Conditional expectation of a functional of an Itô's semimartingale under its equivalent martingale measure

Consider a probability filtered space $(\Omega, \mathcal F, \mathbb F, \mathbb P)$, where $\mathbb F = (\mathcal F_t)_{0\leq t\leq T}$ satisfying the habitual conditions and is generated by $1 d$- Brownian Motion (with $\mathcal F_T = \mathcal F$).

Also, consider a process $X = (X_t)_{t \in [0,T]}$ given by $$X_t= X_0 + \int_0^t \mu_s ~ds +\int_0^t \sigma_s ~dW_s \quad , t \geq 0$$

where $t \in [0,T] \mapsto \mu_t$ and $t \in [0,T] \mapsto \sigma_t \geq 0$ are deterministic and continuous functions.

Suppose that there is a measure $\mathbb Q \sim \mathbb P$ and a $\mathbb Q$-brownian motion $W^{\mathbb Q}$ such that $$X_t= X_0+\int_0^t \sigma_s ~dW_s^{\mathbb Q}\quad , t \leq T$$

I want to evaluate the function $p$ defined as by

$$p(t,x) := \mathbb E^{\mathbb Q} \left [ (X_T-\kappa X_1)^+ | X_t=x \right] \quad \text{for} \ (t,x) \in [0,1]\times(0,\infty)$$

where $\kappa >0$ and $T>1$.

For this, I was claimed to show the following relation (that I failed to demonstrate)

$$p(t,x) = x F(1,\kappa, \int_1^T \sigma_s ^2 ds) \quad \text{if} \ t\in [0,1]$$

where, for $y, K, \gamma^2 >0$ $$F(y,K,\gamma^2) = \mathbb E \left [ (ye^Y -K)^+ \right] \quad \text{with} \ Y \sim \mathcal N(-\gamma^2/2, \gamma^2) \ \text{under} \ \mathbb P$$

I have tried to explore the fact that $( X_T-\kappa X_1)^+ =X_1(X_T/ X_1-\kappa )^+$. It can possibly help those that have some financial mathematical background, to know that the motivation behind this is problem is the pricing of a Forward start option.

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I suspect some typos crept in (for example $p(t,x)$ should depend on $t$ but the formula you suggest does not). Could you check your formulas? – Did Apr 14 '13 at 15:29
@did: There is typo, it's $F(1, \kappa, \int _1 ^T \sigma_s ^2 ds )$ and not $F(1, \kappa, \int _0 ^T\sigma_s ^2 ds)$. – Paul Apr 14 '13 at 16:20
@Did: But I still don't understand why it doesn't depends on $t$ over $[0,1]$. I had not remarked it before you said, thank you. – Paul Apr 14 '13 at 16:26
I suspect there are still some serious problems in this question... What is your source? – Did Apr 14 '13 at 16:27
@Did:An old exam. It's not impossible. – Paul Apr 14 '13 at 16:30

Assume that $\kappa=x=0$, $\sigma_t=1$ and $\mu_t=0$ for every $t\geqslant0$, then $\mathbb Q=\mathbb P$, $(X_t)_{t\geqslant0}$ is a Brownian motion, and $p(t,0)=\mathbb E[(X_T)^+\mid X_t=0]=\mathbb E[(X_{T-t})^+\mid X_0=0]=\sqrt{T-t}\cdot\mathbb E[Z^+]$ where $Z$ is standard normal.

In the same setting, $p(t,0)=\sqrt{T-1+(1-\kappa)^2(1-t)}\cdot\mathbb E[Z^+]$ for every $\kappa$.

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It's a perfect simple reasoning over a particular case in spite of the fact you can not suppose $\kappa =0$ since it's strictly positive. – Paul Apr 14 '13 at 20:32
I am sure you can adapt the example to cover this case. – Did Apr 14 '13 at 20:34
I'm not so sure, but I'll try it. – Paul Apr 14 '13 at 20:54
You missed the condition $x \in (0,+\infty)$ – Paul Apr 14 '13 at 21:01
Actually I did not and, as a matter of fact, this is an aspect of the proposed formula for $p(t,x)$ which makes it rather absurd since $\lim\limits_{x\to0}xF(\cdots)=0$ although $\lim\limits_{x\to0}p(t,x)\ne0$. – Did Apr 14 '13 at 21:48

Note that \begin{align} p(t,x) &:= \mathbb E^{\mathbb Q} \left [ X_1 (X_T/ X_1-\kappa)^+ | X_t=x \right] \\&= \mathbb E^{\mathbb Q} \left [ X_1 \mathbb E^{\mathbb Q} \left [(X_T/ X_1-\kappa)^+ | \mathcal F_1\right] | X_t=x \right] \\&= x \mathbb E^{\mathbb Q} \left [ (X_T/X_1-\kappa )^+ | X_t=x \right] \end{align}

for all $t<1$ and $x \in (0, +\infty)$, since $X$ is a $\mathbb Q$ -martingale. Now, if we include the aditional condition on $\sigma$ that $\sigma_t := \tilde\sigma(X_t) ~X_t$, we can conclude that \begin{align} p(t,x) &= x F(1,\kappa, \int_1^T \tilde\sigma(X_s)^2 ds), \end{align}

where $F$ is defined as in the question BUT under $\mathbb Q$ and not under $\mathbb P$ as it says.

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Why do you replace twice $X_1$ by $X_t=x$ in the identity defining $p(t,x)$? – Did Apr 18 '13 at 9:02
@Did: $X$ is a $\mathbb Q$ -martingale – Paul Apr 18 '13 at 13:28
This is correct. And? – Did Apr 18 '13 at 13:44
See adit to a detailed reasoning. The term $X_1 \mathbb E^{\mathbb Q} \left [(X_T/ X_1-\kappa)^+ | \mathcal F_1\right]$, in 2nd line is a product of martingales, 3rd line is justified by the markov propertie of $X$. – Paul Apr 18 '13 at 13:53
Right, $X$ is a martingale, and? Why the 4th line? Of course, $E[X_1\mid X_t=x]=x$ but here you have $E[X_1Y\mid X_t=x]$ with either $Y=(X_T/X_1-\kappa)^+$ or $Y=E[(X_T/X_1-\kappa)^+\mid X_1]$. How you manipulate this is a mystery to me. – Did Apr 18 '13 at 14:40