The lecture builds the derivatives module as a chain. Each link uses the one before it, so the sheets are meant to be read in order.

The symbols follow the slides. Where Hull or other books differ, the slide version wins (for example \(X\) for the strike, where Hull writes \(K\)).

A derivative is a contract that derives its value from the performance of an underlying asset: stocks, indices, currencies, bonds, interest rates or commodities. The terms are agreed today, but delivery and payment happen at a specified future date. The most common derivatives are options, futures and forwards, and swaps.

The lecture sorts them in two ways. Is the deal conditional (one side may walk away, as with an option) or unconditional (both sides must deliver, as with a future)? And is it standardized and exchange-traded, or tailored over-the-counter? Click a box to see where each product sits.

A call gives its holder the right to buy the underlying for the strike \(X\) at maturity. A put gives the right to sell it for \(X\). The holder is long and has the right. The writer is short and has the obligation if the holder exercises.

The holder exercises only when it pays: a call when \(S_T > X\), a put when \(S_T < X\). That single rule creates the kink in every payoff diagram.

Payoffs add up. Hold a stock and write a call on it, and at maturity you get the stock's value minus the call's payoff. Every strategy in the lecture and the mock exams is a sum of calls, puts, the stock and a bond (cash). Pick a strategy below, or edit the legs and watch the total change.

A protective put is a stock plus a put. It pays \(\max(S_T, X)\): you keep the upside, and the put guarantees at least \(X\). Now hold cash worth \(X\) at maturity plus a call. That also pays \(\max(S_T, X)\): you have the cash, and the call adds the upside above \(X\).

Two portfolios with the same payoff in every state must cost the same today. Otherwise you could buy the cheap one, sell the expensive one and pocket the difference without risk.

The second form is the covered call: stock minus call pays \(\min(S_T, X)\), exactly like cash minus a put. Parity holds for European options on a stock without dividends. For American options it turns into a pair of bounds, which sheet 6 shows.

The practice problems ask you to design a strategy for a market view. Here is the reverse: the yellow band shows a target payoff. Combine positions until your total sits inside it.

A stock trades at 1000. In one year it is worth 1260 with probability 0.55 or 840 with probability 0.45. A call with strike 1050 therefore pays 210 or nothing. The expected payoff is \(0.55 \cdot 210 = 115.5\).

To get a price you would discount that at a risk-adjusted rate. The slide uses 15.5% and gets exactly 100. But where would 15.5% come from? The call is much riskier than the stock, so the stock's own expected return of 7.1% would be the wrong rate. Drag the rate and see how much the answer depends on a number nobody gives you.

Sell one call and buy \(\Delta\) shares. In the up state the portfolio is worth \(1260\Delta - 210\), in the down state \(840\Delta\). There is one \(\Delta\) that makes both values equal. Find it.

With \(\Delta = 0.5\) the portfolio is worth 420 whatever happens. A riskless portfolio must earn the riskless rate, otherwise there is an arbitrage. So today it is worth \(420/1.05 = 400\), and

Turn the hedge around: one call has the same payoff as \(\Delta\) shares financed partly with a loan \(B\). Matching both states gives two equations in two unknowns:

In the example the stock's value moves by \(1260 - 840 = 420\) between the states, the call's by only \(210 - 0 = 210\). That ratio is \(\Delta = 0.5\): the delta effect. In percentage terms it is the other way round. The stock's range is \(420/1000 = 42\%\) of its price, the call's range is \(210/100 = 210\%\): the leverage effect.

A portfolio is an arbitrage if its cost today is zero or negative (you get paid to hold it), its payoff is never negative, and in at least one state it is positive. The law of one price follows: two portfolios with identical payoffs in every state must have the same price.

The call from the previous sheet can be copied for 100. Suppose the market quotes it at a different price. Use the trading desk below to lock in a riskless profit. You win when today's cash flow and both future cash flows are zero or better, and at least one is strictly positive.

The model only makes sense if the riskless return lies between the stock's down and up returns. If the bond always beats the stock, sell the stock short and buy the bond. If the stock always beats the bond, borrow and buy the stock. Either way you earn money in every state.

Start from \(C_0 = \Delta S - B\) and insert \(\Delta\) and \(B\). Step through the slide's derivation one line at a time.

The real-world probability \(p\) of an up move never entered the price. Change it below. The expected payoff of the call changes, and so does the discount rate you would need to reproduce 100. The price itself does not move.

Ask which probability would make the stock's expected return equal to the riskless rate. Solve \(\pi \cdot 1260 + (1-\pi) \cdot 840 = 1000 \cdot 1.05\) and you get \(\pi = 0.5\), the same number as in the formula. Move the pivot below until the beam balances at 1050.

Under \(\pi\), every asset earns the riskless rate in expectation: discounted prices are martingales. That is why \(\pi\) is also called the martingale probability, and why pricing under it is called risk-neutral valuation.

Session 9 builds the same idea from Arrow–Debreu securities: one pays 1 in the up state and 0 otherwise, the other the reverse. Their prices are the state prices, and they are just discounted risk-neutral probabilities:

Any payoff is a bundle of the two pure securities, so its price is the sum of state prices times payoffs. The calculator uses the Session 9 market: a stock at 1050 that moves to 1260 or 840 and a bond at 1000 that pays 1050.

Using \(\pi\) does not assume that investors are risk-neutral. Session 9 shows where it comes from: in an optimum, the price of the pure security for state \(s\) is

Marginal utility is high in bad states, when consumption is low. So \(\pi\) shifts weight from the good state to the bad state compared with \(p\). The more risk-averse people are, the stronger the shift. The sketch uses a simple utility \(U(C) = C^{1-\gamma}/(1-\gamma)\) with consumption 1.26 in the up state and 0.84 in the down state.

With the discount factor \(m_T = 1/(1+R)\) the binomial price is one more instance of the equation that runs through the whole course:

The CAPM, the CCAPM and the APT choose \(\tilde m\) from preferences or factors. Option pricing gets away without that choice because the option can be replicated: in a complete market there is only one \(\tilde m\) consistent with the prices of the stock and the bond.

The slide example: \(S_0 = 1000\), \(u = 1.26\), \(d = 0.84\), \(R = 5\%\) per period, a call with \(X = 1050\) maturing in two years. Since \(\pi = 0.5\) at every node, each step back is a plain average discounted by 1.05. Press the button to walk through the backward induction.

Any binomial problem from the lecture, the practice sets or the mock exams fits into this lab. Pick a preset or set your own numbers. For trees of up to three steps the lab writes out the full working, line by line, the way an exam answer should look.

The price is only fair if the writer can actually hedge. You sold the three-period call below for its tree value. At each node, hold \(\Delta\) shares and a loan \(B\). Then let the stock move, and rebalance. The portfolio pays for itself at every step, and at maturity it is worth exactly what you owe.

Backward induction for a European option collapses into one sum. A terminal node with \(j\) up moves is reached by \(\binom{n}{j}\) paths, each with risk-neutral probability \(\pi^j (1-\pi)^{n-j}\):

The bars in the lab's lower panel are exactly these probabilities. Only the nodes above the strike contribute to the call. As \(n\) grows, the bars turn into a bell curve, which is where the next sheets pick up. For an American option the sum does not work, because the value at each node also depends on whether exercising early beats waiting.

An extra right can never hurt, so \(C^A \ge C^E\) and \(P^A \ge P^E\). The slide example prices a two-year put with \(S_0 = 50\), \(u = 1.2\), \(d = 0.8\), \(X = 52\) and a continuously compounded rate of 5%, so one year grows money by \(e^{0.05}\) and \(\pi = (e^{0.05} - 0.8)/(1.2 - 0.8) = 0.63\).

Exercising a put early hands you the strike now instead of later, and money now earns interest. The cost is the upside you give up if the stock recovers. Deep in the money that upside is small, so the interest wins. In the right panel the European put's value drops below what immediate exercise would pay. An American put can never fall below that line, so it is exercised there.

A call is the other way round. Exercising early means paying the strike sooner. Without dividends, put–call parity shows the call is always worth more alive than dead:

With many steps, the nodes where early exercise is optimal form a region. Its edge is the early-exercise boundary: an American put is exercised as soon as the stock falls below it. The slide lists what moves it: exercise gets more attractive when \(S\) falls, \(r\) rises or \(\sigma\) falls.

Switch to the call and add a dividend yield. Without dividends no node is ever exercised. With a large enough dividend, an exercise region appears at high stock prices: the holder exercises to collect the dividends.

The lecture states the dividend condition with a present value of dividends: early exercise of the call cannot be optimal while the interest saved on the strike exceeds the dividends, that is while \(X - PV(X) > PV(D)\). If the dividends are large enough, exercising just before the ex-dividend date can pay.

Parity used \(C^E\) and \(P^E\). For American options on a stock without dividends we know \(C^A = C^E\) and \(P^A \ge P^E\), which turns the equation into an upper bound. Comparing an American call plus cash \(X\) with an American put plus the stock gives the lower bound:

Session 14 starts from the Session 11 tree with \(p = 0.55\) and measures returns in logs: \(\ln(1260/1000) = 23.1\%\) and \(\ln(840/1000) = -17.4\%\). Log returns over several periods are simply sums of the one-period log returns. That gives two scaling rules for a random walk with identical, independent steps:

The mean grows with time, the volatility with the square root of time, because it is the variances that add: \(\operatorname{Var}(x_1 + \dots + x_t) = t \cdot \sigma^2\) for independent steps.

Francis Galton's board is a binomial tree made of pegs. Every row is one period: a ball bounces right (an up move) or left (a down move). The bin where it lands counts the up moves, which fixes the log return and the final price. Pour enough balls and the bins trace the binomial distribution. Add rows and that distribution becomes the normal curve.

Keep the horizon at one year and split it into \(n\) steps of length \(\Delta t = 1/n\). If each log step were \(\pm\sigma\,\Delta t\), the variance over the year would be \(n\,\sigma^2 \Delta t^2 = \sigma^2/n\), which vanishes as \(n\) grows. Only steps of \(\pm\sigma\sqrt{\Delta t}\) keep the yearly variance at \(n\,\sigma^2\Delta t = \sigma^2\). That is why the CRR model sets

The slide compares the Session 11 tree with a more volatile one: \(u = 1.4\), \(d = 0.6\). Then \(\pi = (1.05 - 0.6)/(1.4 - 0.6) = 0.5625\) and the call with \(X = 1050\) is worth \(0.5625 \cdot 350/1.05 = 187.5\) instead of 100. The reason is the kink: the option holder gains from a bigger up move but cannot lose more than the premium on a bigger down move.

With \(u = e^{\sigma\sqrt{\Delta t}}\) the tree price of a European option converges to the Black–Scholes price as \(n\) grows. It zigzags on the way, because the strike falls between terminal nodes at different spots for odd and even \(n\).

Flip a coin \(n\) times over one year and move \(+\sqrt{\Delta t}\) or \(-\sqrt{\Delta t}\) each time, with \(\Delta t = 1/n\). The variance after one year is \(n \cdot \Delta t = 1\) whatever \(n\) is. As \(n\) grows, the jagged path stops looking like steps and starts looking like a continuous, very rough curve.

The limit is a Brownian motion or Wiener process \(z_t\). The slide lists its properties:

• \(z_0 = 0\);
• \(z_t\) is a continuous function of \(t\);
• its increments are independent and normally distributed: for \(k < s\), \(E(z_s - z_k) = 0\) and \(\operatorname{Var}(z_s - z_k) = s - k\).

Drag the yellow time marker. The cross-section of many paths is normal, and its spread grows like \(\sqrt{t}\).

A smooth curve looks like a straight line under a microscope. A Brownian path does not. Zoom into a quarter of the time window and stretch the vertical axis by \(\sqrt{4} = 2\): the new picture is statistically the same as the old one. That is why a Brownian path has no slope anywhere, and why we need new calculus rules for it.

Cut \([0, T]\) into \(n\) pieces and add up the squared increments of a Brownian path. Each \((\Delta z)^2\) has mean \(\Delta t\), and there are \(n = T/\Delta t\) of them, so the sum tends to \(T\) and its randomness disappears. The sum of absolute increments, in contrast, grows without bound. In the limit the squared increment behaves like the time step itself, which gives the slide's multiplication rules:

An arithmetic Brownian motion \(dX_t = \mu\, dt + \sigma\, dz_t\) adds normally distributed amounts, so \(X_t\) is normal and can become negative. A geometric Brownian motion lets the return follow the Brownian motion:

Both panels start at 100 with the same drift and volatility in euros at the start. Count the paths that end up below zero.

For a smooth function \(D(S, t)\), ordinary calculus would keep only the first-order terms of a Taylor expansion. With a Brownian motion the second-order term \(\tfrac12 D_{SS}\, dS^2\) survives, because \(dS^2 = \sigma^2 S^2 dz^2 = \sigma^2 S^2 dt\) is of order \(dt\). That is Itô's lemma. Step through it for \(D = \ln S\):

The solution is \(S_t = S_0 \exp\{(\mu - \tfrac12\sigma^2)t + \sigma z_t\}\), so \(\ln S_T \sim N\big(\ln S_0 + (\mu - \tfrac12\sigma^2)T,\ \sigma^2 T\big)\). Raise \(\sigma\) and watch the mean and the median drift apart, and the distribution at maturity grow a long right tail.

The expected return \(\mu\) belongs to the real-world measure \(P\). For pricing, Session 15 switches to the martingale measure \(\Pi\) (Girsanov theorem): the volatility stays the same, only the drift changes from \(\mu\) to \(r\).

This is the continuous version of the binomial \(\pi\): under \(\Pi\) the stock earns the riskless rate in expectation, and every derivative is its discounted expected payoff. Volatility matters for the price, the expected return \(\mu\) does not.

With time to maturity \(T - t\) and the continuously compounded rate \(r\):

\(N(\cdot)\) is the standard normal distribution function. The first term is a stock position, the second a loan, exactly like \(\Delta S - B\) in the binomial model. The calculator below writes out each line the way the exam solutions do. The Adidas call from the slide is loaded.

Under the risk-neutral measure the stock grows at \(r\), so \(\ln S_T \sim N\big(\ln S + (r - \tfrac12\sigma^2)(T-t),\ \sigma^2(T-t)\big)\). The probability that the call ends in the money is then exactly \(N(d_2)\). \(N(d_1)\) is the delta: the number of shares in the replicating portfolio. Both are risk-neutral quantities. The real-world chance of exercise depends on \(\mu\) and is a different number.

Session 15 writes every derivative price as a discounted expectation under the martingale measure \(\Pi\): \(D_t = e^{-r(T-t)} E^{\Pi}_t[D_T]\). So you can price the call without any formula. Draw many prices at maturity with drift \(r\), average the payoffs and discount. The average closes in on the Black–Scholes price. Switch to the real-world drift and it misses.

Before maturity the option is worth more than its payoff would be today. That extra is the time value: the chance that the stock still moves the right way. As time passes it melts away, and at maturity the price curve folds into the hockey stick. The call never drops below \(S - Xe^{-r(T-t)}\), which is why an American call on a stock without dividends is never exercised early.

The Session 15 appendix in eleven lines. Combine the option with \(D_S\) shares so that the Brownian motion cancels, require the riskless portfolio to earn \(r\), and you get a partial differential equation that every derivative on the stock must satisfy. Solving it for the call payoff, through the risk-neutral expectation, gives the formula.

The slide lists six inputs and their Greeks. Each small chart below shows the call and the put price as one input varies while the others stay at the base case. The sign of the slope at the yellow point is the sign of the Greek. Drag any yellow point to move the base case and watch the other five charts follow.

Delta is the slope of the price curve: for a small move \(dS\) the option changes by about \(\Delta\, dS\). The straight tangent line is what a delta hedge replicates. The curve bends away from it, so the hedge is only right for small moves. That bend is gamma, \(\partial^2 C/\partial S^2\). It is not on the slide, but it explains why a hedge has to be rebalanced.

The delta of a position is the sum of quantity times delta over its legs, with a minus sign for short legs. A share has a delta of 1. To make the position delta-neutral, trade the opposite number of shares. Load the two exam questions below.

A bank sells a three-month call, receives the Black–Scholes premium and holds \(\Delta\) shares, financed through a bank account at \(r\). Every rebalancing date it adjusts the shares to the new delta. At maturity it pays the option payoff. If it could rebalance continuously, the account would end at exactly zero on every path, whatever the stock does. With discrete rebalancing a small error remains.

Pick a type or let the dice choose. Answers count as correct within a small tolerance, so rounding \(\pi\) to four decimals or reading \(N(\cdot)\) from the table is fine. A decimal comma works too.

Each mock exam opens the derivatives part with true-or-false statements. Here are the mock statements and more in the same style, one at a time.

• Write \(\pi\) with the formula first, then the numbers. Say whether the rate is discrete (\(1+R\)) or continuous (\(e^{r\Delta t}\)).
• Draw the stock tree and the option tree. Label every node with its value.
• For American options, write \(\max(\text{exercise}, \text{continue})\) at every node.
• For Black–Scholes, write \(d_1\), \(d_2\), \(N(d_1)\), \(N(d_2)\) on separate lines before the price.
• For hedges, write the position delta with its sign before deciding to buy or sell shares.
• For a claim that pays \(S_T\) in some states, remember the static replication: the complementary claim is worth \(S_0\) minus its price.