Security price formation and informed trading with constrained short selling

Abstract

Short sale orders account for a substantial portion of trading volume in recent years. This paper develops a sequential trade model with constrained short selling to derive the effect on prices when the market maker can observe short selling in the order flow. The model predicts that market quotes will adjust differently to short sales and regular sales. Furthermore, the model shows that the probability of informed trading is impacted both by the level of short sale constraints and the intensity of actual short sale trades. Simulation evidence confirms that estimates of the probability of informed trade are improved when accounting for past short selling activity. The results demonstrate the information benefits of short selling transparency.

Appendices

Appendix 1

Proof of Proposition I

The decision tree of Fig. 2 gives the following conditional probabilities:

$$\Pr \{ V = \underline{V} |RS\} = \delta \left[ {\frac{{\alpha \mu h + (1 - \alpha \mu )\tfrac{1}{2}\varepsilon h}}{{\alpha \delta \mu h + (1 - \alpha \mu )\tfrac{1}{2}\varepsilon h}}} \right]$$

(9)

$$\Pr \{ V = \underline{V} |SS\} = \delta \left[ {\frac{{\alpha \mu (c_{1} + c_{2} ) + (1 - \alpha \mu )\tfrac{1}{2}\varepsilon c_{1} }}{{\alpha \delta \mu (c_{1} + c_{2} ) + (1 - \alpha \mu )\tfrac{1}{2}\varepsilon c_{1} }}} \right]$$

(10)

It can be shown that $\Pr \{ V = \underline{V} |SS\} - \Pr \{ V = \underline{V} |RS\}$ is equivalent to $\frac{{(1 - \delta )\alpha \mu \tfrac{1}{2}\varepsilon h(1 - \alpha \mu )c_{2} }}{{\left( {\alpha \delta \mu (c_{1} + c_{2} ) + (1 - \alpha \mu )\tfrac{1}{2}\varepsilon c_{1} } \right)\left( {\alpha \delta \mu h + (1 - \alpha \mu )\tfrac{1}{2}\varepsilon h} \right)}}$ which is strictly greater than zero.Additionally, the following probabilities hold:

$$\Pr \{ I|RS\} = \frac{\mu \alpha \delta }{{\mu \alpha \delta + (1 - \mu )\tfrac{1}{2}\varepsilon }}$$

(11)

$$\Pr \{ I|SS\} = \frac{{\mu \alpha \delta (c_{1} + c_{2} )}}{{\mu \alpha \delta (c_{1} + c_{2} ) + (1 - \mu )\tfrac{1}{2}\varepsilon c_{1} }}$$

(12)

Subtracting (11) from (12) gives $\frac{{\alpha \delta \mu (1 - \mu )\tfrac{1}{2}\varepsilon c_{2} }}{{\left( {\alpha \delta \mu (c_{1} + c_{2} ) + (1 - \mu )\tfrac{1}{2}\varepsilon c_{1} } \right)\left( {\alpha \delta \mu + (1 - \mu )\tfrac{1}{2}\varepsilon } \right)}}$ which is also strictly greater than zero. $\square$

Proof of Corollary 1

$$\Pr \{ V = \underline{V} |S\} = \delta \left[ {\frac{{\alpha \mu [h + (1 - h)(c_{1} + c_{2} )] + (1 - \alpha \mu )\tfrac{1}{2}\varepsilon [h + (1 - h)c_{1} ]}}{{\alpha \delta \mu [h + (1 - h)(c_{1} + c_{2} )] + (1 - \alpha \mu )\tfrac{1}{2}\varepsilon [h + (1 - h)c_{1} ]}}} \right]$$

(13)

Computation shows that (10) > (13) > (9), which proves the relation. $\square$

Proof of Proposition 2

The conditional bid quotes at t + 1 are given by

$$b_{t + 1}^{SST} = \frac{{\varPhi_{L}^{*} [\mu (1 - h)(c_{1} + c_{2} ) + (1 - \mu )\tfrac{1}{2}\varepsilon (1 - h)c_{1} ] \cdot \underline{V} + \varPhi_{H}^{*} [(1 - \mu )\tfrac{1}{2}\varepsilon (1 - h)c_{1} ] \cdot \overline{V} + \varPhi_{0}^{*} [\tfrac{1}{2}\varepsilon (1 - h)c_{1} ] \cdot V^{*} }}{{\varPhi_{L}^{*} [\mu (1 - h)(c_{1} + c_{2} ) + (1 - \mu )\tfrac{1}{2}\varepsilon (1 - h)c_{1} ] + \varPhi_{H}^{*} [(1 - \mu )\tfrac{1}{2}\varepsilon (1 - h)c_{1} ] + \varPhi_{0}^{*} [\tfrac{1}{2}\varepsilon (1 - h)c_{1} ]}}$$

(14)

$$b_{t + 1}^{RS} = \frac{{\varPhi_{L}^{*} [\mu h + (1 - \mu )\tfrac{1}{2}\varepsilon h] \cdot \underline{V} + \varPhi_{H}^{*} [(1 - \mu )\tfrac{1}{2}\varepsilon h] \cdot \overline{V} + \varPhi_{0}^{*} [\tfrac{1}{2}\varepsilon h] \cdot V^{*} }}{{\varPhi_{L}^{*} [\mu h + (1 - \mu )\tfrac{1}{2}\varepsilon h] + \varPhi_{H}^{*} [(1 - \mu )\tfrac{1}{2}\varepsilon h] + \varPhi_{0}^{*} [\tfrac{1}{2}\varepsilon h]}}$$

(15)

where²¹

$$\begin{aligned} \varPhi_{L}^{*} & = \Pr \left\{ {\varPsi = L} \right\}\Pr \left\{ {Q^{t} |\varPsi = L} \right\} \\ & = \alpha \delta \left[ {(1 - \mu )\tfrac{1}{2}\varepsilon } \right]^{{B^{t} }} \left[ {\mu h + (1 - \mu )\tfrac{1}{2}\varepsilon h} \right]^{{RS^{t} }} \left[ {\mu (1 - h)(c_{1} + c_{2} ) + (1 - \mu )\tfrac{1}{2}\varepsilon (1 - h)c_{1} } \right]^{{SS^{t} }} \\ & \quad \cdot \,\left[ {\mu (1 - h)c_{3} + (1 - \mu )[(1 - \varepsilon ) + \tfrac{1}{2}\varepsilon (1 - h)(c_{2} + c_{3} )]} \right]^{{N^{t} }} \\ \end{aligned}$$

$$\begin{aligned} \varPhi_{H}^{*} & = \Pr \left\{ {\varPsi = H} \right\}\Pr \left\{ {Q^{t} |\varPsi = H} \right\} \\ & = \alpha (1 - \delta )\left[ {\mu + (1 - \mu )\tfrac{1}{2}\varepsilon } \right]^{{B^{t} }} \left[ {(1 - \mu )\tfrac{1}{2}\varepsilon h} \right]^{{RS^{t} }} \left[ {(1 - \mu )\tfrac{1}{2}\varepsilon (1 - h)c_{1} } \right]^{{SS^{t} }} \left[ {(1 - \mu )[(1 - \varepsilon ) + \tfrac{1}{2}\varepsilon (1 - h)(c_{2} + c_{3} )]} \right]^{{N^{t} }} \\ \end{aligned}$$

$$\begin{aligned} \varPhi_{0}^{*} & = \Pr \left\{ {\varPsi = 0} \right\}\Pr \left\{ {Q^{t} |\varPsi = 0} \right\} \\ & = (1 - \alpha )\left[ {\tfrac{1}{2}\varepsilon } \right]^{{B^{t} }} \left[ {\tfrac{1}{2}\varepsilon h} \right]^{{RS^{t} }} \left[ {\tfrac{1}{2}\varepsilon (1 - h)c_{1} } \right]^{{SS^{t} }} \left[ {(1 - \varepsilon ) + \tfrac{1}{2}\varepsilon (1 - h)(c_{2} + c_{3} )} \right]^{{N^{t} }} \\ \end{aligned}$$

Simplification shows that $b_{t + 1}^{SS} - b_{t + 1}^{RS}$ is equal to

$$\frac{{\varPhi_{L}^{*} \left[ {\tfrac{1}{2}\varepsilon \mu (1 - h)hc_{2} } \right]\left[ {\varPhi_{0}^{*} \left( {\underline{V} - V^{*} } \right) + \varPhi_{H}^{*} (1 - \mu )\left( {\underline{V} - \overline{V} } \right)} \right]}}{{\left\{ {\tfrac{1}{2}\varepsilon (1 - h)c_{1} \left[ {\varPhi_{0}^{*} + \varPhi_{H}^{*} (1 - \mu ) + \varPhi_{L}^{*} (1 - \mu )} \right] + \varPhi_{L}^{*} \mu (1 - h)(c_{1} + c_{2} )} \right\}\left\{ {\tfrac{1}{2}\varepsilon h\left[ {\varPhi_{0}^{*} + \varPhi_{H}^{*} (1 - \mu ) + \varPhi_{L}^{*} (1 - \mu )} \right] + \varPhi_{L}^{*} \mu h} \right\}}}$$

(16)

Since $\overline{V} > V^{*} > \underline{V}$, the numerator is negative and the denominator is positive, so the whole expression is strictly less than zero. $\square$

Proof of Corollary 2

Using the expression for $b_{t + 1}^{*}$ and a similar simplification method as above, it is straightforward to prove that

$$b_{t + 1}^{SS} < b_{t + 1}^{*} < b_{t + 1}^{RS}.$$

$\square$

Appendix 2

Conditional Trade Probabilities by State

Low Signal State ($\varPsi = L$)

$$\begin{aligned} \Pr (B|\varPsi = L) & = (1 - \mu )\tfrac{1}{2}\varepsilon \\ \Pr (RS|\varPsi = L) & = \mu h + (1 - \mu )\tfrac{1}{2}\varepsilon h \\ \Pr (SS|\varPsi = L) & = \mu (1 - h)(c_{1} + c_{2} ) + (1 - \mu )\tfrac{1}{2}\varepsilon (1 - h)c_{1} \\ \Pr (N|\varPsi = L) & = \mu (1 - h)c_{3} + (1 - \mu )[(1 - \varepsilon ) + \tfrac{1}{2}\varepsilon (1 - h)(c_{2} + c_{3} )] \\ \end{aligned}$$

High Signal State ($\varPsi = H$)

$$\begin{aligned} \Pr (B|\varPsi = H) & = \mu + (1 - \mu )\tfrac{1}{2}\varepsilon \\ \Pr (RS|\varPsi = H) & = (1 - \mu )\tfrac{1}{2}\varepsilon h \\ \Pr (SS|\varPsi = H) & = (1 - \mu )\tfrac{1}{2}\varepsilon (1 - h)c_{1} \\ \Pr (N|\varPsi = H) & = (1 - \mu )[(1 - \varepsilon ) + \tfrac{1}{2}\varepsilon (1 - h)(c_{2} + c_{3} )] \\ \end{aligned}$$

Zero Signal State ($\varPsi = 0$)

$$\begin{aligned} \Pr (B|\varPsi = 0) & = \tfrac{1}{2}\varepsilon \\ \Pr (RS|\varPsi = 0) & = \tfrac{1}{2}\varepsilon h \\ \Pr (SS|\varPsi = 0) & = \tfrac{1}{2}\varepsilon (1 - h)c_{1} \\ \Pr (N|\varPsi = 0) & = (1 - \varepsilon ) + \tfrac{1}{2}\varepsilon (1 - h)(c_{2} + c_{3} ) \\ \end{aligned}$$