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. 
I would appreciate any advice. Thanks in advance.
 A: 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 $\kappa$.
A: 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. 
