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We understand that the forward price of a stock S is $K=E^{Q_T}[S_T]$ where $E^{Q_T}$ denotes expectation under the T-forward measure. I have the following derivation that produces incorrect results, but couldn't figure out what is wrong with it. Can anyone help me take a look?

$$\begin{equation} \begin{split} E^{Q_T}[S_T] &= E^{Q}[S_T \frac{P(T,T,T)/P(0,0,T)}{B(T)/B(0)}] \\ &=E^Q[S_T \frac{1}{B(T)P(0,0,T)}] \\ &=E^Q[S_T] \text{ (assuming constant interest rate)} \\ &=S_0e^{rT + 0.5\sigma^2T} \\ \end{split} \end{equation}$$ $$ \text{B(t) denotes money market account and P(t,S,T) denotes a zero coupon bond. } B(t)=e^{\int_0^t r(u)du} $$

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    $\begingroup$ For deterministic interest rate, it is true that $E^{Q_T}(S_T)=E^Q(S_T)$, but why $E^Q(S_T) = S_0e^{rT+0.5\sigma^2T}$? $\endgroup$ – Gordon May 16 '18 at 13:53
  • $\begingroup$ isnt that the mean of lognormal distribution? $\endgroup$ – user1559897 May 16 '18 at 13:55
  • $\begingroup$ What is $S_T$ in your case? $\endgroup$ – Gordon May 16 '18 at 13:56
  • $\begingroup$ a stock that follows geometric Brownian motion. $\endgroup$ – user1559897 May 16 '18 at 13:58
  • $\begingroup$ Can you write out the dynamics for $S_t$, and the solution for $S_T$ specifically? $\endgroup$ – Gordon May 16 '18 at 14:00

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