Stochastic Calculus Flashcards

Question

Are Jointly Markov processes individually Markov?

Answer 1

In general, no (but possible). Modeling a joint process univariately will usually cease to be Markov. The reverse is also true, we can make any univariate non-Markov process into a Markov one by increasing the dimensionality

Answer 2

1. Classic approach: Model yield curve using a single rate process r_t. This is univariate and Markov, giving an SDE representation of dr_t = μ(r_t,t)dt + σ(r_t,t)dW_t 2. HJM approach: Model k forward rates which are assumed jointly Markov with SDE dF(t,T) = σ(t,T,B(t,T))[∂σ/∂T]dt + [∂σ/∂T]dW_t

Answer 3

1. Mean square convergence 2. Almost surely convergence 3. Weak convergence

Answer 4

Martingale: trajectories display no trends or periodicities Ex: PV of asset prices discounted by r in RN measure Submartingale: the process increases on average Ex: PV of asset prices discounted by r in RW measure

Answer 5

A process S_t is a martingale wrt info sets I_t and measure P if for all t>0: * S_t is adapted to the information set * Unconditional forecasts are finite E|S_t| < infinity * Best future forecast of unobserved values is the last observation E_t[S_T] = S_t for all T>t

Answer 6

1. W_t 2. W_t² - t 3. Exp[θW_t - 1/2 θ²t]

Answer 7

1. Transform the submartingales as follows: From B_t to e^-rtB_t and S_t to e^-rtS_t 2. (Preferred) Transform the probability measure If E_t^P[e^-ruS_t+u] > S_t we can find an equivalent probability measure where they are equal

Answer 8

Let {X_n} be iid. Let S_n = X₁+...X_n. If E(X_i) = μ and Var(X_i) = σ² Then (S_n-nμ)/sqrt(n) ~ Normal(0,1) in the limit

Answer 9

Let X be a real valued rv, consider M_t = E^P[X|F_t]. If M_t is a martingale, then we can find an adapted process γ_u such that M_t = M₀ + ∫(0,t) γ_udW_u

Answer 10

If X_t is a right continuous submartingale (increasing) and for all t, E[X_t] < infinity, then X_t admits the decomposition X_t = M_t + A_t Where M_t is a martingale and A_t is an increasing stochastic process

Answer 11

Let Z_t be any martingale wrt I_t and P. Let H_t be any r.v. adapted to I_t and P. Then M_{t_k} = M_t₀ + Σ(1,k)H_{t_i-1}[Z_{t_i}-Z_{t_i-1}] will also be a martingale

Answer 12

Let dZ_t be an infinitesimal stochastic increment with mean 0 wrt I_t and P. Let H_t be any r.v. adapted to I_t. Then M_t = M₀ + ∫(0,t)H_udZ_u will be a martingale

Answer 13

Since x is a r.v, it cannot have 0 variance. Thus E(Δx)²>0 We have 2 approximations for the Taylor expansion with stochastic variables f(x₀+Δx)-f(x₀) = f_xΔx+1/2 f_xxE(Δx)² or f(x₀+Δx)-f(x₀) = f_xΔx+1/2 f_xxE(x^*) Where x^* is the mean square limit of (Δx)²

Answer 14

Rare events have something to do with price discontinuity They are supposed to occur infrequently. As the time step h goes to 0, the probability of occurrence for rare events should also go to 0. But even when the probability goes to 0, the size may not shrink. By definition they have a large size

Answer 15

As h goes to 0, the tails of the normal dist carry less weight, and the size of price changes given by the WP shrink. This ther WP is not suitable for situations where prices can move large distances over an extremely short interval. We need a disturbance term capable of giving large events in short intervals without shrinking the size.

Answer 16

Jump Diffusion dS_t = a(S_t, t)dt + σ₁(S_t, t)dW_t + σ₂(S_t, t)dJ_t

Answer 17

Over a short enough interval, drift (a) and volatility (σ) will be near constant, so we integrate the SDE to get a discrete approximation S_t+h - S_t = a(S_t, t)h + σ(S_t, t)ΔW_t

Answer 18

The proper integration of the SDE involves the Ito Integral ∫(t,t+h) σ(S_u, u)dW_u W_u is a martingale and has unpredictable increments, the sum of which is itself an unpredictable increment. Thus the Ito integral is a martingale E_t[∫(t,t+h) σ(S_u, u)dW_u] = σ(S_t, t)_t[∫(t,t+h) dW_u] = 0

Answer 19

The quadratic variation si _t = ∫(0,t) f(W_s, s)²ds Where the Ito integral is ∫(0,t) f(W_s, s)dW_s We can write it as the sum of increment terms as follows I_t = ∫(0,t) f(W_s, s)dW_s = lim_n>infΣf(W_{t_i}, t_i)(W_{t_i+1}-W_{t_i})

Answer 20

The quadratic variation of an Ito process is _t = ∫(0,t) σ(X_s, s)²ds

Answer 21

The General Ito Integral ∫(0,t) f(S_u, u)dS_u exists if the function f is continuous and non-anticipating. The general Ito integral is a random process that has various expectations, but the martingale property ensures the first moment of the ito integral is 0

Answer 22

I(T) = ∫(0,T) f(t)dW_t ~ Normal(Mean = 0, Variance = ∫(0,T) f(t)²dt)

Answer 23

I(t) = ∫(0,t) W_sdW_s S(t) = ∫(0,t)W_s o dW_s = lim_n>infΣ(W_{t_i}+W_{t_i+1})/2 (W_{t_i+1}-W_{t_i}) S(t) - I(t) = t/2 | Ito Integral is a martingale, Stratonovich integral is not

Answer 24

Let F(S_t, t) depend on a r.v. S and t. Let dS_t = a_tdt + σ_tdW_t Then the SDE for F(S_t, t) is dF_t = (∂F/∂S_t)dS_t + (∂F/∂t)dt +1/2 (∂²F/∂S²)σ_t²dt or dF_t = [∂F/∂S_t a_t + ∂F/∂t + 1/2 (∂²F/∂S²)σ_t²]dt + (∂F/∂S_t)σ_tdW_t | Equality holds in a mean square sense

Answer 25

∫(0,t) F_sdS_u = F(S_t, t) - F(S₀, 0) - ∫(0,t) F_u + 1/2 F_ssσ_u²du

Answer 26

E[∫(0,T) f(t)dW_t x ∫(0,T) g(t)dW_t] =∫(0,T) f(t)g(t)dt

Answer 27

E[∫(0,T) f(X_t, t)dW_t x ∫(0,T) g(X_t, t)dW_t] =∫(0,T) E[f(X_t, t)g(X_t, t)]dt

Answer 28

The strong solution to the SDE dS_t = μS_tdt + σS_tdW_t is S_t = S₀e^{(μ-1/2 σ²)t + σW_t}

Answer 29

(∂V/∂t) + rS(∂V/∂S) + 1/2 σ²S²(∂²V/∂S²) = rV With Greeks: Theta + rS(Delta) +1/2 σ²S²Gamma = rV

Answer 30

dt x dt = 0 dt x dW_t = 0 dS x dt = 0 dW_t x dW_t = dt

Answer 31

* Teh expiration value of the option may depend on some event happening over the option's life, which makes boundary conditions much more complicated than standard BS * Exotic derivative may have random expiration dates * Exotic derivatives can be written on more than one asset

Answer 32

1. Assume S follows a GBM and the typical BS assumptions 2. Build a hedged portfolio Π (short derivative V, long delta S or vice versa) 3. Use Ito's Lemma to get an SDE for change in value dΠ over small time dt 4. See that this special portfolio is free of the random term dZ 5. Conclude that this portfolio is riskless and must earn r over this time interval 6. Replace the dΠ and Π in the equation dΠ/Π = rdt 7. The resulting PDE applies to any derivative written on the underlying asset S

Answer 33

Let G(S,t) be a function of time and stock price S where dS/S = μdt + σdz The process followed by G is dG = [(∂G/∂S)μS + ∂G/∂t + 1/2 (∂²G/∂S²)(σS)²]dt + ∂G/∂S σSdz

Answer 34

(∂V/∂t) + (r-x)S(∂V/∂S) + 1/2 σ²S²(∂²V/∂S²) = rV Where x is thought as some form of 'dividend yield'. For stocks, it is the Dividend yield D. Currency options use foreign interest rate r_f. Commodities use the fraction of value that goes into the carrying cost **-**q

Answer 35

Given information sets {I_t} over a period [0,T] we define a random process ξ_t = exp(∫(0,t) X_udW_u - 1/2∫(0,t) X_u²du where X_t is I_t measurable and W_t is a WP on P We impose the Nosikov Condition on X_t: E(exp(∫(0,t) X_u²du)) < inf

Answer 36

If Nosikov Condition is satisfied, then the process ξ_t will be a square integrable martingale s.t W_t^~ = W_t - ∫(0,t)X_udu is a WP wrt I_t and measure P_T^~ where P_T^~(A) = E^P[1_Aξ_T] and the two WPs are related by dW_t^~ = dW_t - X_tdt

Answer 37

Pricing derivatives have gone on without needing to model true probabilities under P or the risk premium μ. Derivatives can be priced in the risk neutral measure P^~. For forecasting, a decision maker should clearly use the real world probabiliities P and apply expecations in this measure rather than the RN one.

Answer 38

PDE: Use the steps presented to reach the BS PDE and solve it (possibly numerically) Martingale: Find the synthetic probability measure where discounted prices are martingales and take expectations

Answer 39

In trying to find the risk neutral SDE for e^-rtF(S_t, t), we use Ito Lemma and Girsanov to see it is d[e^-rtF(S_t, t)] = e^-rt[-rF + rF_sS_t + F_t + 1/2 F_ss(σS_t)²]dt + F_se^-rtσ_tdW_t^~. Since this is a RN measure, the drift term is 0, giving rF + rF_sS_t + F_t + 1/2 F_ss(σS_t)² = 0, the BS PDE

Answer 40

If f is a convex function and x is an r.v then E(f(x)) >= f(E(x))

Answer 41

1. If you hedge correctly in BS, then all risk is eliminated. With no risk, you don't get any risk premium, so assets will return risk free rates 2. Using the GBM SDE dS/S = μdt + σdz, we see μ cancels out in deriving the BS eqn 3. A price is non arbitrageable in the RW iff it is non arbitrageable in the RN world. But the RN price is always nonarbitrageable bc everything is a martingale, so the RW has no arbitrage.

Answer 42

* In a complete market, derivatives can be replicated by more basic instruments like cash, bonds, and S. In incomplete markets, you cannot achieve this * A complete markes is one for which there exists the same number of linearly independent securities as there are future states of the world * In a complete market you can hedge a derivative with the underlying * In a complete market, the martingale measure is unique, but this is not true for incomplete markets

Answer 43

2 reasons RN pricing doesn't work in practice: Markets are incomplete and dynamic hedging is impossible RW Valuation: You could value an option as the real expectation of the PV of option payoff +/- some multiple of std dev. However this has 2 main problems 1. You need to know real probabilities and real drift which is **very** hard 2. You need to decide on a utility function or a measure of risk aversion

Answer 44

A stochastic process is a SWP if * W₀ = 0 and has continuous sample paths. * Increments are independent, stationary, normally distributed, and variance equals elapsed time

Answer 45

* Markov property: Distribution of W_t given info up to s < t, depends only on W_s * Strong Markov Property: For any s > 0, the r.v. W_t+s - W_t is independent of the filtration up to time t * Martingale

Answer 46

Cov(W_t, W_s) = E(W_tW_s) = min(s,t)

Answer 47

Cov(W_t₁, W_t₂, W_{t_n}) = matrix((Cov(W_t₁, W_t₁), ..., Cov(W_t₁, W_{t_n})), ..., (Cov(W_t₁, W_{t_n}), ..., Cov(W_{t_n}, W_{t_n})) = Matrix((t₁, ..., t₁), ..., (t₁, ..., t_n))

Answer 48

_t = lim_n>infΣ(W_{t_i+1} - W_{t_i})² = t

Answer 49

Let X_t be an Ito process. If V(X_t, t) satisfies the PDE: ∂V/∂t + 1/2 σ²∂²V/∂X² + μ∂V/∂X - rV = 0 with boundary condition V(X_T, T) = Ψ(T) then the solution to the PDE is V(X_t, t) = E[e^{(-∫(t,T) r(u)du} x V(X_T, T)|F_t]

Answer 50

Let X_t, Y_t be Ito processes. The SDE for the product X_tY_t is given by d(X_tY_t) = X_tdY_t = Y_tdX_t + dX_tdY_t

Answer 51

d(F(W_t, t) = F'(W_t)dW(t) + [F'(t) + 1/2 ∂²/∂W² F(W(t))]dt

Answer 52

Let dS/S = μdt + σdW(t) and F = ln(S). Then d(ln(S)) = (μ - σ²/2)dt + σdW(t)

Answer 53

Let dS/S = μdt + σdW(t) and F = e^-r(t)S. Then d(e^-r(t)S)/e^-r(t)S = (μ-r(t))dt + σdW(t)

Answer 54

Let dP/P = μ_Pdt + σ_PdW_P and dM/M = μ_Mdt + σ_MdW_M. Assume Corr(W_P, W_M) = ρ. Then d(PM)/PM = (μ_P + μ_M + σ_Pσ_Mρ)dt + σ_PdW_P + σ_MdW_M

Answer 55

Let dS/S = μdt + σdW(t) and F = 1/S. Then d(1/S)/(1/S) = dF/F = -[(μ-σ²)dt + σdW(t)]

Answer 56

Let dS/S = μdt + σdW(t) and F = Sⁿ. Then dF/F = n x [(μ + 1/2 (n-1)σ²)dt + σdW(t)] | Can use this to arrive at the correct SDE for inverse GBM

Answer 57

* d(∫(0,t) f(s)dW(s)) = f(t)dW(t) * d(∫(0,t) f(s)ds) = f(t)dt * E[∫(0,t) f(W_u, u)dW_u∫(0,t) g(W_u, u)dW_u] = ∫(0,t) E[f(W_u, u)g(W_u, u)]du * E[∫(0,s) f(W_u, u)dW_u∫(s,t) f(W_u, u)dW_u] = 0