Let the price process $S_{t}, 0 \leq t \leq T$, be a diffusion, and savings account be $\beta_{t}$ such that the Equivalent Martingale Measure $Q$ exists. Let $C_{T}=g\left(X_{T}\right)$ be the claim at time $T$, for a bounded function $g$. Show that this claim is attainable, and find a replicating self-financing portfolio for this claim.

My attempt:

The claim is attainable as $\frac{S_t}{\beta_{t}}$ is a $Q$ -martingale and an admissible strategy exists. Therefore $\frac{V(t)}{\beta(t)}$ is also a Q martingale.

The Self replicating portfolio can be found as $V_{t}=a_{t} s_{t}+b_{t} e^{r t}$

$d V_{t}=a_{t} d s_{t}+r b_{t} e^{r t} d t$ is self financing.

let $\left.b_{t}=\left(V_{t}-a_{t} s_{t}\right)\right) e^{-r t}$ then we have

$d v_{t}=a_{t} d s_{t}+r\left(V_{t}-a_{t} s_{t}\right) d t$

$d V_{t}=a_{t} d s_{t}+r V_{t} d t-r a_{t} s_{t} d t$

I am not sure how to carry on further for this question.


With EMM $Q$, associated $Q$-Brownian motion $W$, filtration ${\cal F}$, and

$$d\beta_t = r_t \beta_t dt, \; \beta_t ={\rm e}^{\int_0^t r_u du},$$

consider martingale:

$$ M_t =E\left[{\rm e}^{-\int_0^T r_u du} C_T | {\cal F}_t\right]. $$

By martingale representation theorem, there is a process $N_t$ such that

$$ M_t = M_0 + \int_0^t N_u dW_u, $$

where $$ M_0=E\left[{\rm e}^{-\int_0^T r_u du} C_T\right].$$

With given

$$ d(\beta_t^{-1} S_t) = \sigma_t \beta_t^{-1} S_tdW_t$$

under $Q$, we have:

$$ dM_t = N_t dW_t = a_t \sigma_t \beta_t^{-1} S_t dW_t = a_t d(\beta_t^{-1} S_t) $$

for $a_t$ chosen to be

$$ a_t := \frac{N_t\beta_t}{\sigma_t S_t }.$$

Strategy $a_t$ and $b_t:= \beta_t^{-1}(M_t-a_tS_t)$

$$\Pi_t := b_t\cdot \beta_t + a_t \cdot S_t = \beta_t M_t$$

is admissible (under $Q$, $\beta_t^{-1}\Pi_t$ is a martingale) and self-financing as

$$ d(\beta_t^{-1}\Pi_t) = dM_t = a_t d(\beta_t^{-1} S_t). $$

We also note that:

$$ \Pi_T = \beta_T M_T = C_T. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.