Level 5 — MasteryMarket Microstructure

Market Microstructure

⏱ 90 minutes60 marksprintable — key stays hidden on paper

Chapter: 6.3 Market Microstructure Level: 5 — Mastery (cross-domain: probability/math + coding + proof/derivation) Time limit: 90 minutes Total marks: 60

Instructions. Answer all three questions. Show all derivations. Where code is requested, write clean pseudocode or Python. Use $...$ notation for mathematics. State every assumption.

Question 1 — Price Impact & Limit Order Book Depth (20 marks)

A limit order book (LOB) for a stock has a discrete price grid with tick size $\delta = \$ 0.01 $. On the **ask side**, the resting quantity available at the$ k$-th price level above the best ask is

q_k = Q_0\, e^{-\lambda k}, \qquad k = 0,1,2,\dots

where the best ask is at price $a_0 = \$ 100.00 $, level$ k $sits at price$ a_k = a_0 + k\delta $,$ Q_0 = 5000 $shares, and$ \lambda = 0.20$.

A market buy order of size $M$ shares "walks the book," consuming levels from $k=0$ upward.

(a) Derive a closed-form expression for the cumulative depth $D(n) = \sum_{k=0}^{n} q_k$ available through level $n$ , and hence the minimum number of levels $n^\*$ that must be swept to fill an order of $M = 12000$ shares. (6)

(b) Define the volume-weighted average execution price (VWAP) of the marketable buy order of size $M$ as

P_{\text{exec}}(M) = \frac{1}{M}\sum_{k=0}^{n^\*} a_k\, x_k,

where $x_k$ is the quantity taken from level $k$ (with $x_{n^\*}$ possibly partial). Derive the price impact $I(M) = P_{\text{exec}}(M) - a_0$ as an explicit function, and evaluate it numerically for $M=12000$ . (8)

(c) In the continuous-depth limit ( $\delta \to 0$ , density $\rho(p) = (Q_0/\delta)\,e^{-\lambda (p-a_0)/\delta}$ shares per dollar), show that price impact grows approximately as $I(M)\propto \ln(\cdot)$ or a power law of $M$ for small $M$ , and identify the regime. Comment on what this implies about market depth as a liquidity measure. (6)

Question 2 — Bid–Ask Spread Decomposition & Adverse Selection (22 marks)

Adopt a Glosten–Milgrom-style single-trade model. A market maker (MM) quotes a bid $B$ and ask $A$ around an unknown true value $V$ . With probability $\pi$ the incoming trader is informed (knows $V$ ), and with probability $1-\pi$ is a noise trader who buys or sells with equal probability $\tfrac12$ . The prior distribution of $V$ is: $V = V_H$ (high) or $V = V_L$ (low), each with prior probability $\tfrac12$ . Let $\mu = \tfrac12(V_H+V_L)$ and $\Delta = V_H - V_L > 0$ .

(a) Using Bayes' theorem, show that the competitive (zero-expected-profit) ask price equals the expected value of $V$ conditional on a buy order arriving:

A = \mathbb{E}[V \mid \text{buy}].

Derive $A$ explicitly in terms of $\pi$ , $\mu$ , and $\Delta$ . (8)

(b) By symmetry derive the bid $B = \mathbb{E}[V\mid\text{sell}]$ , and hence show the adverse-selection component of the spread is

S_{\text{adv}} = A - B = \frac{2\pi\,\Delta}{1+\pi}\cdot\frac{1}{2}\cdot 2 \;\; \text{(simplify to closed form).}

Evaluate $S_{\text{adv}}$ for $\pi = 0.30$ , $\Delta = \$ 2.00 $. Prove that$ \partial S_{\text{adv}}/\partial \pi > 0$ and interpret. (8)

(c) Real spreads also contain an order-processing component $c$ (per share, fixed) and an inventory component. Suppose the MM adds a symmetric processing charge $c=\$ 0.02 $. Write the observed half-spread and full spread. Then design a **short algorithm** (pseudocode) that estimates$ \pi $from a stream of trades and mid-price revisions, using the fact that the post-trade midpoint revision after a buy equals$ A - \mu_{\text{prior}}$. State clearly the estimator and one source of bias. (6)

Question 3 — Iceberg Orders, Latency & a Detection Simulation (18 marks)

A trader posts an iceberg (hidden) order at the best bid: total size $T = 100{,}000$ shares, displayed peak $v = 2{,}000$ shares. Each time the displayed peak is fully executed, a new peak of $v$ shares is automatically reposted at the same price with a fresh time-priority timestamp; hidden reserve replenishes until exhausted.

(a) A competing latency-arbitrage bot observes each fill at the bid. Model detection: the bot infers a hidden order is present if it observes $r$ consecutive immediate re-displays of size exactly $v$ at the same price with no price move. If any single "innocent" (non-iceberg) reload occurs with probability $p=0.15$ per event independently, compute the probability of a false positive for a detection threshold of $r=4$ , and choose the smallest $r$ giving false-positive probability below $1\%$ . (6)

(b) The bot sits in a co-location rack with round-trip latency $\ell_{\text{co}} = 40\,\mu s$ ; a remote competitor has $\ell_{\text{rem}} = 900\,\mu s$ . A new displayed peak lasts on average $\tau = 500\,\mu s$ before being hit. Explain quantitatively which participant can reliably react within the peak's lifetime, and compute the latency advantage margin (as a fraction of $\tau$ ) enjoyed by the co-located bot. (4)

(c) Write a simulation (Python-style pseudocode) that, given the fill stream, (i) reconstructs how many iceberg reloads occurred, (ii) estimates total hidden size $T$ , and (iii) outputs the fraction of $T$ that was never displayed. Then compute that hidden fraction for the given numbers and comment on the microstructure trade-off between information leakage and execution speed for iceberg orders. (8)

End of paper.

Answer keyMark scheme & solutions

Question 1

(a) Cumulative depth (6)

Geometric series:

D(n)=\sum_{k=0}^{n}Q_0 e^{-\lambda k}=Q_0\frac{1-e^{-\lambda(n+1)}}{1-e^{-\lambda}}. \quad(3)

Total book depth as $n\to\infty$ : $D(\infty)=Q_0/(1-e^{-\lambda})$ . With $Q_0=5000,\ \lambda=0.2$ : $1-e^{-0.2}=0.181269$ , so $D(\infty)=27{,}584$ shares (>12000, so fillable).

Solve $D(n)\ge M=12000$ : $1-e^{-0.2(n+1)}\ge M(1-e^{-\lambda})/Q_0 = 12000(0.181269)/5000=0.435046$ . $e^{-0.2(n+1)}\le 0.564954 \Rightarrow -0.2(n+1)\le\ln0.564954=-0.571015$ $\Rightarrow n+1\ge 2.855 \Rightarrow n\ge1.855 \Rightarrow n^\*=2.$ (3) (Levels $k=0,1,2$ are swept, with $k=2$ partial.)

(b) Price impact / VWAP (8)

Full levels $k=0,1$ give $D(1)=Q_0(1+e^{-0.2})=5000(1.818731)=9093.65$ shares (round: use exact). Cumulative:

$q_0=5000$ , $q_1=5000e^{-0.2}=4093.65$ , $D(1)=9093.65$ .
Remaining to fill at $k=2$ : $x_2 = M - D(1)=12000-9093.65=2906.35$ shares. (Full $q_2=5000e^{-0.4}=3351.60$ , so partial — consistent.) (2)

Cost:

\text{Cost}=\sum a_k x_k = a_0 x_0 + a_1 x_1 + a_2 x_2,

with $a_k=100+0.01k$ . $=100(5000)+100.01(4093.65)+100.02(2906.35)$ $=500000+409406.9+290693.1$ ... compute precisely:

$100.01\times4093.65=409406.94$
$100.02\times2906.35=290693.13$
Sum $=1{,}200{,}100.07$ . (2)

$P_{\text{exec}}=1{,}200{,}100.07/12000=100.008339$ .

I(M)=P_{\text{exec}}-a_0=\$0.008339\approx0.83\text{ cents}. \quad(2)

Closed form: $I(M)=\delta\cdot\dfrac{\sum_k k\,x_k}{M}$ = tick times the volume-weighted average level index. Here weighted level $=I/\delta=0.8339$ . (2)

(c) Continuous limit & interpretation (6)

Continuous depth density with exponential decay: to fill $M$ shares, sweep price up to $p^\*$ where

M=\int_{a_0}^{p^\*}\rho(p)\,dp=\frac{Q_0}{\lambda\delta}\Big(1-e^{-\lambda(p^\*-a_0)/\delta}\Big).

Let $u=(p^\*-a_0)$ . Then $e^{-\lambda u/\delta}=1-\frac{\lambda\delta M}{Q_0}$ , so

p^\*-a_0=-\frac{\delta}{\lambda}\ln\!\Big(1-\frac{\lambda\delta M}{Q_0}\Big). \quad(2)

For small $M$ (i.e. $\lambda\delta M/Q_0\ll1$ ): expand $-\ln(1-\epsilon)\approx\epsilon+\epsilon^2/2$ , giving

p^\*-a_0\approx \frac{\delta M}{Q_0}\Big(1+\tfrac12\frac{\lambda\delta M}{Q_0}\Big),

so the marginal impact is linear then convex; the average impact $I(M)=\tfrac12(p^\*-a_0)$ to leading order $\approx \delta M/(2Q_0)$ , i.e. linear in $M$ for small orders (linear-impact regime), becoming logarithmically divergent as $M\to Q_0/(\lambda\delta)$ (book near exhaustion). (2)

Interpretation: market depth $\approx Q_0/\lambda$ (or the coefficient $Q_0/(\lambda\delta)$ ) is the reciprocal of the linear price-impact slope. A deeper book (larger $Q_0$ , smaller $\lambda$ ) yields smaller impact per share — depth and impact are inversely related; depth is a valid resiliency/liquidity measure only in the small-order linear regime, and breaks down near exhaustion. (2)

Question 2

(a) Competitive ask (8)

A "buy" is a marketable order lifting the ask. Likelihoods:

If $V=V_H$ : informed traders (prob $\pi$ ) buy for sure; noise (prob $1-\pi$ ) buy w.p. $\tfrac12$ . $P(\text{buy}\mid V_H)=\pi\cdot1+(1-\pi)\cdot\tfrac12=\tfrac{1+\pi}{2}.$
If $V=V_L$ : informed sell (never buy); noise buy w.p. $\tfrac12$ . $P(\text{buy}\mid V_L)=\pi\cdot0+(1-\pi)\cdot\tfrac12=\tfrac{1-\pi}{2}.$ (3)

Posterior after a buy (equal priors $\tfrac12$ ):

P(V_H\mid\text{buy})=\frac{\tfrac12\cdot\frac{1+\pi}{2}}{\tfrac12\cdot\frac{1+\pi}{2}+\tfrac12\cdot\frac{1-\pi}{2}}=\frac{1+\pi}{2}.

(Denominator $=\tfrac12$ .) (2) Zero-profit competitive ask:

A=\mathbb E[V\mid\text{buy}]=\frac{1+\pi}{2}V_H+\frac{1-\pi}{2}V_L=\mu+\frac{\pi}{2}\Delta. \quad(3)

(b) Bid, spread, monotonicity (8)

By symmetry $P(V_H\mid\text{sell})=\frac{1-\pi}{2}$ , so

B=\mathbb E[V\mid\text{sell}]=\frac{1-\pi}{2}V_H+\frac{1+\pi}{2}V_L=\mu-\frac{\pi}{2}\Delta. \quad(3)

Adverse-selection spread:

S_{\text{adv}}=A-B=\pi\Delta. \quad(2)

(The messy form in the paper simplifies to this clean result.) Numeric: $\pi=0.30,\Delta=2 \Rightarrow S_{\text{adv}}=0.30\times2=\$ 0.60. $**(1)**$ \partial S_{\text{adv}}/\partial\pi=\Delta=2>0$: strictly increasing. Interpretation: more informed flow ⇒ MM faces greater adverse selection ⇒ must widen the spread to break even; spread is pure compensation for expected losses to better-informed counterparties. (2)

(c) Full spread + estimator (6)

Observed quotes add symmetric processing charge $c$ : $A_{\text{obs}}=\mu+\tfrac{\pi\Delta}{2}+c$ , $B_{\text{obs}}=\mu-\tfrac{\pi\Delta}{2}-c$ . Half-spread $=\tfrac{\pi\Delta}{2}+c$ ; full spread $=\pi\Delta+2c=0.60+0.04=\$ 0.64.$ (2)

Estimator idea: post-trade midpoint revision after a buy is $A-\mu=\tfrac{\pi\Delta}{2}$ ; after a sell $\mu-B=\tfrac{\pi\Delta}{2}$ . So average absolute mid-revision $\hat r=\widehat{\tfrac{\pi\Delta}{2}}$ , giving $\hat\pi=2\hat r/\Delta$ (with $\Delta$ estimated from realized value moves or long-run price variance). (3)

def estimate_pi(trades, Delta):
    # trades: list of (sign, mid_before, mid_after), sign=+1 buy,-1 sell
    revisions = [sign*(mid_after - mid_before)
                 for (sign, mid_before, mid_after) in trades]
    r_hat = mean(revisions)          # = pi*Delta/2 in expectation
    return 2*r_hat/Delta

One source of bias: inventory-driven quote skew and order-processing cost $c$ contaminate the mid-revision (transitory bounce), so naive $\hat r$ overstates the permanent adverse-selection component unless one uses long-horizon (permanent) price impact rather than the immediate next mid. (1)

Question 3

(a) False positive & threshold (6)

Independent innocent reloads, each prob $p=0.15$ . False positive at threshold $r$ = probability of $r$ consecutive innocent reloads $=p^r$ . (2) For $r=4$ : $p^4=0.15^4=0.00050625\approx0.0506\%$ . Already $<1\%$ . (2) Smallest $r$ with $p^r<0.01$ : $r\ln p<\ln0.01\Rightarrow r>\ln0.01/\ln0.15=4.605/1.897=2.427\Rightarrow r=3$ . Check $0.15^3=0.003375<0.01$ ✓; $0.15^2=0.0225>0.01$ . So smallest $r=3$ . (Given $r=4$ has FP $\approx0.05\%$ .) (2)

(b) Latency (4)

Peak lifetime $\tau=500\,\mu s$ .

Co-located bot: $\ell_{\text{co}}=40