Stochastic Calculus

Question 1

Three general heading of derivatives

Answer 1

1. Futures and Forwards
2. Options
3. Swaps

Question 2

3 main differences between futures and forwards

Answer 2

• Organized exchange vs OTC
• No default risk vs default risk
• Mark to market each day vs settled at maturity

Question 3

List 5 groups of underlying assets

Answer 3

1. Stocks
2. Currencies
3. Interest rates (not asset, take a position on direction via Treasuries)
4. Indexes
5. Commodities

Question 4

4 uses of arbitrage-free prices

Answer 4

• New financial product, no market data available
• Risk management, no data for hypothetical ‘worst case scenarios’
• Marking to market, may need current value of illiquid asset
• Mispricing, if market prices differ we want to take advantage of it

Question 5

Arbitrage Theorem

w/ 3 assets, 2 states, 1 step

Answer 5

1. Positive ψ1 and ψ2 found satisfying
   vec(1, St, Ct) = mx((1+r, S1(t+1), C1(t+1)), (1+r, S2(t+1), C2(t+1))) * vec(ψ1, ψ2)
2. Solution to this system is positive constants ψ1, ψ2

ψi = premium for insurance paying $1 in state i = State (i) price

Question 6

Consequence 1 of no arbitrage opportunities: What measure do we get?

Answer 6

Existence of risk adjusted probabilities
Q1=(1+r)ψ1

By the matrix equation we have ΣQi = 1

Question 7

Consequence 2 of no arbitrage opportunities: What asset value behavior do we force?

Answer 7

Discounted expected value of any asset is a martingale

St=[Q1xS1(t+1)+Q2xS2(t+1)]/(1+r)

Question 8

Consequence 3 of no arbitrage opportunities: What returns do we expect?

Answer 8

The expected risk adjusted return of any asset is risk free rate

(Rearrange consequence 2 for 1+r)

Question 9

Consequence 4 of no arbitrage opportunities: What bounds does the risk free rate have?

Answer 9

Total risk free rate return (1+r) is bounded above and below by the gross returns in the most favorable and least favorable future states

Ψ1R1+Ψ2R2=1 => R1<1+r<R2

If r was greater/less than all possible returns, we would buy/short bonds to short/buy as much S as possible

Question 10

Discounting expected value of assets at r in real world measure results in what?

Answer 10

A submartingale
St<[P1xS1(t+1)+P2xS2(t+1)]/(1+r)

Risky assets in the real world need some risk premium

Question 11

Our First SDE: RW dynamics of stock prices

Answer 11

dSt = μtStdt + σtStdWt

μ creates predicted movement, σ is the unpredictable shock over time dt

Question 12

Can no arbitrage opportunities condition exist in a world with stochastic interest rate?

Neftci 17

Answer 12

Yes, but it is not analytically tractable. Transformation from RN to forward measure reduces complexity

Question 13

Describe the setting for normalization under the Forward Measure and how we get to forward risk adjusted probabilities

Answer 13

In a 2 step model with 4 possible states
Existence of ψ^ij established
Transforming to forward measure changes risk adjusted probabilities to
π^ij = ψ^ij/B_t1

Question 14

What are the steps to use forward measure for pricing?

Answer 14

Because ψ^ij>0, π^ij>0
We multiple expected asset price by corresponding entry of B_t1
B is thought of as the new discount rate
C_t1 = B_t1E[π^ijC^ij]

Question 15

What are 3 important consequences of the forward measure and how does it work better than RN?

Answer 15

Forward adjusted probs satisfy Σπ^ij = 1
Under classic RN measure, forward interest rate is biased, but unbiased in π
You do not need to model a bivariate process in π like you would in RN

Question 16

What is the approximating sun for Riemann Integral
∫f(s)ds over [0,T] and the common explanation of it

Answer 16

Partition [0,T] into n intervals 0=t₀<t₁…<t_n=T
The approximating sum is Σf((t_i+t_i+1)/2)(t_i-t_i+1)

Riemann suggests integral converges to sum of n disjoint rectangles

Question 17

Riemann Integral in stochastic world
What happens/why?

Answer 17

Riemann integration fails! In stochastic environments, the functions vary too much because they involve the Wiener process

Question 18

Classical Total Differential

Also breaks down in stochastic world

Answer 18

Let f(S_t, t) be a function of 2 variables. The total differential is defined as
df = (∂f/∂S_t)dS_t + (∂f/∂t)dt

Question 19

Taylor Expansion of a function around a point

Also breaks down in stochastic world

Answer 19

f(x) = Σ(1/n!)fⁿ(a)(x-a)ⁿ

Question 20

Taylor Expansion for functions of two variables

Answer 20

dV = (∂V/∂S)dS + (∂V/∂t)dt + 1/2(∂²V/∂S²)dS² + 1/2(∂²V/∂t²)dt² + (∂²V/∂S∂t)dSdt + …

Question 21

Name and describe two popular approaches for pricing derivatives

Answer 21

1. Equivalent Martingale Measure
   Notion of no arbitrage gives a measure where assets are martingagles, which are easy to take expectations of
2. Solving the Pricing PDE
   Construct a risk free portfolio and get a PDE implied by no arbitrage opportunities

Question 22

Give the formula for the Stochastic Total Differential

Answer 22

Let F(S_t, r_t, t) be a fn of 2 s.r.v. Then the total differential is
dF = F_sdS_t + F_rdr_t + F_tdt + 1/2 F_ss(dS_t)² + 1/2 F_rr(dr_t)² + F_srdS_tdr_t

Question 23

List 4 Statistical facts about distribution tails under/vs Normal

Neftci 5

Answer 23

1. A dist. with heavier tals than normal means a higher probability of extreme events
2. A normal dist has most ovservations occurring around the center
3. Heavy tailed dists have a more sudden passage from ordinary to extreme events
4. Middle region of a heavy tail dist has relatively less weight than the normal dist

Question 24

What is a Markov Process (in words and math)

Answer 24

A discrete process X₁, …X_t,…
with pdf F(x₁, …x_n) is Markov if
P(X_t+s<=x_t+s|x₁, …x_t) = P(X_t+s<=x_t+s|x_t)

Question 25

Are Jointly Markov processes individually Markov?

Answer 25

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

Question 26

Describe the interest rate model that is univariately Markov and the one that is jointly Markov

Answer 26

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

Question 27

Name three concepts of convergence for pricing derivatives

Answer 27

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

Question 28

Define and give basic examples of Martingale and Submartingale

Neftci 6

Answer 28

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

Question 29

Define a Martingale and give the 3 conditions for it

Answer 29

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

Question 30

Examples of Martingales

Let W_t be a Wiener Process

Answer 30

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

Question 31

2 approaches for finding prob measure where discounted asset prices are martingales

Answer 31

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

Question 32

Donsker Theorem: describe the limit distribution of a sum of IID variables

Answer 32

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

Question 33

Martingale Representation Theorem

Answer 33

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

Question 34

Doob Meyer Decomposition theorem

Answer 34

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

Question 35

The First Stochastic Integral (Discrete)

Uses Doob Meyer to represent a new martingale

Answer 35

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_tk = M_t0 + Σ(1,k)H_ti-1[Z_ti-Z_ti-1]
will also be a martingale

Question 36

First Stochastic Integral (Continuous Doob-Meyer)

Answer 36

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

Question 37

Two approximations of second order Taylor polynomial in stochastic environment

Neftci 7

Answer 37

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

Question 38

What are rare events

Neftci 8

Answer 38

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

Question 39

Problems with characterizing rare events with the Wiener Process

Answer 39

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.

Question 40

What is a model for rare events?

Answer 40

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

Question 41

What is the finite difference approximation of SDE representation?

Neftci 9

Answer 41

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

Question 42

Ito Integral and Properties

Answer 42

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

Question 43

Quadratic Variation fo the Ito Integral

Answer 43

The quadratic variation si <I,I>_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_ti, t_i)(W_ti+1-W_ti)

Question 44

Quadratic variation of an Ito process

Answer 44

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

Question 45

Existence of Ito integral of a general random function f(S_u, u)

Answer 45

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

Question 46

Distribution of Ito integral

Answer 46

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

Question 47

Connection btw Ito and Statonovich integrals

Answer 47

I(t) = ∫(0,t) W_sdW_s
S(t) = ∫(0,t)W_s o dW_s = lim_n>infΣ(W_ti+W_ti+1)/2 (W_ti+1-W_ti)
S(t) - I(t) = t/2

Ito Integral is a martingale, Stratonovich integral is not

Question 48

State Ito Lemma

Nefci 10

Answer 48

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

Question 49

Integral form of Ito Lemma

Answer 49

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

Question 50

Give the formula for the general Ito Isometry of two Stochastic Integrals

Answer 50

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

Question 51

Ito Isometry 2

A r.v. is involved

Answer 51

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

Question 52

Solution to GBM SDE

Neftci 11

Answer 52

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

Question 53

The PDE for Derivative Prices

This PDE only holds under BS assumptions

Neftci 12

Answer 53

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

Question 54

Simple rules in Ito Calculus

Answer 54

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

Question 55

Major difference between BS PDE and exotic PDE

Neftci 13

Answer 55

• 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

Question 56

Seven steps to derive BS PDE

Answer 56

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

Question 57

Ito Lemma when S follows GBM

Answer 57

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

Question 58

BS PDE for Dividend paying stock, currency options, commodities options

Answer 58

(∂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

Question 59

Setting of Girsanov Theorem: Novikov Condition

Neftci 14

Answer 59

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

Question 60

Statement of Girsanov Theorem

Answer 60

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

Question 61

Which measure (RN or RW) should we use for pricing options vs forecasting outcomes?

Answer 61

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.

Question 62

Pricing: Martingale approach vs PDE approach

Neftci 15

Answer 62

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

Question 63

Connection btw Martingale and PDE approaches

Answer 63

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

Question 64

Jensen’s Inequality

QFIQ 113 17

Answer 64

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

Question 65

Why does RN valuation work?

QFIQ 113 17

Answer 65

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.

Question 66

What is meant by Complete and Incomplete market?

Answer 66

• 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

Question 67

Can options be priced in RW measure?

Answer 67

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

Question 68

What is a Weiner Process

Chin

Answer 68

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

Question 69

Basic properties of SWP

Chin

Answer 69

• 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

Question 70

Covariance of 2 SWP

Chin

Answer 70

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

Question 71

What is the Covariance matrix for a set of SWP

Chin

Answer 71

Cov(W_t1, W_t2, W_tn) =
matrix((Cov(W_t1, W_t1), …, Cov(W_t1, W_tn)), …, (Cov(W_t1, W_tn), …, Cov(W_tn, W_tn)) =
Matrix((t₁, …, t₁), …, (t₁, …, t_n))

Question 72

What is the quadratic variation of a SWP

Chin

Answer 72

<W,W>_t = lim_n>infΣ(W_ti+1 - W_ti)² = t

Question 73

Feynman Kac Theorem

Chin

Answer 73

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]

Question 74

Ito Product Rule

Answer 74

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

Question 75

Applying Ito Lemma to a function of a WP F(W_t, t)

Answer 75

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

Question 76

SDE for Logarithm of GBM process

Answer 76

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

Question 77

SDE for PV of a GBM process

Answer 77

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)

Question 78

SDE for product of two GBM processes

Answer 78

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

Question 79

SDE for inverse of a GBM

Answer 79

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

Question 80

SDE for power of a GBM

Answer 80

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

Question 81

Useful tricks for stochastic calculus

Answer 81

• 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