Market Microstructure

Question 1 (14)

(a) Mid = $(100.00+100.05)/2 = 100.025$. Spread = $100.05-100.00 = 0.05 = 5$ cents. (2) — mid is midpoint of best bid/ask; spread is best ask − best bid.

(b) Buy 1,500 walks the ask side:

• 300 @ 100.05 = 30,015
• 600 @ 100.10 = 60,060
• remaining 600 @ 100.15 = 60,090

Total cost = $30015+60060+60090 = 150{,}165$ for 1,500 shares. VWAP $= 150165/1500 = 100.11$. (3) Price impact vs mid = $100.11 - 100.025 = 0.085 = 8.5$ cents. (2)

(c) 900-share level at 100.15 partially consumed (600 of 900 taken), leaving 300 @ 100.15. New best ask = $100.15$. Best bid unchanged $100.00$. New mid $=(100.00+100.15)/2 = 100.075$. New spread $=0.15 = 15$ cents. (3) — award marks for ask, mid, spread.

(d) A 1,500 sell walks the bid side: 500@100.00, 800@99.95, 200@99.90 → VWAP $=(50000+79960+19980)/1500 = 149940/1500 = 99.96$, impact $=100.025-99.96=0.065=6.5$ cents. Different from buy impact (8.5c) because depth is asymmetric. (2) Top-3 bid depth $=500+800+1200=2500$; top-3 ask depth $=300+600+900=1800$. Bid-heavy (more resting buy volume), so sells absorb into deeper liquidity → smaller impact. (2)

Question 2 (12)

(a) Post-trade mid reversion/continuation measures the permanent (information) price move a trade causes: informed buys push the "true" value up. The average one-minute post-trade mid move of $2.4$ cents in the trade direction = permanent impact = adverse-selection half-component = 2.4 cents. (4) — must link permanent price move to information content.

(b) Half-spread = order-processing + inventory + adverse-selection. $4.0 = 1.0 + \text{Inv} + 2.4 \Rightarrow \text{Inv} = 0.6$ cents. (2)

(c) More informed flow → adverse-selection ↑ (bigger permanent moves); order-processing ≈ unchanged (technology/cost driven); inventory ↑ slightly (harder to offload against informed flow / more risk). Net: quoted spread widens. (4) — 1 mark each component + net.

(d) If the MM quotes a spread below $2\times$ adverse-selection, then on average the permanent price move against them exceeds the round-trip spread earned; informed traders systematically pick them off, giving negative expected profit. Competition drives spread down only to the point where processing + inventory profit covers adverse-selection loss; the floor is set by adverse selection. (2)

Question 3 (12)

(a) Minimum spread now $=0.01$ (one tick), down from $0.05$. Likely changes: (i) narrower quoted spreads / more price levels populated; (ii) reduced displayed depth at each price level (liquidity fragmented across many ticks), and more frequent quote flickering / smaller size at best. (4) — 2 for min spread, 2 for any two depth effects.

(b) 10,000 / 500 = 20 slices; each slice ~3 s to be consumed → $20 \times 3 = 60$ s ≈ 60 seconds minimum. Detection risk: repeated identical 500-share refills at the same price level reveal a hidden iceberg (pattern/refill signature), allowing others to front-run. (4) — 2 for 60 s, 2 for detection.

(c) Under strict round-lot regime the odd 50 shares fall below the 100-share lot and cannot be routed as a standard lot; the order may be rejected/truncated to 1,000 (nearest round lot) or routed to a separate odd-lot facility. Exchanges provide odd-lot handling to serve retail/fractional interest and improve accessibility & liquidity for small trades. (2)

(d) With a smaller tick, price-improvement costs only 1 cent, so a competitor can jump the queue by posting one tick better very cheaply, eroding the value of time priority earned by a large resting order. Large passive providers lose queue position frequently, reducing their fill probability. (2)

Question 4 (12)

(a) Matched volume at price $P$ = $\min(\text{cumulative buy demand} \ge P,\ \text{cumulative sell supply} \le P)$.

• $P=98$: buy(≥98)=1300, sell(≤98)=300 → matched $=300$
• $P=99$: buy(≥99)=900, sell(≤99)=700 → matched $=700$
• $P=100$: buy(≥100)=500, sell(≤100)=1000 → matched $=500$
• $P=101$: buy(≥101)=200, sell(≤101)=1400 → matched $=200$

Max matched volume = 700 at $P=99$. (6) — 1 per correct line, 2 for identifying max.

(b) Matched volume = 700. At $P=99$: demand 900, supply 700 → imbalance = $900-700 = 200$ shares excess buy demand (unfilled buy). (3)

(c) A call auction aggregates all orders to a single clearing price, giving a robust price-discovery point from consolidated liquidity and reducing the ability of a single fast/informed trader to pick off stale quotes (batching removes continuous-time speed advantage), lowering adverse selection at the open. (3)

Question 5 (10)

(a) Opportunity lasts $150\,\mu s$. Firm B (40 μs < 150) captures it; Firm A (250 μs > 150) arrives after it disappears and captures nothing. (3)

(b) Dark pool: orders are hidden (no pre-trade display), so a 1,000,000-share sell does not signal intent to the lit market → less information leakage and less price impact / walking the book. Disadvantage: uncertain/low fill probability (execution not guaranteed) and possible interaction with predatory flow / lack of pre-trade transparency. (4) — 2 for impact+leakage, 2 for disadvantage.

(c) Latency-advantaged traders can update/cancel and pick off a slow provider's stale quote right after news, so the slow provider disproportionately trades when the price is about to move against them — this is exactly adverse selection, widening the spread they must quote to survive. (3)

[
  {"claim":"Q1b VWAP of 1500-share buy is 100.11","code":"cost=300*100.05+600*100.10+600*100.15; vwap=cost/1500; result = abs(vwap-100.11)<1e-9"},
  {"claim":"Q1b price impact is 8.5 cents","code":"cost=300*100.05+600*100.10+600*100.15; vwap=cost/1500; mid=100.025; result = abs((vwap-mid)-0.085)<1e-9"},
  {"claim":"Q1d sell VWAP is 99.96","code":"cost=500*100.00+800*99.95+200*99.90; vwap=cost/1500; result = abs(vwap-99.96)<1e-9"},
  {"claim":"Q2b inventory component is 0.6 cents","code":"inv=4.0-1.0-2.4; result = abs(inv-0.6)<1e-9"},
  {"claim":"Q4 max matched volume is 700 at price 99","code":"buy={98:1300,99:900,100:500,101:200}; sell={98:300,99:700,100:1000,101:1400}; m={p:min(buy[p],sell[p]) for p in buy}; best=max(m,key=lambda p:m[p]); result = (best==99 and m[99]==700)"}
]