What you will learn, and how to learn it
You do not need calculus or linear algebra to begin. The first chapter explains quantities, units, functions and growth. Later chapters reuse these ideas to develop derivatives, integrals, vector geometry, uncertainty, estimation and computational methods. Stochastic models and derivative pricing bring those strands together. The final chapter introduces transforms, dependence, heavy tails and hidden-state estimation.
Each lesson has a question to investigate, an adjustable visual, a plain-language explanation, mathematical notation, a worked example and a calculation table. Make a prediction before moving a control. Explain the result in your own words, answer the conceptual question, and work the numerical check before marking a lesson understood. That mark records your own assessment, not a certification.
This is a substantial, self-contained path through the core mathematical tools used in quantitative finance. It is not a complete graduate mathematics degree or a production modelling library. Full measure-theoretic proofs, general stochastic differential equation existence theorems, functional analysis and specialist numerical schemes require further depth. Source links support verification and continued study; they are not required to operate the lessons.
Examples are original educational constructions. Models, independence assumptions, finite grids and numerical approximations are stated beside the experiments. Financial examples explain mathematics rather than recommend trades. Reproducible samples let you compare inputs; changing a seed gives another sample, not a better forecast.
Apply the chapters: nine guided projects with solutions
Use these projects after completing the corresponding chapter. Show the units, state the assumptions, calculate the result and explain what would make your answer unreliable. Each project combines ideas rather than asking for a formula in isolation.
1. Growth and returns: explain a round trip
Start with 100, gain 20%, then lose 20%. Calculate the final value, cumulative simple return and sum of log returns. Explain why the arithmetic average of the two simple returns is misleading for terminal wealth.
Solution. The final value is 100×1.2×0.8=96. The cumulative return is −4%. The log returns sum to ln(1.2)+ln(0.8)=ln(0.96)≈−0.040822. An arithmetic average of zero ignores compounding on different bases. A 25% gain, not 20%, is needed to recover a 20% loss.
2. Calculus: defend a local approximation
For f(x)=x² at x=2, approximate the change for h=0.1 and h=1 using the tangent. Compare with exact changes, then include the second-order term.
Solution. The derivative is 4, so tangent changes are 0.4 and 4. Exact changes are 0.41 and 5. The omitted term is h². Since f″=2, the Taylor correction ½f″h²=h² makes both answers exact for this quadratic. For a general function, higher derivatives can still leave a remainder.
3. Linear algebra: diagnose a redundant hedge system
Suppose exposures of two hedge instruments are (1,2) and (2,4). Can they replicate target exposure (1,0)? Explain the answer geometrically before solving equations.
Solution. Both columns lie on the same line. Their combinations have second coordinate twice the first, whereas the target does not. The matrix has rank one and the target is outside its column space. A numerical inverse cannot create the missing direction. Add an independent hedge instrument or accept and quantify residual exposure.
4. Optimisation: compare two portfolio objectives
Two assets each have 20% volatility, correlation zero and expected returns of 8% and 4%. Find the long-only minimum-variance weight and calculate its expected return. Then explain why a required expected return of 7% changes the allocation.
Solution. Variance is 0.04[w²+(1−w)²], minimised at w=0.5. Volatility is √0.02≈14.14%, and expected return is 6%. A 7% target requires 4%+4%w=7%, giving w=0.75 and volatility √0.025≈15.81%. The extra return target changes the feasible set; it does not invalidate the earlier optimum.
5. Probability: interpret a credit warning
Among 10,000 borrowers, 1% default. A signal catches 90% of defaults and flags 10% of nondefaults. Calculate how many are flagged and the probability of default among them. Explain the role of the base rate.
Solution. There are 100 defaults and 9,900 nondefaults. The signal flags 90 defaults and 990 nondefaults, for 1,080 flags. The conditional default probability is 90/1,080≈8.33%. The signal is informative, but most flagged borrowers do not default because the nondefault population is much larger. This is a hypothetical classification model, not an estimated policy threshold.
6. Statistics: separate observation spread from estimator uncertainty
Independent normal observations have known standard deviation 2. A sample of 100 has mean 0.3. Calculate the normal 95% confidence interval for the population mean. What changes if the observations are strongly serially correlated?
Solution. The standard error is 2/√100=0.2 and the interval is 0.3±1.96×0.2, or [−0.092,0.692]. This is uncertainty about the mean, not a range containing 95% of observations. Serial dependence changes the variance of the sample mean; the independent-data formula can understate uncertainty under positive dependence.
7. Numerical methods: choose an error budget
You estimate the integral of x² on [0,1] using independent uniform Monte Carlo draws. Determine the theoretical standard error with 1,000 draws and the sample size needed to reduce it by half. Compare with the one-trapezoid result.
Solution. Since Var(U²)=4/45, the Monte Carlo standard error is √(4/45,000)≈0.009428. About 4,000 draws halve it. A single trapezoid gives 0.5, a deterministic error of 1/6. Sampling error and discretisation error are different; more Monte Carlo draws cannot correct a biased payoff implementation.
8. Pricing: reproduce a claim without forecasting
A stock costs 100 today and will become 120 or 80. A call has strike 100. The risk-free gross return is 1.05. Find the hedge, price and risk-neutral up probability. Check both terminal states.
Solution. Payoffs are 20 and 0. The stock holding is (20−0)/(120−80)=0.5. Cash today is (0−0.5×80)/1.05≈−38.0952. Price is 50−38.0952≈11.9048. The up probability is (1.05−0.8)/(1.2−0.8)=0.625. At expiry the borrowed balance is 40: 60−40=20 in the up state and 40−40=0 in the down state. No actual up probability was required.
9. Advanced tools: test a tail assumption
For a Pareto loss with minimum one and shape two, calculate 99% VaR, expected shortfall and whether the mean and variance exist. Explain why blindly applying the ordinary finite-variance CLT would be unsafe.
Solution. VaR is (1−0.99)^(−1/2)=10 and ES is 2×10/(2−1)=20. The mean is 2, but variance is infinite because shape is not above two. A finite dataset still has a finite sample variance; that does not prove the population variance exists. Normal standard-error reasoning needs its assumptions checked.