Cameron Martin Girsanov - discounted asset price under Q

[vegan]

Hi

Core reading 53 of chapter 14 (Cameron Martin Girsanov theorem) calculates the expected value of the discounted asset price under Q as shown below.

\[E_q[e^{-rt}S_t | F_u] =e^{-rt}S_uE_Q[exp{(r-0.5sigma^{2})(t-u)+sigma(*Z_t - *Z_u)}]\]

\[e^{-rt}S_uE_Q[exp{(r-0.5sigma^{2})(t-u)+0.5sigma^{2}(t -u)}\]


Note: \[*Z_t\] denotes standard brownian motion under Q


Please could someone explain how \[sigma(*Z_t - *Z_u)\] from the first line, simplifies to \[0.5sigma^{2}(t -u)\] of the second line?




Thanks,

 

[Oxymoron]

Expectation under lognormal distribution = Mean + .5*Sigma^2

From Brownian Motion:
E[Zt - Zu] = 0
Var[Zt - Zu] = t-u = E[ (Zt-Zu)^2] - (E[Zt-Zu])^2 = E[(Zt-Zu)^2 + 0

Back to equation:
Mean of the equation you wrote down is simply exp(r*t) which cancels out with exp(-rt) outside bracket

EQ[exp(r−0.5sigma2)(t−u)+sigma(∗Zt−∗Zu)]

= .5*(first term is deterministic and has 0 sigma + 2nd term is sigma^2 * (t-u))

 

[vegan]

Hi thanks for your help, but I am a bit confused by some parts of the answer you've set out.

I'm not sure why the log normal distribution is relevant as my understanding is it only applies to St rather than the standard brownian motion Zt. In the St case, St is assumed to be log normally distributed with mean of exp(logS0 +("mu" -0.5sigma^2)t+0.5(sigma^2)t); so I am confused where your expectation under th log normal distribution has come from; or why it would apply to the question.

I agree that Zt-Zu has mean 0 and variance t-u.

But my understanding is my equation in the original post, only considers the expectation (mean) as opposed to any variance. Therefore the expectation of the second term , i.e. E(exp(sigma (Zt -Zu))) should equal exp(sigma*E (Zt-Zu)=exp (sigma*0)=0. Therefore I am unable to understand why the notes give an answer of exp (0.5sigma^2 (t-u)).

As for the first term in the expectation; my understanding is this is just E (exp (r-0.5sigma^2)(t-u)) which just produces itself, as E (exp (r-0.5sigma^2)(t-u)). If that's right, I don't have any issues with the first term.

Clearly, I am misunderstanding something. Is there anyone that could help me out please?

Thanks

 

[Oxymoron]

But my understanding is my equation in the original post, only considers the expectation (mean) as opposed to any variance. Therefore the expectation of the second term , i.e. E(exp(sigma (Zt -Zu))) should equal exp(sigma*E (Zt-Zu)=exp (sigma*0)=0. Therefore I am unable to understand why the notes give an answer of exp (0.5sigma^2 (t-u)).

E[exp(Zt-Zu)] is not exp(E[Zt-Zu]).

You can approximately visualize this from a simple Taylor series expansion:
LHS: E[1 + (Zt-Zu)/1! + (Zt-Zu)^2/2! + ....]
= E[1] + E[Zt-Zu] + (1/2!)*E[(Zt-Zu)^2] + (1/3!)*E[(Zt-Zu)^3] + .....
= 1 + 0 + .5*(t-u) + 0 + 0 + 0 + ..... = 1 + .5*(t-u)

RHS: exp(E[Zt-Zu]) = exp(0) = 1

 

[vegan]

Thanks so much, that's excellent!

 

[satyam]

Explanation with MGF

Thanks so much, that's excellent!

Consider the MGF of Normal dist, MGF[X(t)] = E [exp(tX)] = exp [mean*t+0.5*variance*t^2] from page 11 of the Tables.

Here, equate X = Zt - Zu, and mean=0 and variance = t-u, also note that "sigma" is there in place of t.

Then, E[exp(sigma*(Zt-Zu)] = exp[mean*sigma+0.5*variance*sigma^2]
= exp[0*sigma+0.5*(t-u)*sigma^2] = exp[.5*(t-u)*sigma^2] = as required.

Hope it helps further.

 

[ActEder]

Do we need to know the 5-step proof (chapter 14 &15)?