# Is independence preserved in this special setting under a change of measure?

This is a question due to the answer of Did in this post Independent increments of $X_t:=\int_0^t\phi(s) dW_s$. Precisely, we assume that the dynamics of a stock prices follows

$$dS_t=S_t(\phi_tdW_t)$$

where $\phi_t$ is a deterministic function, and assume bounded. Hence $S_t=\mathcal{E}(\phi\bullet W)=S_0e^{\int_0^t\phi_sdW_s-\frac{1}{2}\int_0^t\phi_s^2ds}$. For simplicity, $S_0=1$ and, regarding the question above, $X_t:=\int_0^t\phi_sdW_s$. Let $K>0$ be a constant, and we want to find the arbitrage free price of an European call option with strike $K$ and maturity $T$ at time $t$, i.e.

$$\pi_t:=E[(S_T-K)^+|\mathcal{F}_t]$$

As usual, with $A=\{S_T>K\}$ write this as

$$\pi_t=E[S_T\mathbf1_A|\mathcal{F}_t]-KE[\mathbf1_{A}|\mathcal{F}_t]$$

The second term can be easily calculated, using the above question, i.e.

$$KE[\mathbf1_{A}|\mathcal{F}_t]=KP[\frac{S_T}{S_t}>\frac{K}{S_t}|\mathcal{F}_t]$$

since $\frac{S_T}{S_t}$ is independent of $\mathcal{F}_t$ and $\frac{K}{S_t}$ is $\mathcal{F}_t$ measurable, we get

$$KP[\frac{S_T}{S_t}>\frac{K}{S_t}|\mathcal{F}_t]=KP[Y>\frac{K}{x}]|_{x=S_t}$$

where $Y$ is lognormal distributed with mean $-\frac{1}{2}\int_t^T\phi_s^2 ds$ and variance $\int_t^T\phi_s^2ds$. To calculate the first conditional expectation, $E[S_T\mathbf1_A|\mathcal{F}_t]$, I would define a new measure (on $\mathcal{F}_T$ through

$$\frac{dQ}{dP}:=S_T$$

since $S_T$ is a $P$ Martingale and $E[S_0]=1$ (if $S_0\not=1$, just use the measure $\frac{dQ}{dP}:=\frac{S_T}{S_0}$). Hence the density process is given by $Z_t:=E[\frac{dQ}{dP}|\mathcal{F}_t]=S_t$. Using Bayse, we get

$$S_tE[\frac{S_T}{S_t}\mathbf1_A|\mathcal{F}_t]=S_tE_Q[\mathbf1_A|\mathcal{F}_t]$$.

If we know that under $Q$, $\frac{S_T}{S_t}$ is still independent of $\mathcal{F}_t$, we could conclude in the same way as above. If this is not the case, how would you then calculate the first conditional expectation?

-

To calculate the first conditional expectation, $\mathbb E[S_T\mathbf 1_A|\mathcal{F}_t]$, I would...
...simply note (as you already explained yourself) that $S_T=YS_t$ where the distribution of $Y$ is lognormal with known parameters, and $Y$ is independent of $\mathcal F_t$.
Since $A=[S_T\gt K]=[YS_t\gt K]$, this remark yields the identity $$\mathbb E[S_T\mathbf 1_A|\mathcal{F}_t]=\mathbb E[YS_t\mathbf 1_{YS_t\gt K}|\mathcal{F}_t]=G(S_t),$$ where the function $G$ is defined by $$G(x)=\mathbb E(Yx\mathbf 1_{xY\gt K})=x\mathbb E(Y\mathbf 1_{Y\gt K/x}).$$
ok, this seems easier. However I guess the question about the independence can be answered with yes (if we assume that $\mathcal{F}_t$ is generated by $W_t$. Using Girsanov, we can find a new Brownian motion (w.r.t to $Q$). Since the filtration generated by these Brownian Motions are the same, the result should still be valid. Is this correct? –  math Feb 10 '13 at 14:46