Market Microstructure

Question 1 (10)

(a) (4)

• Best bid = 100.02, best ask = 100.05. (1)
• Mid-price $= (100.02 + 100.05)/2 = \mathbf{100.035}$. (1)
• Quoted spread = 100.05 - 100.02 = \mathbf{\0.03}$. (1)
• In bps: $\dfrac{0.03}{100.035}\times 10^4 \approx \mathbf{3.0\ \text{bps}}$. (1)

(b) (3)

• Cumulative depth = total resting quantity available from the best price up to a stated price. (1)
• Ask side to 100.06 inclusive: $200 + 400 = \mathbf{600\ \text{shares}}$. (2)

(c) (3)

• Fill 200 @ 100.05 (exhausts level 1), remaining 300 @ 100.06. (1)
• VWAP $= \dfrac{200(100.05) + 300(100.06)}{500} = \dfrac{20010 + 30018}{500} = \dfrac{50028}{500} = \mathbf{100.056}$. (1)
• Level 100.06 had 400; 300 consumed, 100 remain. New best ask = 100.06 (100 shares). (1)

Question 2 (10)

(a) (5)

• Order sweeps levels until filled. Let $k$ be the level where $Q$ is exhausted; for full levels $1..k-1$ take $q_i$, at level $k$ take residual $r = Q - \sum_{i<k} q_i$. (2)
• Total cash paid $C = \sum_{i=1}^{k-1} p_i q_i + p_k r$. (1)
• VWAP $= C/Q$. (1)
• Slippage (cost vs mid) $= Q\cdot(\text{VWAP} - m) = C - mQ$; per-share impact $= \text{VWAP} - m$. (1)

(b) (5)

• With continuous linear depth $p(x)=m+\lambda x$, cost to buy $Q$: $C=\int_0^Q (m+\lambda x)\,dx = mQ + \tfrac12\lambda Q^2$. (3)
• Slippage $= C - mQ = \tfrac12 \lambda Q^2$. (1)
• Average impact per share $= \dfrac{\tfrac12\lambda Q^2}{Q} = \tfrac12 \lambda Q$ (half the marginal impact $\lambda Q$). (1)

Question 3 (12)

(a) (6, 2 each)

1. Order-processing cost — fixed operational/clearing/exchange costs the dealer recovers per round-trip. (2)
2. Inventory-holding cost — dealers demand compensation for price risk of holding unwanted positions and skew quotes to manage inventory. (2)
3. Adverse-selection cost — the risk of trading against better-informed counterparties; the dealer widens the spread to offset expected losses to informed flow. (2)

(b) (6)

• Model: observed price $p_t = m_t + c\,q_t$, with efficient price $m_t$ a random walk, $q_t=\pm1$ i.i.d. equal prob, independent of $m$. (1)
• $\Delta p_t = \Delta m_t + c(q_t - q_{t-1})$. (1)
• Assume $\Delta m_t$ white noise, independent of $q$'s. Then $\text{Cov}(\Delta p_t,\Delta p_{t-1}) = c^2\,\text{Cov}(q_t-q_{t-1},\,q_{t-1}-q_{t-2}).$ Only overlapping term is $-c^2\,\text{Var}(q_{t-1}) = -c^2\cdot 1$. (2)
• So $\text{Cov}(\Delta p_t,\Delta p_{t-1}) = -c^2 \Rightarrow c = \sqrt{-\text{Cov}}$. Full spread $S = 2c = \mathbf{2\sqrt{-\text{Cov}(\Delta p_t,\Delta p_{t-1})}}$. (2)

Question 4 (10)

(a) (3)

• Some counterparties know $V$; they buy only when $V=V_H$ (asset undervalued) and sell only when $V=V_L$. (1)
• A buy order is therefore informative: it raises the conditional expected value. (1)
• Posting a single price = mid would systematically lose to informed traders, so the dealer must set ask > mid and bid < mid → positive spread, even with no processing/inventory cost. (1)

(b) (7)

• Prior $P(V_H)=P(V_L)=\tfrac12$. On a buy: informed buys iff $V_H$; uninformed buys w.p. $\tfrac12$. (1)
• $P(\text{buy}\mid V_H) = \alpha\cdot1 + (1-\alpha)\tfrac12 = \tfrac{1+\alpha}{2}$. (1)
• $P(\text{buy}\mid V_L) = \alpha\cdot0 + (1-\alpha)\tfrac12 = \tfrac{1-\alpha}{2}$. (1)
• Bayes: $P(V_H\mid \text{buy}) = \frac{\tfrac12\cdot\tfrac{1+\alpha}{2}}{\tfrac12\cdot\tfrac{1+\alpha}{2}+\tfrac12\cdot\tfrac{1-\alpha}{2}} = \frac{1+\alpha}{2}.$ (2)
• Ask $= \mathbb{E}[V\mid\text{buy}] = \tfrac{1+\alpha}{2}V_H + \tfrac{1-\alpha}{2}V_L$. (1)
• Numeric: $\alpha=0.4$: $\tfrac{1.4}{2}(101)+\tfrac{0.6}{2}(99)=0.7(101)+0.3(99)=70.7+29.7=\mathbf{100.4}$. (1)

(By symmetry bid = 99.6, spread = 0.8.)

Question 5 (10)

(a) (4)

• Iceberg = large order that displays only a small "peak" quantity; when the peak fills, a new peak is automatically refreshed from the hidden reserve. (2)
• Detection (any two): (i) Repeated refills at the same price — trades keep executing at a level whose displayed size seemingly never depletes. (ii) Post-fill instant reappearance of a same-size quote at that price. Also acceptable: pinging with small orders, timestamp regularity of refresh. (2)

(b) (6)

• Executable volume at each price = $\min(\text{cum buy}, \text{cum sell})$:
  • 50.0: $\min(900,200)=200$
  • 50.1: $\min(700,450)=450$
  • 50.2: $\min(500,500)=500$
  • 50.3: $\min(300,750)=300$ (3)
• Maximum executable volume $=500$ at price 50.2. (2)
• Rule: single-price auction chooses the price that maximizes matched volume (tie-breaks by minimum imbalance / reference price). Clearing price $=\mathbf{50.2}$, matched qty $=\mathbf{500}$. (1)

Question 6 (8)

(a) (3) Co-location places the trader's servers physically inside/adjacent to the exchange's matching engine, minimizing signal propagation and network hops → lower round-trip latency. At the top of book, order priority is time-based, so being microseconds faster wins the queue, captures fleeting price edges, and avoids being picked off on stale quotes.

(b) (3) Since \0.006 < $0.01$, the spread cannot narrow below one tick — it is **binding at$$0.01$**. The excess (would-be) competition instead expresses itself through queue length: many traders post at the same best price and compete on time priority, so speed determines fills rather than price.

(c) (2) A larger tick reduces the incentive to "step ahead" by an economically trivial amount, thickening displayed depth/queues and reducing quote-flickering (lower message/adverse-selection load) — improving displayed liquidity stability.

[
  {"claim":"Q1c VWAP for 200@100.05 + 300@100.06 over 500 = 100.056","code":"vwap=(200*100.05+300*100.06)/500; result = (vwap==100.056)"},
  {"claim":"Q1a spread in bps approx 3.0","code":"bps=(0.03/100.035)*10000; result = abs(bps-3.0)<0.05"},
  {"claim":"Q2b slippage for linear depth is (1/2)*lambda*Q^2","code":"from sympy import symbols,integrate; m,lam,x,Q=symbols('m lam x Q',positive=True); C=integrate(m+lam*x,(x,0,Q)); slip=C-m*Q; result = (slip.simplify()==(lam*Q**2/2))"},
  {"claim":"Q4b ask price with alpha=0.4,VH=101,VL=99 equals 100.4","code":"a=0.4; VH=101; VL=99; ask=(1+a)/2*VH+(1-a)/2*VL; result = (abs(ask-100.4)<1e-9)"},
  {"claim":"Q5b max matched volume is 500 at price 50.2","code":"buy=[900,700,500,300]; sell=[200,450,500,750]; vols=[min(b,s) for b,s in zip(buy,sell)]; result = (max(vols)==500 and vols.index(500)==2)"}
]