Financial Engineering for Quants reference sheet

This sheet follows the fourteen units of the course in order. Each unit starts with the big idea in a few sentences, then gives the definitions, formulas, models and theorems you will look up most often, each with the assumptions it needs. When you want the full explanation, a derivation or more examples, follow the lesson links under each topic. The numerical examples use small constructed inputs, not market data, chosen so that every number can be checked by hand or with a few lines of code.

Time is measured in years. The current date is $t$, a maturity is $T$ and the time remaining is $\tau = T - t$. The asset price $S_t$ is ex-dividend, $K$ is a strike, $r$ an interest rate, $q$ a continuous dividend yield and $\sigma$ an annualized volatility. Rates are continuously compounded unless a topic states another compounding convention, and one basis point is $1\,\text{bp} = 0.0001$. The money-market account $B_t$ is a positive numeraire, $D(t,T)$ is a deterministic discount factor, and $P(t,T)$ is the time $t$ price of a zero-coupon bond paying 1 at $T$, which is random once interest rates are random. Prices, cash flows and profit and loss (PnL) are in currency units and are written as plain numbers.

The physical measure $\mathbb{P}$ describes how prices actually move and is the one used for forecasts and risk. A pricing measure $\mathbb{Q}$ is equivalent to $\mathbb{P}$, makes discounted traded prices martingales and need not be unique, and $\mathbb{Q}^{T}$ is the $T$-forward measure, whose numeraire is the bond maturing at $T$. Expectations name the measure and the information they condition on, as in $\mathbb{E}^{\mathbb{Q}}[X \mid \mathcal{F}_t]$, and a Brownian motion under $\mathbb{Q}$ is written $W^{\mathbb{Q}}_t$. The functions $\Phi$ and $\varphi$ are the standard normal distribution function and density, $(x)^{+} = \max(x, 0)$ is the positive part and $\mathbf{1}_{A}$ is the indicator of an event $A$. Vectors are plain letters with transpose $^{\top}$, $\mathbf{1}$ is also the vector of ones, and a hat marks an estimate. A loss $L$ is positive for an adverse outcome, so $L = -\text{PnL}$ wherever a topic converts one into the other.

An equals sign means exact under the stated hypotheses, and $\approx$ marks a Taylor truncation, an asymptotic formula, a grid result or a statistical estimate. Every model statement on this page is a mathematical result under its assumptions, not a claim about real markets, and nothing here is investment advice.

Mathematical financial models

A contract is a schedule of dated cash flows that depend on the state of the world, and its price is a separate number assigned to that schedule. The absence of arbitrage ties the prices of different contracts together, while discount factors and returns record the time value of money and the growth of wealth under stated conventions. The unit also collects the calculus, linear algebra, optimization and probability that the later units use.

Contracts, payoffs and valuation

In a finite model the possible states are $\Omega = \{\omega_1, \ldots, \omega_m\}$, and a contract is the list of amounts $C_{t_1}(\omega), \ldots, C_{t_n}(\omega)$ it pays at fixed dates in each state $\omega$. Positive amounts are receipts of the holder and negative amounts are payments, so the counterparty holds the negative of the same list. A payoff function is a rule, such as the call payoff $X = (S_T - K)^{+}$, while a present value is one number assigned to the whole rule. Changing a term of the contract, such as the strike, gives a different object with its own value.

For terminal claims in a one-period model, the state price $\pi_j$ is the time 0 price of one currency unit paid in state $\omega_j$ and nowhere else. State prices define the valuation operator

It is linear, $\Pi(aX + bY) = a\,\Pi(X) + b\,\Pi(Y)$, and when every $\pi_j > 0$ it is strictly positive, so $X \ge 0$ with $X \ne 0$ gives $\Pi(X) > 0$. The sum of the state prices is the price of a sure unit, the one-period discount factor. With constructed state prices $0.45$ and $0.5$ and a stock that ends at $80$ or $120$, a sure unit costs $0.95$ and the stock $96$. The call with strike $100$ pays $(0, 20)$ and costs $10$, and the call with strike $110$ pays $(0, 10)$ and costs $5$.

Lessons Financial contracts as mathematical objects

Arbitrage and the law of one price

Take $n$ securities with time 0 prices $c \in \mathbb{R}^{n}$ and an $m \times n$ payoff matrix $A$ whose row $j$ lists their payoffs in state $\omega_j$, where every state has positive probability. A portfolio $\theta \in \mathbb{R}^{n}$ costs $c^{\top}\theta$ and pays $A\theta$. Trading is frictionless and any real holding is allowed, including short sales and borrowing.

• Arbitrage. A portfolio with $c^{\top}\theta \le 0$ and $A\theta \ge 0$ in every state, and with $c^{\top}\theta < 0$ or $A\theta \ne 0$. It costs nothing or less, never loses later, and gains either today or in at least one state.
• Law of one price. If $A\theta = A\phi$ then $c^{\top}\theta = c^{\top}\phi$. No arbitrage implies it, because buying the cheaper and selling the dearer of two equal payoffs is an arbitrage. The converse fails. A market with one security that pays $(1, 0)$ and costs $0$ satisfies the law of one price, yet buying that security is an arbitrage.
• Replication and linear pricing. Under the law of one price, a claim $X$ with $A\theta = X$ for some $\theta$ has the well defined price $c^{\top}\theta$, whatever replicating portfolio is used, and this price is linear on the attainable claims.
• Dominance. Under no arbitrage, attainable claims with $X \ge Y$ in every state satisfy $\Pi(X) \ge \Pi(Y)$, with strict inequality when $X \ne Y$.

A stock costs $100$ and pays $80$ or $120$, and a bond costs $100$ and pays $105$ in both states. The call payoff $(0, 20)$ is replicated by $\Delta = \frac{1}{2}$ share and $b = -\frac{8}{21}$ bond, a borrowing, at the cost $50 - \frac{800}{21} = \frac{250}{21} \approx 11.9048$. At a quoted call price of $14$, selling the call and buying the replicating portfolio receives $\frac{44}{21} \approx 2.0952$ today and leaves no obligation later. If short sales of the call are not allowed, that trade is unavailable, and a constraint of this kind turns the price equality into an inequality.

Lessons Arbitrage and the law of one price

Discount factors, compounding and forward rates

A deterministic discount function gives the time 0 value $D(0,T) > 0$ of one currency unit paid at $T$, with $D(0,0) = 1$. A fixed schedule of amounts $C_k$ paid at times $t_k$ is worth

A rate is a way of reporting the factor, so one discount factor gives different rates under different compounding conventions. For $T > 0$ and $D = D(0,T)$,

The time $T$ is a year fraction under a stated day count, such as actual/365 or actual/360, and a different day count gives a different rate for the same factor. For $T_1 < T_2$ with accrual fraction $\delta$ between them, the simple forward rate and, where $D$ is differentiable, the instantaneous forward rate are

With $D(0,1) = 0.96$ and $D(0,2) = 0.90$, the forward rate for the second year is $F(0;1,2) = \frac{1}{15} \approx 0.0667$ and the continuous spot rate is $z(2) \approx 0.0527$, while the same $D(0,2)$ reports as $0.0556$ simple, $0.0541$ annually compounded and $0.0534$ semiannually compounded. A bond paying $5$ after one year and $105$ after two is worth $5(0.96) + 105(0.90) = 99.3$. If instead $D(0,2) = 0.98$, the forward rate is about $-0.0204$, and the curve has to allow negative forward rates.

Lessons Time value of money and discount functions

Returns and self-financing portfolios

For a positive price $S_t$ and a cash dividend $\operatorname{Div}_{t+1}$ paid at $t + 1$,

Gross returns multiply over consecutive periods, $1 + R_{0,T} = \prod_{t}(1 + R_{t,t+1})$, so log returns add. This is exact algebra under a consistent reinvestment convention and needs no assumption about distributions. For fixed initial weights, a portfolio’s simple return is the weighted sum of its components’ simple returns, but its log return is not the weighted sum of their log returns. A gain of 10 percent followed by a loss of 10 percent gives the gross return $1.1 \times 0.9 = 0.99$ and the log return $\ln 0.99 \approx -0.01005$.

A strategy holds $\theta_t$ shares and $\beta_t$ units of the cash account $B_t$ after trading at $t$, chosen with the information available at $t$, so its wealth is $V_t = \theta_t S_t + \beta_t B_t$. It is self-financing when every rebalance is paid for from inside the portfolio,

Sales, borrowing (a negative $\beta_t$) and the dividend on the shares held over the last period fund the purchases, and no outside cash enters or leaves. Start with 2 shares at $100$ and 1 cash unit worth $50$. At $t = 1$ the share is worth $102$ and pays a dividend of $3$, and the cash unit is worth $51$, so the wealth is $2(102) + 51 + 2(3) = 261$. Buying a third share at $102$ without outside cash needs $\beta_1 = (261 - 306)/51 = -\frac{15}{17}$, a borrowing of $45$. If the price then rises to $110$ and the cash account grows by 2 percent, the wealth is $3(110) - 45(1.02) = 284.1$.

Lessons Returns and self-financing portfolio accounting

Gradients, Hessians and Taylor expansions

For a value function $V$ of the risk factors $x \in \mathbb{R}^{d}$, the gradient $\nabla V(x)$ lists the first-order price sensitivities, in currency units per unit of each factor, and the Hessian $H(x)$ has the entries $H_{ij} = \partial_i\partial_j V(x)$. The change along a direction $h$ has the first-order rate $\nabla V(x)^{\top}h$. If $V$ is twice continuously differentiable on an open set containing the segment from $x$ to $x + h$, then $H$ is symmetric there and

where, with $\Delta H(s) = H(x + sh) - H(x)$,

The remainder follows from the one-variable expansion of $g(s) = V(x + sh)$, since $g''(s) = h^{\top}H(x + sh)\,h$. It is $o(\lVert h \rVert^2)$ as $h \to 0$ and is exactly zero when $V$ is quadratic. For $V(x) = x_1^2 + 2x_1x_2 + 3x_2^2$ at $x = (1, 2)$ with $h = (0.1, -0.2)$, the value is $17$, the gradient is $(6, 14)$, the linear term is $-2.2$ and the quadratic term is $0.09$, which gives exactly $V(x + h) = 14.89$.

A kink breaks the hypothesis. The call payoff $(s - K)^{+}$ has slope $0$ below $K$ and $1$ above it, so it has no Taylor expansion at $s = K$, and an expansion at a nearby point fails once the move crosses $K$.

Lessons Multivariable calculus and Taylor expansions

Covariance matrices, projection and hedging

For a random vector $R$ with finite second moments and mean $\mu$, the covariance matrix $\Sigma = \mathbb{E}[(R - \mu)(R - \mu)^{\top}]$ is symmetric and positive semidefinite, because $w^{\top}\Sigma w = \operatorname{Var}(w^{\top}R) \ge 0$ for every $w$. By the spectral theorem, a real symmetric matrix has the form $\Sigma = Q\Lambda Q^{\top}$ with orthonormal eigenvectors $q_j$ as the columns of $Q$ and real eigenvalues $\lambda_j$, which are nonnegative here, and

A zero eigenvalue is a combination with zero variance, and then $\Sigma$ is singular.

The minimum-error hedge of a centered target $X$ with centered instruments $Y$ minimizes $\mathbb{E}[(X - h^{\top}Y)^2]$. Its solutions are those of the normal equations $\Sigma_Y h = \operatorname{Cov}(Y, X)$, which always have a solution, and $h^{\top}Y$ is the orthogonal projection of $X$ onto the span of $Y$. When $\Sigma_Y$ is invertible the hedge is unique, $h = \Sigma_Y^{-1}\operatorname{Cov}(Y, X)$, and the residual variance is $\operatorname{Var}(X) - \operatorname{Cov}(X, Y)\,h$. When $\Sigma_Y$ is singular, the pseudoinverse $\Sigma_Y^{+} = Q\operatorname{diag}(\lambda_j^{+})Q^{\top}$, with $\lambda_j^{+} = 1/\lambda_j$ for $\lambda_j > 0$ and $0$ otherwise, gives the solution of smallest norm.

With $\Sigma_Y = \begin{pmatrix} 4 & 2 \\ 2 & 4 \end{pmatrix}$ the eigenvalues are $6$ and $2$, and the equally weighted combination has variance $3$. For $\operatorname{Cov}(Y, X) = (3, 1)$ and $\operatorname{Var}(X) = 5$, the hedge is $h = (\frac{5}{6}, -\frac{1}{6})$ with residual variance $\frac{8}{3}$. Two identical instruments with $\Sigma_Y = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ and $\operatorname{Cov}(Y, X) = (2, 2)$ admit every $h$ with $h_1 + h_2 = 2$, and the pseudoinverse picks $h = (1, 1)$.

Lessons Linear algebra for covariance and hedging

Convex optimization and duality

Consider minimizing $f(x)$ subject to $g_i(x) \le 0$ for $i = 1, \ldots, p$ and $Ax = b$, with optimal value $p^{*}$. The Lagrangian and the dual function are

• Weak duality. For every $\lambda \ge 0$ and every feasible $x$, $q(\lambda, \nu) \le f(x)$, so the dual optimum satisfies $d^{*} \le p^{*}$. No convexity is needed.
• KKT conditions. Stationarity $\nabla f(x^{*}) + \sum_i \lambda_i^{*}\nabla g_i(x^{*}) + A^{\top}\nu^{*} = 0$, primal feasibility, $\lambda^{*} \ge 0$ and complementary slackness $\lambda_i^{*}\,g_i(x^{*}) = 0$.
• Sufficiency. If $f$ and the $g_i$ are convex and differentiable, any point satisfying the KKT conditions is a global minimizer and its multipliers solve the dual.
• Necessity. Multipliers need a constraint qualification. Slater’s condition, a point with $Ax = b$ and $g_i(x) < 0$ for every nonaffine $g_i$, gives strong duality $d^{*} = p^{*}$ for a convex problem, and KKT multipliers exist at every minimizer. Without it they can fail to exist. Minimizing $x$ subject to $x^2 \le 0$ has the minimizer $0$, where $1 + 2\lambda \cdot 0 \ne 0$ for every $\lambda$.

A constraint is binding, or active, when $g_i(x^{*}) = 0$, and a positive multiplier forces it to bind. Minimizing $\frac{1}{2}(x - 3)^2$ subject to $x - 1 \le 0$ gives the Lagrangian minimizer $x = 3 - \lambda$ and the dual function $q(\lambda) = 2\lambda - \frac{1}{2}\lambda^2$. Its maximum $q(2) = 2$ equals $f(1) = 2$, so $x^{*} = 1$ and $\lambda^{*} = 2$, with the constraint binding. With the bound $x \le 4$ instead, $x^{*} = 3$ and $\lambda^{*} = 0$.

Lessons Convex optimization and duality

Probability spaces and Bayes’ theorem

A probability space $(\Omega, \mathcal{F}, \mathbb{P})$ has a sigma algebra $\mathcal{F}$ of events, which contains $\Omega$ and is closed under complements and countable unions, and a probability measure with $\mathbb{P}(\Omega) = 1$ that adds over countably many disjoint events. Information that reveals which block of a partition $\{B_1, \ldots, B_k\}$ of $\Omega$ has occurred corresponds to the sigma algebra of all unions of blocks. For $\mathbb{P}(B) > 0$,

with $\mathbb{P}(B) = \mathbb{P}(B \mid A)\,\mathbb{P}(A) + \mathbb{P}(B \mid A^{c})\,\mathbb{P}(A^{c})$ by the law of total probability. Conditioning on an event of probability zero needs regular conditional distributions, beyond these formulas.

The base rate matters. If a default has prior probability $0.02$ and a warning signal fires with probability $0.90$ before a default and $0.10$ otherwise, then $\mathbb{P}(+) = 0.018 + 0.098 = 0.116$ and $\mathbb{P}(D \mid +) = \frac{9}{58} \approx 0.155$. With the prior $0.2$ the same signal gives about $0.692$.

Lessons Probability spaces and conditional information

Probability and financial econometrics

Prices and returns are modeled as random variables, and the questions about them are about distributions, conditional expectations and dependence. Estimation, regression and time-series models turn finite data into statements about those objects, and each statement holds only under its own conditions, such as integrability, exogeneity or stationarity. The measure here is the physical one, $\mathbb{P}$, unless a topic says otherwise.

Distributions and change of variables

The distribution function $F_X(x) = \mathbb{P}(X \le x)$ is nondecreasing and right-continuous with limits $0$ and $1$. A density $f_X$ satisfies $F_X(x) = \int_{-\infty}^{x} f_X(u)\,du$, and it exists only for an absolutely continuous law, not for laws with atoms. If $Y = g(X)$ with $g$ strictly monotone and differentiable with $g' \ne 0$ on the support of $X$, then on the image of that support

and $f_Y = 0$ elsewhere. For a one-to-one differentiable map of a random vector, the absolute value of the Jacobian determinant of the inverse map takes the place of the derivative.

If $Z \sim N(\mu, \sigma^2)$ with $\sigma > 0$ and $S = S_0e^{Z}$ with $S_0 > 0$, then $S$ is lognormal. With $u = \ln(s/S_0)$ its density for $s > 0$ is

Completing the square gives $\mathbb{E}[e^{aZ}] = e^{a\mu + a^2\sigma^2/2}$ for every real $a$, and therefore

while the median is $S_0e^{\mu}$. For $S_0 = 100$, $\mu = 0$ and $\sigma = 0.2$, the median is $100$, the mean is about $102.02$ and the variance about $424.76$. Doubling $\sigma$ to $0.4$ leaves the median at $100$ and raises the mean to about $108.33$.

Lessons Random variables and distribution transformations

Expectation, conditional expectation and variance

The expectation $\mathbb{E}[X]$ exists when $X$ is integrable, $\mathbb{E}\lvert X \rvert < \infty$, and the variance $\operatorname{Var}(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2$ needs $\mathbb{E}[X^2] < \infty$. A Cauchy variable has no mean, and no sample average estimates one. For integrable $X$ and a sigma algebra $\mathcal{G} \subseteq \mathcal{F}$, the conditional expectation $Y = \mathbb{E}[X \mid \mathcal{G}]$ is the $\mathcal{G}$-measurable random variable with $\mathbb{E}[\mathbf{1}_{A}X] = \mathbb{E}[\mathbf{1}_{A}Y]$ for every $A \in \mathcal{G}$. It exists and is unique up to null sets. For a partition into blocks $B_i$ with $\mathbb{P}(B_i) > 0$, it equals $\mathbb{E}[X\mathbf{1}_{B_i}]/\mathbb{P}(B_i)$ on $B_i$.

The tower property says that if $\mathcal{H} \subseteq \mathcal{G}$, then $\mathbb{E}\big[\mathbb{E}[X \mid \mathcal{G}] \mid \mathcal{H}\big] = \mathbb{E}[X \mid \mathcal{H}]$, and in particular $\mathbb{E}\big[\mathbb{E}[X \mid \mathcal{G}]\big] = \mathbb{E}[X]$. If $\mathbb{E}[X^2] < \infty$, the law of total variance splits the variance into a within-group part and a between-group part,

Four equally likely states pay $X = (0, 2, 4, 10)$, and the information reveals whether the state is one of the first two or one of the last two. Then $\mathbb{E}[X] = 4$ and $\mathbb{E}[X \mid \mathcal{G}] = (1, 1, 7, 7)$. The conditional variances are $1$ and $9$, so the within-group part is $5$, the between-group part is $9$ and $\operatorname{Var}(X) = 14$.

Lessons Expectation conditional expectation and variance

Joint distributions, correlation and copulas

For $X$ and $Y$ with finite second moments, $\operatorname{Cov}(X, Y) = \mathbb{E}[XY] - \mathbb{E}[X]\,\mathbb{E}[Y]$, and when both standard deviations are positive the correlation $\rho = \operatorname{Cov}(X, Y)/(\sigma_X\sigma_Y)$ lies in $[-1, 1]$. Independence implies zero covariance, but not conversely. If $X$ is $-1$, $0$ or $1$ with probability $\frac{1}{3}$ each and $Y = X^2$, then $\operatorname{Cov}(X, Y) = \mathbb{E}[X^3] = 0$, yet $\mathbb{P}(X = 0, Y = 0) = \frac{1}{3}$ while $\mathbb{P}(X = 0)\,\mathbb{P}(Y = 0) = \frac{1}{9}$.

A random vector is jointly Gaussian when $X = \mu + LZ$ with $Z \sim N(0, I)$, and then $\Sigma = LL^{\top}$ and every combination $a^{\top}X \sim N(a^{\top}\mu, a^{\top}\Sigma a)$. For jointly Gaussian vectors, zero correlation implies independence, but two Gaussian variables that are not jointly Gaussian can be uncorrelated and dependent. A portfolio with weights $w$ and return covariance $\Sigma$ has variance $w^{\top}\Sigma w$, which for two assets is $w_1^2\sigma_1^2 + w_2^2\sigma_2^2 + 2w_1w_2\rho\,\sigma_1\sigma_2$. Equal weights on volatilities $0.2$ and $0.3$ give the variance $0.0475$ when $\rho = 0.5$ and $0.0175$ when $\rho = -0.5$.

A copula is a joint distribution function on $[0,1]^2$ with uniform marginals. By Sklar’s theorem, every joint distribution function has a copula $C$ with

and $C$ is unique when the marginals are continuous. The copula carries the dependence and the marginals carry the scales, so one set of marginals is compatible with many joint laws.

Lessons Multivariate distributions and dependence

Likelihood, estimation and limit theorems

For observations $x_1, \ldots, x_n$ that are independent with density $f_\theta$, the likelihood is $L(\theta; x) = \prod_j f_\theta(x_j)$, a function of the parameter, and the maximum likelihood estimator maximizes the log-likelihood $\ell(\theta; x) = \sum_j \ln f_\theta(x_j)$. For independent $N(\mu, \sigma^2)$ data, $\hat{\mu} = \bar{x}$ and $\hat{\sigma}^2_{\text{MLE}} = \frac{1}{n}\sum_j (x_j - \bar{x})^2$, which is biased, while $s^2 = \frac{1}{n-1}\sum_j (x_j - \bar{x})^2$ is unbiased. An estimator is consistent when it converges in probability to the true parameter as $n \to \infty$.

• Law of large numbers. For independent, identically distributed $X_j$ with $\mathbb{E}\lvert X \rvert < \infty$, the sample mean $\bar{X}_n$ converges to $\mathbb{E}[X]$, in probability and almost surely. With finite variance, Chebyshev’s inequality gives $\mathbb{P}(\lvert \bar{X}_n - \mu \rvert > \varepsilon) \le \sigma^2/(n\varepsilon^2)$.
• Central limit theorem. If moreover $0 < \sigma^2 < \infty$, then $\sqrt{n}\,(\bar{X}_n - \mu)/\sigma$ converges in distribution to $N(0, 1)$.

The standard error of $\bar{X}$ is $\sigma/\sqrt{n}$. For normal data with known $\sigma$, the interval $\bar{X} \pm z_{0.975}\,\sigma/\sqrt{n}$ with $z_{0.975} \approx 1.96$ covers $\mu$ with probability exactly $0.95$. With $s$ in place of $\sigma$, normal data call for the $t$ quantile with $n - 1$ degrees of freedom, and for other data the central limit theorem gives approximate coverage. The probability describes the procedure over repeated samples, not one computed interval. For the sample $(1, 2, 3, 4)$ with known $\sigma = 2$, $\bar{x} = 2.5$, the standard error is $1$ and the interval is about $[0.540, 4.460]$, while $\hat{\sigma}^2_{\text{MLE}} = 1.25$ and $s^2 = \frac{5}{3}$.

Lessons Statistical estimation and likelihood

Least squares regression and identification

Ordinary least squares chooses $\hat{\beta}$ to minimize $\lVert y - Xb \rVert_2^2$. The minimizers solve the normal equations $X^{\top}X\hat{\beta} = X^{\top}y$, and when $X$ has full column rank

The fitted vector $X\hat{\beta}$ is the orthogonal projection of $y$ onto the column space of $X$, so the residuals are orthogonal to every column. This is algebra and holds for any data. Interpreting $\hat{\beta}$ as a model parameter needs identifying assumptions. In the model $y = X\beta + \varepsilon$ with exogeneity $\mathbb{E}[\varepsilon \mid X] = 0$, the error $\hat{\beta} - \beta = (X^{\top}X)^{-1}X^{\top}\varepsilon$ has conditional mean zero, and with $\Omega = \operatorname{Var}(\varepsilon \mid X)$

Under homoskedasticity, $\Omega = \sigma^2I$, this reduces to $\sigma^2(X^{\top}X)^{-1}$. Under heteroskedasticity the sandwich form stays valid, and replacing $\Omega$ by the diagonal matrix of squared residuals gives the heteroskedasticity-consistent sandwich estimator, which is consistent under regularity conditions.

If the true model is $Y = \alpha + \beta X + \gamma Z + \varepsilon$ with $\mathbb{E}[\varepsilon \mid X, Z] = 0$ and $Z$ is left out, the slope of $Y$ on $X$ alone targets $\beta + \gamma\operatorname{Cov}(X, Z)/\operatorname{Var}(X)$, the omitted variable bias. With $\beta = 1$, $\gamma = 2$, $\operatorname{Cov}(X, Z) = 0.5$ and $\operatorname{Var}(X) = 1$, it targets $2$. For the data $x = (0, 1, 2, 3)$ and $y = (1, 2, 2, 5)$, the fitted line is $\hat{y} = 0.7 + 1.2x$, with residuals $(0.3, 0.1, -1.1, 0.7)$ and residual sum of squares $1.8$.

Lessons Regression and statistical identification

Stationarity and ARMA models

A process is weakly stationary when its mean is constant, its variance is finite and $\operatorname{Cov}(X_t, X_{t-h}) = \gamma(h)$ depends only on the lag $h$. The autocorrelation is $\rho(h) = \gamma(h)/\gamma(0)$. Let $\varepsilon_t$ be white noise with variance $\sigma_\varepsilon^2$. The AR(1) equation $X_t = \phi X_{t-1} + \varepsilon_t$ with $\lvert \phi \rvert < 1$ has a unique causal stationary solution,

If the shocks have conditional mean zero and constant conditional variance given the past, then $\mathbb{E}[X_{t+h} \mid \mathcal{F}_t] = \phi^{h}X_t$ and the forecast error variance is $\sigma_\varepsilon^2(1 - \phi^{2h})/(1 - \phi^2)$, which rises toward $\gamma(0)$ as $h$ grows. An ARMA($p$, $q$) model is $\Phi(L)X_t = \Theta(L)\varepsilon_t$ with the lag operator $L$, $\Phi(z) = 1 - \phi_1z - \cdots - \phi_pz^{p}$ and $\Theta(z) = 1 + \theta_1z + \cdots + \theta_qz^{q}$. When the two polynomials have no common root, it has a causal stationary solution exactly when every root of $\Phi$ lies outside the unit circle, and it is invertible when every root of $\Theta$ does.

The random walk $X_t = X_{t-1} + \varepsilon_t$ with $X_0 = 0$ has a unit root. Its variance $t\sigma_\varepsilon^2$ grows without bound, so it is not stationary, while its difference $\Delta X_t = \varepsilon_t$ is. With $\phi = 0.6$ and $\sigma_\varepsilon^2 = 4$, the autocovariances are $6.25$, $3.75$ and $2.25$ at lags $0$, $1$ and $2$. From $X_t = 10$ the two-step forecast is $3.6$ with error variance $5.44$, and the random walk with the same shocks has variance $20$ at $t = 5$.

Lessons Stationarity autocovariance and ARMA models

Conditional volatility and GARCH

Write the shock as $\varepsilon_t = \sqrt{h_t}\,z_t$, where the $z_t$ are independent with mean $0$ and variance $1$, $z_t$ is independent of $\mathcal{F}_{t-1}$, and the conditional variance $h_t = \operatorname{Var}(\varepsilon_t \mid \mathcal{F}_{t-1})$ is known at $t - 1$. The ARCH($q$) model sets $h_t = \omega + \sum_{j=1}^{q}\alpha_j\varepsilon_{t-j}^2$ with $\omega > 0$ and $\alpha_j \ge 0$. The GARCH(1,1) model adds persistence,

with $\omega > 0$, $\alpha \ge 0$ and $\beta \ge 0$. When $\alpha + \beta < 1$, the stationary solution has the finite unconditional variance $\bar{h} = \omega/(1 - \alpha - \beta)$, and for $k \ge 1$

The sum $\alpha + \beta$ measures persistence. At $\alpha + \beta = 1$ the unconditional variance is infinite, although a strictly stationary solution can still exist, since strict stationarity of GARCH(1,1) holds exactly when $\mathbb{E}[\ln(\alpha z_t^2 + \beta)] < 0$. With $\omega = 0.1$, $\alpha = 0.1$, $\beta = 0.8$, $h_0 = 1$ and $\varepsilon_0 = 2$, the long-run variance is $\bar{h} = 1$, the next variance is $h_1 = 1.3$ and the forecast is $\mathbb{E}[h_2 \mid \mathcal{F}_0] = 1.27$. If the next shock turns out to be $\varepsilon_1 = 0$, the realized $h_2$ is $1.14$.

Lessons Conditional volatility and GARCH

State space models and the Kalman filter

A linear Gaussian state space model has a latent state equation and an observation equation,

with $w_t \sim N(0, Q)$ and $v_t \sim N(0, R)$ independent of each other, over time and of the initial state $x_0 \sim N(m_0, P_0)$. The Kalman filter alternates a prediction step, the first two lines below, and an update step.

Here $y_t - Hm_t^{-}$ is the innovation, $S_t$ its covariance and $K_t$ the Kalman gain. Under the Gaussian assumptions $m_t$ and $P_t$ are the exact conditional mean and covariance of $x_t$ given $y_1, \ldots, y_t$. Without normality, the same recursion still gives the best linear estimator in mean square. In the scalar case with $H = 1$ and $P_t^{-} > 0$, the gain tends to $0$ as $R \to \infty$ and to $1$ as $R \to 0$. With $A = H = 1$, $Q = 0.25$, $R = 1$, $m_0 = 0$ and $P_0 = 1$, the observation $y_1 = 2$ gives $P_1^{-} = 1.25$, $K_1 = \frac{5}{9}$, $m_1 = \frac{10}{9}$ and $P_1 = \frac{5}{9}$, and then $y_2 = 0$ gives $K_2 = \frac{29}{65}$ and $m_2 = \frac{8}{13}$.

Lessons State space models and the Kalman filter

Cash flows instruments and curves

Bonds, forwards, futures, swaps and currency and commodity contracts are schedules of dated cash flows. With deterministic discount factors and frictionless trading, their values follow from discounting and from replication, and every formula depends on stated conventions for compounding, day counts, settlement and quotation. Discounting is deterministic throughout this unit except where futures meet random interest rates.

Bond prices, accrued interest and yield

A default-free bond with fixed amounts $C_k$ at times $t_k$, coupons plus the principal at maturity, has the full price $V = \sum_k C_k\,D(0, t_k)$ for settlement at time 0. Quotes are usually clean prices, and between coupon dates the full (dirty) price is the clean price plus accrued interest,

where $c$ is the coupon per period and the days are counted under the bond’s day count. A full price of $99.3$ with a coupon of $5$ that is 40 percent accrued has $\text{AI} = 2$ and a clean price of $97.3$.

On a coupon date with annual payments $C_1, \ldots, C_n$ after $1, \ldots, n$ years, the yield to maturity is the $y > -1$ that solves

It is the internal rate of return of the cash flows. If every $C_k \ge 0$ and some $C_k > 0$, the right side decreases strictly from $+\infty$ to $0$ on $(-1, \infty)$, so every positive price has exactly one yield. Cash flows of mixed sign can have several. The yield is a summary of the price, not a forecast of the return, and the return over the life of the bond equals $y$ only if the bond is held to maturity and every coupon is reinvested at $y$. The bond paying $5$ and $105$ is worth $98.1666$ at $y = 0.06$, and a bond whose coupon rate equals its yield is priced at par on a coupon date.

Lessons Bond cash flows and yield mathematics

Bootstrapping a discount curve

Order default-free instruments by maturity so that instrument $k$ pays $C_{kj}$ at the curve nodes $T_j$ for $j \le k$ and costs $V_k$. The pricing equations $V_k = \sum_{j \le k} C_{kj}D_j$ form a triangular system, solved in order when every $C_{kk} \ne 0$,

A zero-coupon bond paying $100$ in one year at price $95$ gives $D_1 = 0.95$. Annual 4 percent coupon bonds with two and three years to maturity at prices $97$ and $96$ give