CHANGE OF MEASURE. D.Majumdar

Transcription

1 CHANGE OF MEASURE D.Majumdar

2 We had touched upon this concept when we looked at Finite Probability spaces and had defined a R.V. Z to change probability measure on a space Ω. We need to do the same thing when we shift from Real World Probability Measure P to Risk Neutral Probability Measure Q in financial markets. We had initially defined Z(ω) = Q(ω) P(ω) But when Ω is uncountably infinite we will have P(ω) = Q(ω)=0 for every ω Ω. Z(ω) = Q(ω) P(ω) = 0 0 no longer makes sense. But we can re write 6.1 as Z(ω)* P(ω) = Q(ω) 6.2 This eq. is still meaningful with both sides = 0, but we still have no information about the relationship between Z, P and Q because for any value of Z this eq. will always hold. If we forget for a moment that P(ω) = Q(ω)=0, then this eq. tells us:- 1. If Z > 1, then P < Q 2. If Z < 1, then P > Q This gives us an idea on how we should adjust our probabilities while moving from P to Q. But rather than doing this as ω by ω we have to do this set by set.

3 THEOREM 1 Let (Ω,F,P) be a Probability space and let Z be an a.s. nonnegative R.V {P(Z>0) = 1} with E[Z] = 1.For A F, we define Q(A) = A Z ω dp(ω)..(6.3). Then Q(A) is a probability measure. Remember in order for Q(A) to be a Probability Measure it needs to satisfy 1. Q(Ω) = 1 ڂ) Q 2. If A 1, A 2,. Are disjoint sets, then i=1 A i ) = σ i=1 Q(A i ) Proof :- Q(Ω) = Ω Z ω dp(ω) = E[Z] = 1 from above For rest of the proof refer to Shreve. Furthermore, if X is a nonnegative R.V., then E Q X = E[XZ].6.4 If Z is a.s. strictly +ve we also have, E[Y] = E Q [ Y Z ].6.5 For every non-negative R.V. Y

4 EQUIVALENT PROBABILITY MEASURE Definition :- Let Ω be a nonemepty set and F a σ-algebra of subsets of Ω. Two probability measures P and Q on (Ω, F) are said to be equivalent if they agree which sets in F have probability 0. Based on Theorem 1 in the previous slide, P(Z>0) = 1, P and Q are equivalent Probability Measures. Let s say A Fand P(A) = 0 Q(A) = A Z ω dp(ω) = Q(A) = Ω I A (ω)z ω dp(ω) = 0 {The R.V. I A Z under P is 1 if A happens a.s. 0. I A = 0 otherwise } Let s say B Fand Q(B) = 0. Then we can also say 1 Z I B = 0 a.s. under Q probability E Q I B Z = 0 = E[I B] -> P(B) = 0 From 6.5 This shows that P and Q agree which sets have Probability 0. As the sets with Probability 1 are complements of the sets with Probability 0, P and Q agree which sets have Probability 1 as well. As P and Q are equivalent, we do not need to state which measure we mean when we say an event happens a.s.

5 In a financial model we refer to computations done under the actual measure as computations in the real world and computations done under the risk neutral measure as computations in the risk neutral world. The question that comes to mind is whether prices computed in the risk neutral world are appropriate for the real world in which we live and have our actual PnL? There is only 1 world in the models. There is a single sample space Ω representing all possible future states of the financial markets, and there is a single set of asset prices, modelled by random variables. We sometimes work in this world assuming that probabilities are given by an empirically estimated actual probability measure and sometimes assuming that they are given by risk-neutral probabilities, but we do not change our view on the world of possibilities. A hedge that works a.s. under 1 assumption of probabilities, will work a.s. under the other assumption as well, since they are equivalent probability measures.

6 EXAMPLE Let P be the uniform measure on [0,1] and we have another measure =[ Q[a,b a b 2ωdω = a 2 b 2, 0 a b We can transform this into the Lebesgue Integral Q[a,b]= [a,b] 2ωdP(ω)..2.2 We can do this because P(d ω) = d ω, as P is an uniform measure Remember B[0,1] is the σ algebra generated by the closed intervals, 2.2 is valid for all closed intervals [a,b] [0,1], hence it is valid for all Borel subsets of [0,1] Q(B) = B 2ωdP(ω) for every Borel set B R. This is similar to eq. 6.3, when Z(ω) = 2 ω. Z(ω) > 0 a.s. {ω [0,1], P(0) = 0} E[Z] = Ω 0 = 2ωdP(ω) 1 2ωdω = 1 Eq. 6.4 tells us that if X is a nonnegative R.V., then E Q X = E[XZ] 0 1 X ω dq ω 0 = 1 X ω 2ωdω. From this we can say dq(ω) = 2ωdω = 2ωdP(ω).6.6

7 When we have P, Q and Z as in Theorem, we can say Ω X ω dq ω = Ω X ω Z ω dp(ω), Ω Y ω dp ω = Ω Y ω Z ω dq(ω) We can formally write Z ω = dq ω dp(ω) Definition :- Let (Ω, F,P) be a probability space, let Q be another probability measure on (Ω, F), that is equivalent to P, and let Z be an a.s. positive R.V. that relates P and Q via eq Then Z is called the Radon-Nikodym derivative of Q w.r.t. P, and we write, Z = dq dp

8 EXAMPLE Let X be a Standard Normal R.V. defined on some probability space (Ω, F, P). When X is a standard normal R.V. we know :- μ X B = P X B = B φ x dx for every Borel subset B of R..(6.7) Where φ x form = 1 2π e x2 2 is the standard normal density. This can be reduced to the b P{X b} = φ x dx for every b R..6.8 We are also aware that E[X] = 0 and Var[X] = 1 Let us define another R.V. Y = X + θ, where θ is a +ve constant. We have E[Y] = θ, Var[Y] = 1. We would like to change to a new probability measure Q, on the same sample space Ω, such that Y is a standard normal random variable. Which means E Q Y = 0, Var Q Y = 1. We would like to achieve this by redistributing the probabilities. We want to change the distribution of Y, without changing the R.V. Y. In financial modelling, risk neutral probability measure changes the distribution of asset prices without changing the asset price themselves.

9 Let s define a R.V., Z(ω) = e θx ω 1 2 θ2 for all ω Ω. In order for Z to be a Radon Nikodym Derivative it needs to satisfy the following 2 properties:- 1. Z ω > 0 for all ω Ω. Z is an exponential hence this is obvious. 2. E[Z] = 1 Proof E[Z] = e θx 1 2 θ2 * 1 2π e x2 2 dx = 1 2π e θx 1 2 θ2 x2 2 dx = 1 2π e 1 2 x+θ 2 dx Let x + θ = y, dx = dy E[Z] = 1 2π e 1 2 y 2 dy = 1 So now we can create a new Probability Measure Q Q(A) = A Z ω dp ω for all A F.6.9 Now we need to show that Y is a standard normal R.V, and we can do that with the help of the Radon Nikodym Derivative

10 Q(Y b) = ω:y ω b Z ω dp ω Ω = I ω:y ω b Z ω dp ω Ω = I ω:x ω b θ e θx ω 1 2 θ2 dp ω Now this is a function in X which is a standard normal variable and hence we can transform this into I x b θ e θx 1 2 θ π b 2π e x2 e 1 2 y 2 dy = Q(Y b) 2 dx = 1 2π b θ e θx 1 2 θ2 e x2 2 dx = 1 2π b θ e 1 2 x+θ 2 dx = This shows that Y is a standard normal random variable under probability measure Q. Theorem(Radon Nikodym) Let P and Q be equivalent probability measures defined on (Ω,F). Then there exists and a.s. +ve R.V. Z such that E[Z] = 1 and Q(A) = A Z ω dp ω for all A F In Financial Modelling we are interested in Risk Neutral probabilities, and these must b equivalent to the actual probability measure. How can such probability measures Q arise? As we have seen, we begin with P and an a.s. +ve R.V Z and constructed the equivalent probability measure Q. This is the only way to find a probability measure Q equivalent to P. The proof is beyond the scope of Shreve s book.

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