} \right) =\left( \frac{3.5}{\sigma _{z}\sigma _{u}} ,.\, 494\,\frac{\sigma _{z}}{\sigma _{u}}\right)$.

Summarizing, we have that

• 1. the initial slope of g is bigger than the initial slope of f; 

• 2. g is unimodal, it has one critical point which is a maximum occuring at $A=\frac{2}{\sigma _{u}\sigma _{z}};$

• 3. the intersection point of f and g occurs to the right of the maximum of g,  at roughly $A^{*}\simeq \frac{3.5}{\sigma _{u}\sigma _{z}};$

• 4. $h=0$ at two points.

Then it follows that there is one point $\left( {\tilde{A}},{\tilde{\lambda }} \right) ,$ where $\lambda <\lambda _{HS}$ for $A<{\tilde{A}},$ and $\lambda >\lambda _{HS}$ for $A>{\tilde{A}}.$$\square$

1.4 Appendix D: Proof of Proposition 2

Recall that the conditional variance in Tighe (1989) is given by

$$\begin{aligned} var ({\tilde{z}}|{\tilde{y}}) = \frac{1}{3} \sigma _{z}^2 \end{aligned}$$

(20)

Computing the conditional variances in our model and in the Holden and Subrahmanyam (1994) model, we obtain respectively

$$\begin{aligned} var({\tilde{z}}|{\tilde{y}})= & {} \frac{\left( 3\left( 1+A\lambda ^* \sigma _{u}^{2}\right) ^{2}+2\left( 1+A\lambda ^* \sigma _{u}^{2}\right) \right) \sigma _{z}^{2}-(\lambda ^*)^{2}\left( 2\left( 1+A\lambda ^* \sigma _{u}^{2}\right) +1\right) ^{2}\sigma _{u}^{2}}{ \left( 2\left( 1+A\lambda ^* \sigma _{u}^{2}\right) +1\right) ^{2}}, \end{aligned}$$

(21)

$$\begin{aligned} var({\tilde{z}}|{\tilde{y}})= & {} \frac{[(3+A\lambda ^{HS} \sigma _{u}^{2})^{2}-4]\sigma _{z}^{2}-(\lambda ^{HS})^{2}(3+A\lambda ^{HS} \sigma _{u}^{2})^{2}\sigma _{u}^{2}}{(3+A\lambda ^{HS} \sigma _{u}^{2})^{2}}. \end{aligned}$$

(22)

We begin first by showing that (20) is less than (21). Indeed, combining (21) with (15) and after some simplifications, the problem becomes equivalent to show that

$$\begin{aligned} \frac{1}{3}<\frac{ 3+5A\lambda ^* \sigma _{u}^{2}+2(\lambda ^*) ^{2}A^{2}\sigma _{u}^{4}}{\left( 2\left( 1+A\lambda ^* \sigma _{u}^{2}\right) +1\right) ^{2}}. \end{aligned}$$

(23)

Plugging the lower bound, $\lambda ^T$, (the upper bound, $\frac{\sigma _z}{2\sigma _u}$) of $\lambda ^*$ found in Lemma in the numerator (denominator) of the right hand side of (23), we obtain,

$$\begin{aligned} \frac{ 3+5A\sigma _{u}^{2}\left( \frac{\sqrt{2}}{3}\frac{\sigma _{z}}{\sigma _{u}}\right) +2\left( \frac{\sqrt{2}}{3}\frac{\sigma _{z}}{ \sigma _{u}}\right) ^{2}A^{2}\sigma _{u}^{4}}{\left( 2\left( 1+A\left( \frac{ \sigma _{z}}{2\sigma _{u}}\right) \sigma _{u}^{2}\right) +1\right) ^{2}}< \frac{ 3+5A\lambda \sigma _{u}^{2}+2\lambda ^{2}A^{2}\sigma _{u}^{4}}{\left( 2\left( 1+A\lambda \sigma _{u}^{2}\right) +1\right) ^{2}}. \end{aligned}$$

The left hand side simplifies to

$$\begin{aligned} \frac{1}{3}+\frac{\left( \left( 15\sqrt{2}-18\right) A\sigma _{u}\sigma _{z}+\sigma _{z}^{2}\sigma _{u}^{2}A^{2}\right) }{9\left( 3+A\sigma _{z}\sigma _{u}\right) ^{2}}>\frac{1}{3}. \end{aligned}$$

This completes the first part of the proof. It remains to show that the conditional variance in our model is less than the conditional variance in Holden and Subrahmanyam model (1994). Indeed, combining (21) with (15) and combining (22) with (19), the problem reduces to showing that

$$\begin{aligned} \frac{1+A\lambda ^{*}\sigma _{u}^{2}}{3+2A\lambda ^{*}\sigma _{u}^{2} }\le \frac{A\lambda ^{HS}\sigma _{u}^{2}+1}{A\lambda ^{HS}\sigma _{u}^{2}+3} . \end{aligned}$$

(24)

In order to complete the proof, we need to show first the following result.

Lemma 4

$\lambda ^{HS}\in \left[ \frac{\sigma _{z}}{4\sigma _{u}},\frac{ \sigma _{z}}{2\sigma _{u}}\right]$.

Proof

Consider the equation

$$\begin{aligned} g(\lambda )=A^{2}\sigma _{u}^{4}\lambda ^{4}+6A\sigma _{u}^{2}\lambda ^{3}+9\lambda ^{2}-2A\sigma _{z}^{2}\lambda -2\frac{\sigma _{z}^{2}}{\sigma _{u}^{2}}, \end{aligned}$$

(25)

such that $g(\lambda ^{HS})=0$. Note that

$$\begin{aligned} g\left( \frac{\sigma _{z}}{4\sigma _{u}}\right)= & {} A^{2}\sigma _{u}^{4}\left( \frac{\sigma _{z}}{4\sigma _{u}}\right) ^{4}+6A\sigma _{u}^{2}\left( \frac{\sigma _{z}}{4\sigma _{u}}\right) ^{3}+9\left( \frac{ \sigma _{z}}{4\sigma _{u}}\right) ^{2}-2A\sigma _{z}^{2}\left( \frac{\sigma _{z}}{4\sigma _{u}}\right) -2\frac{\sigma _{z}^{2}}{\sigma _{u}^{2}} \\= & {} \frac{1}{256}A^{2}\sigma _{z}^{4}-\frac{13}{32}\frac{A}{ \sigma _{u}}\sigma _{z}^{3}-\frac{23}{16}\frac{\sigma _{z}^{2}}{\sigma _{u}^{2}}= \frac{1}{256}\sigma _{z}^{2}\frac{A^{2}\sigma _{z}^{2}\sigma _{u}^{2}-104A\sigma _{z}\sigma _{u}-368}{\sigma _{u}^{2}}. \end{aligned}$$

Consider the numerator $A^{2}\sigma _{z}^{2}\sigma _{u}^{2}-104A\sigma _{z}\sigma _{u}-368=0$. It has the solutions $A=4\frac{ 13+8\sqrt{3}}{\sigma _{z}\sigma _{u}}=\frac{107.\, 43}{\sigma _{z}\sigma _{u}}$ or $\ A=4\frac{13-8\sqrt{3}}{\sigma _{z}\sigma _{u}}=$$- \frac{3.\, 425\,6}{\sigma _{z}\sigma _{u}}.$ Now, the derivative

$$\begin{aligned} \frac{d\left( \frac{1}{256}\sigma _{z}^{2}\frac{A^{2}\sigma _{z}^{2}\sigma _{u}^{2}-104A\sigma _{z}\sigma _{u}-368}{\sigma _{u}^{2}} \right) }{dA}= \frac{1}{128}\frac{\sigma _{z}^{3}}{\sigma _{u}} \left( A\sigma _{z}\sigma _{u}-52\right) . \end{aligned}$$

So for $0\le A<\frac{52}{\sigma _{z}\sigma _{u}}$, the derivative is negative and $g\left( \frac{\sigma _{z}}{4\sigma _{u}}\right)$ (at this value only) is decreasing in A for $0\le A<\frac{52}{\sigma _{z}\sigma _{u}}$. As also for $A=0,$$g\left( \frac{\sigma _{z}}{4\sigma _{u}}\right) =9\left( \frac{\sigma _{z}}{4\sigma _{u}}\right) ^{2}-2\frac{\sigma _{z}^{2} }{\sigma _{u}^{2}}$$=-\frac{23}{16}\frac{\sigma _{z}^{2}}{\sigma _{u}^{2}}$ is negative, we can conclude that $\frac{\sigma _{z}}{4\sigma _{u}}$ is a lower bound for $\lambda$. Similarly, it follows from what we had before,

$$\begin{aligned}&A^{2}\sigma _{u}^{4}\left( \frac{\sigma _{z}}{2\sigma _{u}}\right) ^{4}+6A\sigma _{u}^{2}\left( \frac{\sigma _{z}}{2\sigma _{u}}\right) ^{3}+9\left( \frac{\sigma _{z}}{2\sigma _{u}}\right) ^{2}-2A\sigma _{z}^{2}\left( \frac{\sigma _{z}}{2\sigma _{u}}\right) -2\frac{\sigma _{z}^{2}}{\sigma _{u}^{2}} \\&\quad = \frac{1}{16}\sigma _{z}^{2}\frac{A^{2}\sigma _{z}^{2}\sigma _{u}^{2}-4A\sigma _{z}\sigma _{u}+4}{\sigma _{u}^{2}}= \frac{1}{16 }\sigma _{z}^{2}\frac{\left( A\sigma _{z}\sigma _{u}-2\right) ^{2}}{\sigma _{u}^{2}}>0. \end{aligned}$$

Thus, by the Intermediate Value Theorem, the proof of Lemma 3 is complete. $\square$

We turn now to show that (24) holds. Since the expressions in (24) are both monotonically increasing in $\lambda$, plugging the lower bound of Lemma 4 in the right hand side of (24), and plugging the upper bound of Lemma 4 in the left hand side of (24), and then subtracting the resulting expressions from each others, we obtain

$$\begin{aligned} \frac{1}{2}\frac{2+A\sigma _{z}\sigma _{u}}{3+A\sigma _{z}\sigma _{u}}-\frac{ A\sigma _{z}\sigma _{u}+4}{A\sigma _{z}\sigma _{u}+12}=-\frac{1}{2}\frac{ \left( A\sigma _{z}\sigma _{u}\right) ^{2}}{\left( 3+A\sigma _{z}\sigma _{u}\right) \left( A\sigma _{z}\sigma _{u}+12\right) }<0. \end{aligned}$$

(26)

This completes the proof of Proposition 2. $\square$