Asymmetric Market Making with Adverse Selection

-q$, a noise trader buys (prob $\frac{1}{2}$) or sells (prob $\frac{1}{2}$). Since noise trades are independent of $V$ and $E[V]=0$:

$\pi_{\text{noise}} = \frac{1-q}{2}\,a + \frac{1-q}{2}(-b) = \frac{1-q}{2}(a - b)$

*Informed trader contribution.* With probability $q$, an informed trader arrives.

• Informed buy (when $V > a$): Market maker sells, P&L $= a - V < 0$.

$\pi_{\text{inf,ask}} = q\int_a^{\infty}(a - v)\,f(v)\,dv = q\bigl[a(1 - \Phi(a/\sigma)) - \sigma\,\phi(a/\sigma)\bigr]$

where we used the identity $E[V \cdot \mathbf{1}_{V > a}] = \sigma\,\phi(a/\sigma)$ for $V \sim N(0,\sigma^2)$.

• Informed sell (when $V < b$): Market maker buys, P&L $= V - b < 0$.

$\pi_{\text{inf,bid}} = q\int_{-\infty}^{b}(v - b)\,f(v)\,dv = q\bigl[-\sigma\,\phi(b/\sigma) - b\,\Phi(b/\sigma)\bigr]$

using $E[V \cdot \mathbf{1}_{V < b}] = -\sigma\,\phi(b/\sigma)$.

Combining all terms:

$\pi(a,b) = \frac{1-q}{2}(a-b) + q\bigl[a(1-\Phi(a/\sigma)) - \sigma\phi(a/\sigma)\bigr] + q\bigl[-\sigma\phi(b/\sigma) - b\,\Phi(b/\sigma)\bigr]$

This separates as $\pi(a,b) = g(a) + h(b)$ where:

$g(a) = a\!\left[\frac{1-q}{2} + q(1 - \Phi(a/\sigma))\right] - q\sigma\,\phi(a/\sigma)$

$h(b) = -b\!\left[\frac{1-q}{2} + q\,\Phi(b/\sigma)\right] - q\sigma\,\phi(b/\sigma)$

The separability is key: the optimal ask and bid can be chosen independently.

Part 2: First-order conditions and the monopolist's problem.

Differentiate $g(a)$ using $\frac{d}{da}\Phi(a/\sigma) = \frac{1}{\sigma}\phi(a/\sigma)$ and $\frac{d}{da}\phi(a/\sigma) = -\frac{a}{\sigma^2}\phi(a/\sigma)$:

$g'(a) = \frac{1-q}{2} + q(1-\Phi(a/\sigma)) - \frac{qa}{\sigma}\phi(a/\sigma) + \frac{qa}{\sigma}\phi(a/\sigma) = \frac{1-q}{2} + q(1-\Phi(a/\sigma))$

The terms $-\frac{qa}{\sigma}\phi(a/\sigma)$ (from the product rule on $a \cdot (1-\Phi)$) and $+\frac{qa}{\sigma}\phi(a/\sigma)$ (from differentiating $-q\sigma\phi(a/\sigma)$) cancel exactly. Since $\frac{1-q}{2} > 0$ and $q(1-\Phi(a/\sigma)) > 0$ for all finite $a$, we have $g'(a) > 0$ for all $a$.

Similarly, $h'(b) = -\frac{1-q}{2} - q\,\Phi(b/\sigma) < 0$ for all $b$, so $h$ is decreasing in $b$ (making $b$ more negative increases profit).

Critical observation: With perfectly inelastic noise demand, $\pi(a,b)$ is strictly increasing in $a$ and strictly decreasing in $b$. There is no interior maximum -- a monopolist market maker would set the spread infinitely wide. This is a fundamental result: when noise traders are price-insensitive, the monopolist extracts unlimited rents.

Part 3: The competitive equilibrium (Glosten-Milgrom zero-profit condition).

In practice, Bertrand competition among market makers drives profit to zero. The competitive ask satisfies $a^{*} = E[V \mid \text{someone buys at } a^{*}]$ and the competitive bid satisfies $b^{*} = E[V \mid \text{someone sells at } b^{*}]$.

The probability someone buys at the ask is:

$P(\text{buy}) = \frac{1-q}{2} + q(1 - \Phi(a/\sigma))$

The expected value of $V$ conditional on a buy:

$E[V \mid \text{buy}] = \frac{q \cdot E[V \cdot \mathbf{1}_{V>a}] + \frac{1-q}{2} \cdot E[V]}{P(\text{buy})} = \frac{q\,\sigma\,\phi(a/\sigma)}{\frac{1-q}{2} + q(1-\Phi(a/\sigma))}$

The zero-profit condition $a = E[V \mid \text{buy}]$ gives:

$a\left[\frac{1-q}{2} + q(1-\Phi(a/\sigma))\right] = q\,\sigma\,\phi(a/\sigma)$

This is equivalent to $g(a) = 0$. Note $g(0) = -q\sigma\phi(0) < 0$ and $g(a) \to +\infty$ as $a \to \infty$ (since $g'(a) > 0$). By the intermediate value theorem and strict monotonicity, there is a unique $a^{*} > 0$ solving $g(a^{*}) = 0$.

By the symmetry of $N(0, \sigma^2)$, the bid satisfies $b^{*} = -a^{*}$, giving a symmetric spread.

Concavity in the competitive setting. To verify this is a proper equilibrium, consider the market maker's profit when deviating from the competitive quotes. If a market maker undercuts (sets $a < a^{*}$), she attracts more informed adverse flow and $g(a) < 0$: she loses money. If she sets $a > a^{*}$, she is overcut by competitors and gets no flow. Thus $a^{*}$ is the unique equilibrium ask in the competitive game. Formally, the second-order condition on $g$ evaluated at the zero-profit point confirms it is a crossing from below (since $g'(a^{*}) > 0$, $g$ crosses zero from negative to positive), ensuring uniqueness.

Numerical example. With $\sigma^2 = 2$ and $q = 0.3$, solve $g(a) = 0$ numerically. The condition is:

$a\left[0.35 + 0.3(1-\Phi(a/\sqrt{2}))\right] = 0.3\sqrt{2}\,\phi(a/\sqrt{2})$

Numerical solution gives $a^{*} \approx 0.52$, so the competitive spread is approximately $a^{*} - b^{*} \approx 1.04$, or about $0.74\sigma$.

Answer:

The expected P&L is: $\pi(a,b) = \frac{1-q}{2}(a-b) + q[a(1-\Phi(a/\sigma)) - \sigma\phi(a/\sigma)] + q[-\sigma\phi(b/\sigma) - b\,\Phi(b/\sigma)]$

This separates into $g(a) + h(b)$ with $g'(a) > 0$ and $h'(b) < 0$, so a monopolist sets an infinite spread. Under competition, the unique equilibrium is the zero-profit condition $a^{*} = E[V|\text{buy at }a^{*}]$, $b^{*} = E[V|\text{sell at }b^{*}]$, with $b^{*} = -a^{*}$ by symmetry. The zero-profit ask solves $a[\frac{1-q}{2} + q(1-\Phi(a/\sigma))] = q\sigma\phi(a/\sigma)$, which has a unique positive root.