Questions & explanations
1. Give an example of an arbitrage opportunity in a simple two-state model and show how it violates the existence of a risk-neutral measure.
Consider a stock priced at $100 that can go to $110 or $90 in one year. A risk-free bond yields 5%. If a call option with strike $100 is priced at $5, there is arbitrage. You can buy the call and sell the stock short. The cost is $5 - $100 = -$95 (you receive $95). In the up state, you exercise the call to buy stock at $100 and cover the short, losing $10 on the stock but gaining $10 from the call? Actually, let's correct: The arbitrage is: sell the stock for $100, buy the call for $5, invest $95 at 5% to get $99.75. In up state, call pays $10, you buy stock for $110? Wait, need a proper example. Better: Suppose stock is $100, up to $120, down to $80. Risk-free rate 0%. A call with strike $100 costs $12. Then you can sell the call for $12 and buy a portfolio that replicates it cheaper. The replicating portfolio: buy 0.5 shares for $50 and borrow $40? Actually, the no-arbitrage price is $10. So $12 is too high; you sell the call and buy the replicating portfolio, making $2 profit risk-free. This violates the existence of a risk-neutral measure because under any risk-neutral measure, t
2. Use RK4 with h=0.2 to approximate y(0.4) for y' = t - y, y(0)=1.
We have t0=0, y0=1, f(t,y)=t-y. h=0.2. Step 1: k1 = f(0,1)=0-1=-1. k2 = f(0.1, 1+0.1*(-1)=0.9) = 0.1-0.9=-0.8. k3 = f(0.1, 1+0.1*(-0.8)=0.92) = 0.1-0.92=-0.82. k4 = f(0.2, 1+0.2*(-0.82)=0.836) = 0.2-0.836=-0.636. y1 = 1 + 0.2/6*(-1 + 2*(-0.8) + 2*(-0.82) + (-0.636)) = 1 + 0.03333*(-1 -1.6 -1.64 -0.636) = 1 + 0.03333*(-4.876) = 1 - 0.1625 = 0.8375. So y(0.2)≈0.8375. Next step: t1=0.2, y1=0.8375. k1 = f(0.2,0.8375)=0.2-0.8375=-0.6375. k2 = f(0.3, 0.8375+0.1*(-0.6375)=0.77375) = 0.3-0.77375=-0.47375. k3 = f(0.3, 0.8375+0.1*(-0.47375)=0.790125) = 0.3-0.790125=-0.490125. k4 = f(0.4, 0.8375+0.2*(-0.490125)=0.739475) = 0.4-0.739475=-0.339475. y2 = 0.8375 + 0.2/6*(-0.6375 + 2*(-0.47375) + 2*(-0.490125) + (-0.339475)) = 0.8375 + 0.03333*(-0.6375 -0.9475 -0.98025 -0.339475) = 0.8375 + 0.03333*(-2.904725) = 0.8375 - 0.0968 = 0.7407. So y(0.4)≈0.7407. Exact solution: y(t)=t-1+2e^{-t}, so y(0.4)=0.4-1+2e^{-0.4}= -0.6+2*0.6703=0.7406. Very close.
3. How does a key exchange work using braid groups?
In a braid group key exchange, Alice and Bob agree on a public braid P. Alice picks a secret braid A and sends A*P*A^{-1} to Bob. Bob picks a secret braid B and sends B*P*B^{-1} to Alice. Then Alice computes A*(B*P*B^{-1})*A^{-1} and Bob computes B*(A*P*A^{-1})*B^{-1}. Because braid groups are non-commutative, these two results are equal only if A and B commute, which they don't in general. Actually, the standard protocol uses the commutator: both compute A*B*P*B^{-1}*A^{-1}? Wait, correct: Alice sends A*P*A^{-1}, Bob sends B*P*B^{-1}, then Alice computes A*(B*P*B^{-1})*A^{-1} = (A*B)*P*(B^{-1}*A^{-1}) and Bob computes B*(A*P*A^{-1})*B^{-1} = (B*A)*P*(A^{-1}*B^{-1}). These are not equal unless A and B commute. So the protocol is flawed. Actually, the Anshel-Anshel-Goldfeld key exchange uses different operations. The point is that non-commutative groups allow new key exchange methods, but many have been attacked.
4. Give an example where Value-at-Risk (VaR) violates subadditivity, but Expected Shortfall (ES) does not.
Consider two independent bonds each with 4% chance of default, losing $100 if default, else $0. VaR at 95% confidence for each bond is $0 (since 95% of the time no loss). The portfolio of both bonds has a 0.04*0.04=0.16% chance of both defaulting, but a 7.68% chance of exactly one default. At 95% confidence, the loss is $100 (since 95% of the time loss ≤ $100? Actually 95.84% of the time loss ≤ $100? Wait, need careful: probability of loss $0 is 0.96^2=0.9216, loss $100 is 2*0.04*0.96=0.0768, loss $200 is 0.0016. So 95% VaR is $100 (since 92.16%+7.68%=99.84% >95%). So VaR portfolio=$100 > sum of individual VaRs=$0, violating subadditivity. ES at 95% for each bond is $80 (since 5% tail average loss is $100*0.04/0.05=$80). For portfolio, ES is (0.0768*100+0.0016*200)/0.05= (7.68+0.32)/0.05=160, which is ≤ 80+80=160, so subadditive.
5. What does Cybenko's theorem say about a neural network with one hidden layer and a sigmoidal activation function?
Cybenko's theorem states that a feedforward neural network with one hidden layer containing a finite number of neurons, each using a sigmoidal activation function (an S-shaped function that squashes input to a range, like the logistic function), can approximate any continuous function on a compact set arbitrarily well. This means for any continuous function and any desired accuracy, there exists such a network whose output is within that accuracy. The theorem guarantees that the network is a universal approximator, but does not tell how to find the weights or how many neurons are needed. It applies to functions from a compact subset of R^n to R. The result is foundational for neural network theory, showing that a single hidden layer is sufficient in principle.
6. Give an example of the typical set for a fair coin toss. What fraction of sequences are typical for large n?
For a fair coin (p=0.5), entropy H=1 bit. The typical set contains sequences with about n/2 heads and n/2 tails. For large n, almost all sequences are typical. For example, with n=1000, sequences with 500±√(n) heads are typical. The fraction of sequences that are typical approaches 1. The number of typical sequences is about 2^{1000}, which is huge but still a tiny fraction of all 2^{1000} sequences? Actually, all sequences are equally likely, so the typical set is almost all sequences. Wait: For p=0.5, all sequences have the same probability, so the typical set includes all sequences. The AEP still holds because the probability of each sequence is exactly 2^{-n}. So the typical set is the whole space.
7. Compare the use of local volatility versus stochastic volatility models for pricing exotic options in an emerging market. Which is more suitable and why?
Local volatility models make volatility a function of the underlying asset price and time, so they can fit the market prices of plain vanilla options exactly. But they assume volatility will behave in the future as it did in the past, which may not hold in emerging markets where volatility patterns change. Stochastic volatility models allow volatility to have its own random process, capturing the uncertainty better. For exotic options that are sensitive to the volatility path, stochastic volatility is more suitable because it reflects the dynamic nature of emerging markets. However, local volatility is simpler and faster, so it might be used for quick pricing when the market is stable.
8. What is the limitation of pole placement when not all states are measurable?
Pole placement requires full state feedback, meaning you need to measure or estimate all states. If some states are not measurable, you cannot directly implement the control law u = -Kx. In that case, you must design an observer to estimate the states from the output. The observer itself has dynamics, and the combined system (controller + observer) has poles that are the union of the controller poles and observer poles. The separation principle says you can design the controller and observer independently, but the observer poles must be faster (typically 2-5 times faster) than the controller poles to ensure good estimation. This adds complexity and may require additional sensors.
9. Compare the use of a GARCH model versus a stochastic volatility model for forecasting cryptocurrency volatility. Which is more appropriate and why?
GARCH models assume that volatility depends on past squared returns and past volatility, so they are good at capturing volatility clustering. They are simpler to estimate and work well for daily data. Stochastic volatility models let volatility have its own random process, independent of returns, which can capture more complex dynamics. For cryptocurrencies, which have sudden jumps and regime changes, stochastic volatility models may be more flexible because they can incorporate jumps in volatility. However, they are harder to estimate. In practice, GARCH is often used for short-term forecasts, while stochastic volatility is better for pricing options with longer maturities.
10. Why is it challenging to model cryptocurrency returns using traditional financial models like the normal distribution?
Cryptocurrency returns have very heavy tails, meaning extreme gains or losses happen much more often than a normal distribution predicts. They also show volatility clustering and negative skewness, where large drops are more common than large rises. Traditional models assume returns are normally distributed, which underestimates the risk of big moves. For example, Bitcoin has seen daily drops of over 20% multiple times, which would be virtually impossible under a normal model. Therefore, models like Student's t-distribution or jump-diffusion models are used to better capture these fat tails. Failing to account for this can lead to poor risk management and mispriced options.
11. Give an example of a system that is controllable but not observable, and explain why.
Consider a system with A=[[0,1],[0,0]], B=[[0],[1]], C=[[1,0]]. The controllability matrix is [B AB] = [[0,1],[1,0]] with rank 2, so controllable. The observability matrix is [C; CA] = [[1,0];[0,1]] with rank 2, so observable. That's both. For a non-observable example, take A=[[0,1],[0,0]], B=[[0],[1]], C=[[1,0];[0,0]]? Actually, let C=[[1,0]] and D=0, but that's observable. A simple non-observable system: A=[[0,1],[0,0]], C=[[0,1]]. Then observability matrix = [[0,1];[0,0]] rank 1 < 2, so not observable. The state x1 does not affect the output. This system is still controllable because B affects both states. So you can control the states but cannot see x1 from the output.
12. How do you calibrate a stochastic volatility model to market data? Describe the steps.
Calibration involves finding model parameters that minimize the difference between model and market option prices. Steps: 1) Collect market prices of liquid options (e.g., European calls) across strikes and maturities. 2) Compute implied volatilities from these prices. 3) Choose a stochastic volatility model (e.g., Heston). 4) Use an optimization algorithm (e.g., Levenberg-Marquardt) to adjust model parameters (κ, θ, ξ, ρ, initial variance) to minimize the sum of squared errors between model and market implied volatilities. 5) Validate the calibrated model by pricing out-of-sample options. The goal is to fit the entire implied volatility surface as closely as possible.