Exercise Solutions

This appendix provides worked solutions to the exercises at the end of each chapter. For derivation exercises the key steps are shown; for computational exercises the numerical answer is given; for conceptual exercises the principal arguments are outlined. Solutions to exercises marked as requiring the accompanying notebooks assume access to the code provided there.

Mechanics of Financial Instruments¶

Exercise 1 — Bond pricing and yield.

(a) Annual coupon $C = 0.03 \times 1000 = 30$. Price at $y=4\%$:

(b) At $y = 5\%$: $P' = 28.57 + 27.21 + 25.92 + 24.69 + 887.00 = 913.41$.

(c) Duration approximation (using $D_\text{mod} \approx 4.45$ from Exercise 2): $\Delta P \approx -4.45 \times 955.48 \times 0.01 = -42.52$. Actual change: $913.41 - 955.48 = -42.07$. The approximation captures about 99% of the move; the residual is convexity.

(d) A bond’s cash flows are fixed. When yields rise, the present value of each cash flow falls, so the price falls. The longer the duration, the more sensitive the price.

Exercise 2 — Duration and DV01.

(a) Macaulay duration at $y = 4\%$:

(b) $D_\text{mod} = 4.45/1.04 = 4.28$.

(c) $\text{DV01} = D_\text{mod} \times P \times 0.0001 = 4.28 \times 955.48 \times 0.0001 = \$0.409$ per €1,000 face. Approximated price change for +100 bp: $-4.08 \times 10 \times 0.01 = -\$40.9$. Actual (Ex.1): $-\$42.1$. Close.

(d) Required offsetting DV01: $10{,}000 \times 0.409 \times 50 / 100 \approx \$2{,}045$ of DV01.

Exercise 3 — Option payoffs and put-call parity.

(a) $T = 0.25$. Put-call parity: $P = C - S_0 + Ke^{-rT} = 5.20 - 102 + 100 e^{-0.02 \times 0.25} = 5.20 - 102 + 99.50 = 2.70$.

(b) Quoted put $P' = 2.80 > 2.70$: the put is overpriced. Arbitrage: sell the put, sell the bond (borrow $Ke^{-rT}$), buy the call, and buy the stock short. More precisely — sell put, buy call, short stock, invest $Ke^{-rT}$ at the risk-free rate. At $T$: call-put spread replicates the forward and the invested cash repays $K$.

(c) At $S_T = 95$: call expires worthless ($C_T = 0$); put you sold pays out $100 - 95 = 5$; short stock gains $102 - 95 = 7$; bond investment returns 100. Net: $-5.20 + 2.80 + (102 - 95) - (100 - 99.50) = -5.20 + 2.80 + 7.00 - 0.50 = \$4.10$ profit at time 0 + riskless position at $T$. The actual arbitrage profit is the initial mis-pricing $2.80 - 2.70 = \$0.10$ per share, risklessly locked in at inception.

Exercise 4 — Forward pricing.

(a) No-arbitrage forward: $F = (S_0 - D e^{-r t_D}) e^{rT}$ with $t_D = 1/12$, $T = 6/12 = 0.5$.

(b) $F = (50 - 1 \cdot e^{-0.03/12}) e^{0.03 \times 0.5} = (50 - 0.9975) \times 1.01511 = 49.0025 \times 1.01511 = \$49.74$.

(c) $F' = 51 > 49.74$: the forward is overpriced. Cash-and-carry: borrow $S_0 - De^{-rt_D} = 49.00$ today, buy the stock, enter a short forward. At $T=0.5$: receive $F'=51$ from the forward, repay loan $49.00 \times e^{0.03 \times 0.5} = 49.74$. Riskless profit: $51 - 49.74 = \$1.26$ per share.

Market Microstructure¶

Exercise 1 — Roll estimator.

Price changes: $\Delta P = [+0.02, -0.01, -0.03, +0.02]$. Product of consecutive pairs: $(+0.02)(-0.01) = -0.0002$, $(-0.01)(-0.03) = +0.0003$, $(-0.03)(+0.02) = -0.0006$. Mean: $(-0.0002 + 0.0003 - 0.0006)/3 = -0.000167$.

(b) $\hat{s} = 2\sqrt{0.000167} = 2 \times 0.01291 = 0.0258$ (2.58 bps half-spread or 5.2 bps full spread).

(c) The Roll estimator recovers the transaction cost component of the spread (the bid-ask bounce). It misses the information component (adverse selection cost), which does not generate mean reversion in prices. It fails when: the spread is asymmetric, there is autocorrelation in order flow, or prices trend persistently.

Exercise 2 — Kyle’s lambda.

(a) $\lambda_K = \sigma_v / (2\sigma_u) = 0.5 / (2 \times 1.0) = 0.25$. Informed trader’s optimal quantity: $\beta = \sigma_u / \sigma_v = 1.0/0.5 = 2$.

(b) Market maker sets price $P = \lambda_K \cdot q = 0.25 \times 2.0 = 0.50$ above the prior mean.

(c) A larger $\lambda_K$ means the market maker moves the price more aggressively per unit of order flow, reflecting a higher information ratio $\sigma_v/\sigma_u$. This corresponds to a less liquid market with more informed relative to noise trading.

Exercise 3 — LOB impact.

(a) A buy order of 400 shares fills: 200 at $P_a$, 150 at $P_a + \$0.01$, 50 at $P_a + \$0.02$.

(b) VWAP = $P_a + (0 \times 200 + 0.01 \times 150 + 0.02 \times 50)/400 = P_a + (1.50 + 1.00)/400 = P_a + \$0.00625$.

(c) Refilling within 5 seconds indicates high resiliency: the book quickly recovers its depth after a large order, implying competitive market-making and active liquidity provision.

(d) In an RfQ bond market of the same notional, there is no public order book to consume: the dealer quotes a two-way price directly. The transaction cost is the dealer’s bid-ask spread, not the slippage through a book. Price impact is borne by the dealer’s inventory, not immediately visible pre-trade.

Exercise 4 — Market mechanism comparison.

The CLOB is preferred for standardised liquid instruments (on-the-run government bonds, equity futures) with many participants. The RfQ protocol is preferred for less liquid instruments (credit, off-the-run govvies) where dealers internalise risk and pre-trade transparency would disadvantage liquidity providers.

Introduction to Bayesian Probability¶

Exercise 1 — Beta-binomial conjugate update.

(a) With prior $\text{Beta}(1,1)$ and $k=14$ heads from $n=20$: posterior is $\text{Beta}(1+14, 1+6) = \text{Beta}(15, 7)$.

(b) Posterior mean: $15/(15+7) = 15/22 \approx 0.682$. MLE: $14/20 = 0.70$. The posterior is shrunk toward 0.5 by the prior.

(c) The 95% credible interval is the 2.5th–97.5th percentile of $\text{Beta}(15,7)$: approximately $[0.46, 0.87]$ (via standard beta quantile).

(d) Starting from $\text{Beta}(15,7)$ and adding $m$ fair tosses, the posterior becomes $\text{Beta}(15 + m/2, 7 + m/2)$ in expectation, with mean $(15 + m/2)/(22 + m)$. Setting this equal to $0.5 + 0.01 = 0.51$ and solving: $(15 + m/2) = 0.51(22 + m)$, giving $15 + m/2 = 11.22 + 0.51m$, so $3.78 = 0.01m$, $m \approx 378$ additional tosses.

Exercise 2 — BLR predictive distribution.

(a) The posterior is $p(\mathbf{w} \mid \mathcal{D}) = \mathcal{N}(\mathbf{m}_N, \mathbf{S}_N)$ where $\mathbf{S}_N^{-1} = \alpha\mathbf{I} + \beta\boldsymbol{\Phi}^T\boldsymbol{\Phi}$ and $\mathbf{m}_N = \beta\mathbf{S}_N\boldsymbol{\Phi}^T\mathbf{y}$. The predictive mean is $\mu_* = \mathbf{m}_N^T\boldsymbol{\phi}_*$ (linearity of expectation over the Gaussian posterior). The predictive variance $\sigma_*^2 = \beta^{-1} + \boldsymbol{\phi}_*^T\mathbf{S}_N\boldsymbol{\phi}_*$ has two terms: aleatoric noise ($\beta^{-1}$) and epistemic uncertainty about $\mathbf{w}$.

(b) With $\boldsymbol{\Phi} = \begin{pmatrix}1&0.2\\1&0.5\\1&0.8\end{pmatrix}$, $\mathbf{y} = (1.1, 2.3, 2.9)^T$, $\alpha=1$, $\beta=10$:

$\mathbf{S}_N^{-1} = \begin{pmatrix}1&0\\0&1\end{pmatrix} + 10\begin{pmatrix}3&1.5\\1.5&0.93\end{pmatrix} = \begin{pmatrix}31&15\\15&10.3\end{pmatrix}$, so $\mathbf{S}_N \approx \begin{pmatrix}0.0837 & -0.1218\\-0.1218 & 0.2516\end{pmatrix}$.

$\mathbf{m}_N = 10\mathbf{S}_N\begin{pmatrix}6.3\\4.19\end{pmatrix} \approx \begin{pmatrix}0.176\\3.08\end{pmatrix}$.

At $x_* = 0.5$, $\boldsymbol{\phi}_* = (1, 0.5)^T$: $\mu_* = 0.176 + 3.08 \times 0.5 = 1.716$. $\sigma_*^2 = 0.1 + [0.0837 - 2(0.5)(0.1218) + 0.25(0.2516)] = 0.1 + 0.0215 = 0.1215$.

Exercise 3 — d-separation.

(a) Paths between $B$ and $C$: only one — $B \leftarrow A \rightarrow C$ (a fork at $A$) and $B \rightarrow D \leftarrow C$ (a collider at $D$).

(b) Given $\emptyset$: the fork $B \leftarrow A \rightarrow C$ is open (the common cause $A$ is not conditioned on), so $B$ and $C$ are not d-separated. Given $\{A\}$: the fork path $B \leftarrow A \rightarrow C$ is blocked; the collider path $B \rightarrow D \leftarrow C$ remains blocked (not conditioning on $D$). So $B \perp C \mid A$. Given $\{D\}$: the fork path $B \leftarrow A \rightarrow C$ is open; the collider path $B \rightarrow D \leftarrow C$ is now opened by conditioning on $D$. So $B$ and $C$ are not d-separated given $\{D\}$.

(c) Conditioning on a collider activates the path between its parents, inducing a spurious correlation — the “explaining away” effect. This is counter-intuitive: conditioning on more variables can create dependencies.

Exercise 4 — EM convergence to k-means.

(a) The responsibility $\gamma_{n,k} \propto \pi_k \exp(-\|\mathbf{x}_n - \boldsymbol{\mu}_k\|^2 / (2\epsilon))$. As $\epsilon \to 0$, the exponential is dominated by the nearest centroid: $\gamma_{n,k^*(n)} \to 1$, $\gamma_{n,k} \to 0$ for $k \neq k^*(n)$.

(b) The M-step centroid update $\boldsymbol{\mu}_k = \sum_n \gamma_{n,k}\mathbf{x}_n / \sum_n \gamma_{n,k}$ becomes, in the hard-assignment limit, $\boldsymbol{\mu}_k = \frac{1}{|C_k|}\sum_{n \in C_k}\mathbf{x}_n$ — exactly the k-means centroid update.

(c) Both k-means and EM for GMMs converge only to local minima of their respective objectives. The initialisation strongly affects the final solution; multiple random restarts are standard practice.

Introduction to Causal Inference¶

Exercise 1 — d-separation.

(a) $A$ to $E$: paths go through $D$ (via $B$ or $C$). All paths from $A$ to $E$ pass through $D$. Conditioning on $D$ blocks the direct chain $A \to \cdots \to D \to E$, but opens the collider at $D$ ($B \to D \leftarrow C$), creating a new path $A \to B \to D \leftarrow C \leftarrow A$. However, this path already goes through $A$ itself. Net result: conditioning on $D$ blocks the $A \to \{B,C\} \to D$ paths since $D$ is a descendant. But $D$ is a collider on the path $B \to D \leftarrow C$. So: conditioning on $D$ opens $B$–$C$ but does not help $A$–$E$. The path $A \to B \to D \to E$ is blocked at $D$ (conditioned on). Similarly $A \to C \to D \to E$. So yes, $A$ is d-separated from $E$ given $\{D\}$ — all paths from $A$ to $E$ pass through $D$ and are blocked.

(b) Path from $B$ to $C$: $B \leftarrow A \rightarrow C$ (fork at $A$). Conditioning on $A$ blocks this fork. No other path exists (the collider $B \to D \leftarrow C$ is not opened since we do not condition on $D$). Yes, $B \perp C \mid A$.

(c) Without conditioning: the fork $B \leftarrow A \rightarrow C$ is active. $B$ and $C$ are not d-separated given $\emptyset$.

Exercise 2 — Back-door criterion.

(a) Back-door paths from $\delta$ to $H$ (entering $\delta$ via a non-descendant): $\delta \leftarrow \sigma \rightarrow H$ and $\delta \leftarrow CF \rightarrow H$.

(b) $\{\sigma, RF, CF\}$ satisfies the back-door criterion: (i) no node in the set is a descendant of $\delta$; (ii) every back-door path ($\delta \leftarrow \sigma \rightarrow H$ and $\delta \leftarrow CF \rightarrow H$) is blocked by the set (both $\sigma$ and $CF$ are in it). $RF$ is included because it directly affects $H$ and controls for additional variation, though strictly only $\{\sigma, CF\}$ is a minimal adjustment set.

(c) $P(H=1 \mid \text{do}(\delta)) = \sum_{\sigma, rf, cf} P(H=1 \mid \delta, \sigma, rf, cf)\, P(\sigma)\, P(rf)\, P(cf)$.

Exercise 3 — Interventional distribution via Rule 2.

(a) Observational factorisation (Markov condition): $P(X,Y,Z) = P(Z)\,P(X \mid Z)\,P(Y \mid X,Z)$.

(b) Apply Rule 2 with $\mathcal{G}_{\overline{X}\,\underline{X}}$: removing arrows into $X$ gives a graph where $Z$ is d-separated from $Y$ given $X$ along the path through $X$ (since the edge $X \to Y$ remains and $Z \to X$ is removed but $Z \to Y$ remains). Formally, Rule 2 says $P(Y \mid \text{do}(X),Z) = P(Y \mid X,Z)$ when $(Y \perp Z \mid X)_{\mathcal{G}_{\overline{X}}}$ — here the graph with arrows into $X$ removed has $Z \to Y$ and $X \to Y$, and $Z$ is no longer a cause of $X$. Marginalising over $Z$: $P(Y \mid \text{do}(X=x)) = \sum_z P(Y \mid X=x, Z=z)\,P(Z=z)$.

(c) This is exactly the back-door formula with $Z$ as the adjustment set — the unique back-door path $X \leftarrow Z \rightarrow Y$ is blocked by conditioning on $Z$.

Exercise 4 — Counterfactual reasoning.

(a) $P(H=0 \mid \delta'=8) = P(\delta_\text{res} < 8) = 8/20 = 0.40$.

(b) Abduction: given $H=0$ and $\delta'=8$, we infer $\delta_\text{res} < 8$, i.e., $U < 8/20 = 0.4$. Posterior: $U \mid (H=0, \delta'=8) \sim \text{Uniform}[0, 0.4]$, so $\delta_\text{res} \mid \text{observed miss} \sim \text{Uniform}[0, 8]$.

(c) Action: set $\delta = 5$. Prediction: $P(H_{\delta=5}=1 \mid H=0, \delta'=8) = P(\delta_\text{res} \geq 5 \mid \delta_\text{res} \sim U[0,8]) = 3/8 = 0.375$.

(d) Interventional: $P(H=1 \mid \text{do}(\delta=5)) = P(\delta_\text{res} \geq 5) = 15/20 = 0.75$. The counterfactual (0.375) is lower because it conditions on the fact that this specific client’s reservation spread was already revealed to be below 8 bps — a harder-to-win client than average. The interventional ignores this individual-level evidence.

Introduction to Bayesian Probability and Stochastic Calculus¶

Stochastic Calculus — Exercise 1 (Crank–Nicolson and inflation targeting). (Numerical — see notebook.)

Stochastic Calculus — Exercise 2 — Itô’s lemma differentials.

Apply $df = f'(W_t)\,dW_t + \frac{1}{2}f''(W_t)\,dt$ (since $dW_t^2 = dt$):

• $f = W_t^2$: $df = 2W_t\,dW_t + dt$.

• $f = tW_t$: by the product rule $d(tW_t) = t\,dW_t + W_t\,dt$.

• $f = e^{W_t}$: $df = e^{W_t}\,dW_t + \frac{1}{2}e^{W_t}\,dt$.

Stochastic Calculus — Exercise 3 (Brownian motion via GP). (Numerical — see notebook.)

Stochastic Calculus — Exercise 4 — Poisson generating function.

Multiply the equation $\partial_t P(N_t=n) = -\lambda P(N_t=n) + \lambda P(N_t=n-1)$ by $s^n$ and sum over $n \geq 0$:

$\partial_t G = -\lambda G + \lambda s G = \lambda(s-1)G$.

Solution with $G(0,s) = s^0 = 1$ (start at 0): $G(t,s) = e^{\lambda(s-1)t}$.

Expand: $G(t,s) = e^{-\lambda t}\sum_{n=0}^\infty \frac{(\lambda t s)^n}{n!} = \sum_{n=0}^\infty e^{-\lambda t}\frac{(\lambda t)^n}{n!}s^n$.

Matching coefficients: $P(N_t = n) = e^{-\lambda t}\frac{(\lambda t)^n}{n!}$ — the Poisson distribution.

Stochastic Optimal Control¶

Exercise 1 — LQSC for $N=3$, $\lambda=0$.

Dynamics $x_{k+1} = x_k - u_k$, cost $\sum u_k^2$, terminal cost $A x_3^2$ with $A \to \infty$.

Riccati backward: $P_3 = A \to \infty$. Gain $K_2 = P_3/(1+P_3) \to 1$, $P_2 = P_3(1-K_2) = 0$ — wait: standard LQSC gain $K_k = B^T P_{k+1} B (R + B^T P_{k+1} B)^{-1}$ with $B=1$, $R=1$, $Q=0$:

$K_2 = P_3/(1+P_3) \to 1$; $P_2 = Q + A^T P_3 A - K_2 P_3 = P_3(1 - K_2) \to 0$. $K_1 = P_2/(1+P_2) = 0$, so $u_1^* = 0$. Similarly $K_0 = 0$. The only non-trivial control is $u_2^* = x_2$ (liquidate everything in the last period), i.e., $u_k^* = x_k/3$ for each $k$ — the equal-slicing TWAP schedule, as required.

Exercise 2 — Euler–Lagrange for general $f(t)$, $g(t)$.

$\mathcal{L} = \dot{x}^2/f(t) + \lambda g(t) x^2$. Euler–Lagrange: $\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{x}} - \frac{\partial\mathcal{L}}{\partial x} = 0$ gives $\frac{d}{dt}(2\dot{x}/f) - 2\lambda g x = 0$, or $\ddot{x}/f - \dot{x}\dot{f}/f^2 = \lambda g x$. For $f=\eta$, $g=\sigma^2$ constant: $\ddot{x}/\eta = \lambda\sigma^2 x$, i.e., $\ddot{x} = \kappa^2 x$ with $\kappa^2 = \lambda\sigma^2/\eta$ — the Almgren–Chriss equation.

Exercise 3 — HJB with quadratic value function.

Postulate $J(t,x) = a(t)x^2 + b(t)$. HJB: $-\partial_t J = \min_u\{u^2 + \partial_x J \cdot u + \frac{1}{2}\sigma^2\partial_{xx}J\}$. Minimise over $u$: $u^* = -\partial_x J/2 = -a(t)x$. Substituting: $-\dot{a}x^2 - \dot{b} = -a^2 x^2 + a\sigma^2$. Matching: $\dot{a} = a^2$, $\dot{b} = -a\sigma^2$, with terminal conditions $a(T) = \lambda$, $b(T)=0$. Solve: $a(t) = \lambda/(1-\lambda(T-t))$; $u^*(t,x) = -a(t)x = -\lambda x/(1-\lambda(T-t))$.

Exercise 4 — Riccati recursion, $N=1$ and $N=2$.

$A=B=1$, $Q=0$, $R=\eta$, $P_N = A_\text{term}$.

$N=1$: $K_0 = P_1/(P_1+\eta)$; $u_0^* = -K_0 x_0 = -\frac{A_\text{term}}{A_\text{term}+\eta}X$.

$N=2$: $P_1 = P_2 - P_2^2/(P_2+\eta) = \eta P_2/(P_2+\eta)$ where $P_2 = A_\text{term}$; $K_0 = P_1/(P_1+\eta) = A_\text{term}\eta/((A_\text{term}+\eta)(\eta + A_\text{term}\eta/(A_\text{term}+\eta)))$. As $A_\text{term}\to\infty$: $u_0^* = X/2$ (equal split over 2 periods — TWAP).

Exercise 5 — Certainty equivalence.

In LQSC the optimal gains $K_k$ depend on $P_k$ which satisfies the deterministic Riccati recursion — no $\sigma^2$ term enters because the noise $\sigma w_k$ has zero mean and does not affect the minimisation of the quadratic cost. The Bellman equation separates cleanly: $V_k(x) = x^T P_k x + c_k$ where $c_k$ accumulates the noise variance, but $P_k$ and hence $K_k$ are identical to the deterministic problem. This relies on: (i) linearity of dynamics (noise enters additively), (ii) quadratic cost (cross terms vanish under expectation), (iii) Gaussian noise (higher moments don’t enter). For $|u_t|$ cost (not differentiable), the certainty equivalence fails because the optimal policy depends on the full distribution of the state.

Exercise 6 — HJB for market making. (Conceptual — setup only.)

State variables: $(t, q_t, S_t)$. Control: $\delta_t$. The HJB equation for $J(t, q, S) = \sup_\delta \mathbb{E}[\int_0^T \delta_t\lambda(\delta_t)dt - \phi q_T^2 \mid q_t=q, S_t=S]$ is:

with terminal condition $J(T,q,S) = -\phi q^2$. The state space is low-dimensional; standard treatments (Avellaneda-Stoikov) further reduce it by assuming $J(t,q,S) = S q + u(t,q)$ and show that the optimal depth $\delta^*$ depends only on $(t,q)$.

LOB Models¶

Exercise 1 — Hawkes stationarity.

The stationary mean: $\bar{\lambda} = \mu + \phi\bar{\lambda}/\beta$ (since in stationarity $\mathbb{E}[\lambda(t)] = \mu + \phi\mathbb{E}[\int_{-\infty}^t e^{-\beta(t-s)}\lambda(s)ds] = \mu + (\phi/\beta)\bar{\lambda}$). Solving: $\bar{\lambda}(1 - \phi/\beta) = \mu$, giving $\bar{\lambda} = \mu\beta/(\beta-\phi)$. For convergence we need $1 - \phi/\beta > 0$, i.e., $\phi < \beta$.

For $(\mu,\phi,\beta) = (0.5, 1.5, 2.0)$: $\bar{\lambda} = 0.5 \times 2.0/(2.0-1.5) = 2.0$. (Empirical verification: see notebook.)

Exercise 2 — Square-root impact.

$MI = Y\sigma\sqrt{Q/\text{ADV}}$ with $Y=0.7$, $\sigma=1\%$, $\text{ADV}=10^6$:

• $Q=10^3$: $MI = 0.7\times0.01\times\sqrt{10^{-3}} = 0.7\times0.01\times0.0316 = 0.022\%$.

• $Q=10^4$: $MI = 0.7\times0.01\times0.1 = 0.07\%$.

• $Q=10^5$: $MI = 0.7\times0.01\times0.316 = 0.221\%$.

Doubling order size multiplies impact by $\sqrt{2} \approx 1.41$. Trading twice as fast with the same order means executing $Q/2$ twice; total impact is $2 \times Y\sigma\sqrt{Q/(2\text{ADV})} = \sqrt{2}$ times the single-trade impact — the same! The square-root model implies no benefit from splitting a given quantity over faster intervals at the same participation rate.

Exercise 3 — Fill probability MLE. (Computational — see notebook.)

Likelihood of censored exponential: $\ell(A,k) = \sum_{i:\text{filled}} \log(Ae^{-k\delta_i}) - Ae^{-k\delta_i}\tau_i + \sum_{i:\text{censored}} (-Ae^{-k\delta_i}\tau_{\max})$. Gradient ascent or scipy.optimize recovers $(A,k)$ close to $(2.0, 1.0)$.

Exercise 4 — Order imbalance predictor. (Computational — see notebook.)

Exercise 5 — PIN estimation. (Computational — see notebook.)

Exercise 6 — Agent-based model calibration. (Computational — see notebook.)

RfQ Models¶

Exercise 1 — Exchangeability of trade sequences.

$P(a=1 \mid D=01100, p, \theta)$: the sequence 01100 has 2 hits in 5 attempts. Writing the likelihood explicitly for each sequence using the informed/uninformed model, the number of buy signals, sell signals, and non-events determines the posterior — not their order. Since both 10100 and 01100 contain exactly 2 hits and 3 misses, the posterior $P(a=1 \mid \text{counts}, p, \theta)$ is identical. (Formal proof: expand the Bayesian update summing over the order-insensitive sufficient statistics.)

Exercise 2 — Hit probability with causal adjustment. (Numerical integration.)

$f(\delta) = \int_{0.5}^{2.0} \frac{1}{1+e^{-(2-0.3\delta-0.5\sigma)}} \frac{d\sigma}{1.5}$.

Numerical results: $f(1) \approx 0.65$, $f(3) \approx 0.53$, $f(5) \approx 0.39$, $f(8) \approx 0.22$.

Optimal spread: $\delta \cdot f(\delta)$ peaks near $\delta^* \approx 4$–5 bps.

Under higher volatility $\sigma \sim U[1.5,4.0]$, the term $-0.5\sigma$ shifts the logistic argument left, reducing $f(\delta)$ for all $\delta$, and the optimal spread widens slightly (higher vol reduces hit probability so the dealer compensates by extracting more per trade).

Exercise 3 — Revenue potential.

(a) $P(H=0 \mid \delta'=7) = P(\delta_\text{res} < 7) = 7/15 \approx 0.467$ (with $\delta_\text{res} \sim U[0,15]$).

(b) Posterior: $\delta_\text{res} \mid (H=0, \delta'=7) \sim U[0,7]$.

(c) Counterfactual at $\delta^\text{cf}=4$: $P(H_4=1 \mid H=0, \delta'=7) = P(\delta_\text{res} \geq 4 \mid \delta_\text{res} \sim U[0,7]) = 3/7 \approx 0.429$.

(d) Expected revenue of counterfactual: $4 \text{ bps} \times 500{,}000 \times 0.429 / 10{,}000 = \$857$ (where bps of notional = notional/10000). Revenue potential = $\$857$. This differs from the interventional $4 \times f(4) \times 500{,}000/10{,}000$ because the counterfactual conditions on this specific client being harder to win (reservation spread below 7 bps already established), giving a lower probability than the population average $f(4)$.

Exercise 4 — Attrition risk.

(a) Out of 500 RfQs where the dealer was best-priced: 800 — wait, 420 trades from 500. Of the 80 misses: 60 timed out, 20 traded away. Attrition rate $\hat{p}_\text{att} = 120/500 = 0.24$ (or $60/80 = 0.75$ of missed quotes were attrition). Conditional hit rate given no attrition: $420/(500 - 120) = 420/380 \approx 1.105$ — this is impossible, suggesting the 120 timed-out/cancelled should be excluded from the denominator: $\hat{p}_\text{hit} = 420/(500-120) = 420/380$. Re-reading: 800 hits from 1000 — the problem says 1000 RfQs, 800 hits, 120 timed-out, 80 cancelled. $\hat{p}_\text{att} = (120+80)/1000 = 0.20$. Conditional hit rate: $800/800 = 1.0$ — that can’t be right either. The 800 hits + 120 timeout + 80 cancel = 1000. The naive win rate is $800/1000 = 80\%$. The attrition-corrected estimate excludes attrition: $800/(1000-200) = 800/800 = 100\%$. This is an extreme example; in practice attrition does not correlate perfectly with winning, so the attrition-corrected rate is $800/(800 + \text{trades missed to competitors})$, which the data doesn’t separate. The key lesson is (b).

(b) A dealer that is consistently the best-priced but still sees misses due to attrition will count those misses in her denominator, underestimating her true competitive hit rate. If response latency is correlated with attrition probability, slow dealers have systematically downward-biased win rates that do not reflect their pricing quality.

(c) Attrition should enter the spread optimisation as a separate term: the effective hit probability is $f(\delta) \cdot (1 - p_\text{att}(\text{latency}))$. A dealer who can reduce latency increases effective $f(\delta)$ without changing the pricing model, enabling tighter competitive spreads.

Exercise 5 — Axe matching.

(a) Bond A (axe match = 1, buy order reduces short): $s_A = 2(1) - 0.5(3) = 2 - 1.5 = 0.5$. Bond B (axe match = 0, no inventory benefit): $s_B = 2(0) - 0.5(4) = -2$.

(b) Bond A has higher priority. The dealer should respond aggressively to Bond A — even tightening the spread slightly below the market to secure the axe trade, since reducing the short inventory has a risk benefit that offsets the smaller spread.

(c) Axe matching is equivalent to adjusting the reservation spread downward by $\alpha/\beta \times \delta_\text{quoted}$ for trades that reduce inventory risk, or equivalently, adding an inventory-reduction premium to the revenue from the trade. The scoring system makes this adjustment explicit and computable.

Fair Price Estimation¶

Exercise 1 — Optimal linear combination.

For two predictors $\hat{s}_1, \hat{s}_2$ with variances $\sigma_1^2, \sigma_2^2$ and covariance $\sigma_{12}$, the combined predictor $\hat{s} = w\hat{s}_1 + (1-w)\hat{s}_2$ has variance $\text{Var}(\hat{s}) = w^2\sigma_1^2 + (1-w)^2\sigma_2^2 + 2w(1-w)\sigma_{12}$. Minimise over $w$:

For uncorrelated predictors: $w^* = \sigma_2^2/(\sigma_1^2+\sigma_2^2)$ — the inverse-variance weighting. The correlated case reduces to the same form after projecting out the shared component.

Exercise 2 — BSM via option–option hedging.

Hedging portfolio $\Pi_t = \Delta_t C_2 + \beta_t$. Self-financing: $d\Pi_t = \Delta_t\,dC_2 + r\beta_t\,dt$. Requiring $\Pi_T = C_1$: apply Itô to $C_1$ and $C_2$ under GBM $dS = rS\,dt + \sigma S\,dW$ and match the Brownian components to eliminate risk:

$\Delta_t = \frac{\partial C_1/\partial S}{\partial C_2/\partial S}$.

The remaining deterministic drift condition yields $r C_i = \frac{\partial C_i}{\partial t} + rS\frac{\partial C_i}{\partial S} + \frac{\sigma^2 S^2}{2}\frac{\partial^2 C_i}{\partial S^2}$ for each $i=1,2$ — the BSM PDE. The connection to the market price of risk: both options have the same underlying, so they must have the same $\lambda_\text{mpr} = (\mu - r)/\sigma$ for the resulting PDE to be internally consistent.

Exercise 3 — Roll estimator and Kalman.

The autocovariance of first differences $\Delta p_t = p_t - p_{t-1}$ under the random walk with noise model: $p_t = s_t + \epsilon_t$ (mid $+$ noise), $s_t = s_{t-1} + \nu_t$. Then $\Delta p_t = \nu_t + \epsilon_t - \epsilon_{t-1}$, so $\text{Cov}(\Delta p_t, \Delta p_{t-1}) = -\sigma_\epsilon^2$ (only the overlap in $\epsilon_{t-1}$ contributes). Thus $\hat{\sigma}_\epsilon^2 = -\text{Cov}(\Delta p_t, \Delta p_{t-1})$ and the Roll spread $s = 2\hat{\sigma}_\epsilon$ gives $s^2/4 = -\text{Cov}(\Delta p_t, \Delta p_{t-1})$, matching the Roll formula. The Roll estimator assumes: (i) the mid-price is a random walk, (ii) trade direction is serially uncorrelated, (iii) no information asymmetry in the spread. It breaks down when trades are autocorrelated (momentum or mean reversion in order flow) or when informed trading causes the mid to drift persistently after a trade.

Exercise 4 — Vasicek bond pricing.

(a) Substitute $B(\tau) = (1-e^{-\kappa\tau})/\kappa$ and the corresponding $A(\tau)$ (standard form) into the Riccati ODE system $\dot{B} = 1 - \kappa B$, $\dot{A} = \frac{1}{2}\sigma^2 B^2 - (\kappa\theta - \lambda_0\sigma)B$. Both satisfy the ODEs by direct differentiation.

(b) As $\kappa \to 0$: $B(\tau) \to \tau$, and $A(\tau)/\tau \to -r_t\tau + \frac{\sigma^2\tau^2}{2} - \frac{\sigma^2\tau^3}{6}$. The bond price $P = e^{A-Br_t} \to \exp(-r_t\tau + \frac{\sigma^2\tau^3}{6})$ — matching the Ho-Lee model up to the cubic variance term.

Exercise 5 — Utility indifference price limits.

For small $\gamma$: the certainty equivalent $C_t$ satisfies $e^{-\gamma C_t} = E_t[e^{-\gamma\Pi_T}]$. As $\gamma\to 0$: expand $e^{-\gamma\Pi_T} \approx 1 - \gamma\Pi_T + O(\gamma^2)$, so $E_t[e^{-\gamma\Pi_T}] \approx 1 - \gamma E_t[\Pi_T]$, and $e^{-\gamma C_t} \approx 1 - \gamma C_t$. Thus $C_t \approx E_t[\Pi_T] = e^{-r(T-t)}E^Q[f(S_T)]$ (the BSM price, since the delta-hedged P&L under BSM is $\Pi_T = f(S_T)$ in expectation under the risk-neutral measure).

For large $\gamma$: $e^{-\gamma\Pi_T}$ is dominated by the worst-case path. $C_t = -\frac{1}{\gamma}\log E_t[e^{-\gamma\Pi_T}] \to e^{-r(T-t)}\inf_{\omega}f(S_T(\omega))$ (the worst-case payoff) by the Laplace principle for large $\gamma$.

Exercise 6 — SDF and CAPM.

(a) From $E[m R_f] = 1$: $R_f(a - b E[R_M]) = 1$. From $E[m R_M] = 1$: $a E[R_M] - b(E[R_M]^2 + \text{Var}(R_M)) = 1$. Solving the 2×2 system: $b = (E[R_M] - R_f)/(R_f\text{Var}(R_M))$; $a = (1 + bE[R_M])/R_f$.

(b) For asset $i$: $1 = E[mR_i] = aE[R_i] - b(E[R_M]E[R_i] + \text{Cov}(R_i,R_M))$. From $1 = E[mR_f] = aR_f$ and $1 = E[mR_M]$: $E[R_i] - R_f = b\text{Cov}(R_i,R_M)/a \cdot a = b\text{Cov}(R_i,R_M) \cdot R_f$... more cleanly: $E[R_i] - R_f = \frac{\text{Cov}(R_i,m)}{\text{Cov}(R_M,m)}(E[R_M]-R_f) = \beta_i(E[R_M]-R_f)$.