I'm trying to price a forward start option with payoff $\Big(\dfrac{S_{T_2}}{S_{T_1}}-1\Big)^+$ with Heston Monte Carlo.
Heston Model: $$ dS_t = rS_tdt + \sqrt{v_t}S_tdW_t^1$$ $$ dv_t = \kappa(m-v_t) + \alpha\sqrt{v_t}dW_t^2$$ The Heston parameter I use is: $\{v_0=0.01,\kappa=1,m=0.01,\alpha=1.5,\rho=-0.6\}$.
I have implemented the QE scheme to do the simulation and get $N$ simulated paths ($M$ time step): $S_k^i$ and $v_k^i$ ($i=1...N$, $k=1...M$).
Assuming $r=q=0$, the obvious way to calculate the price is: $$p = AVG_{i=1...N}\Big[\Big(\dfrac{S_{k(T_2)}^i}{S_{k(T_1)}^i}-1\Big)^+\Big]$$
However, this is another way I can understand this question:
\begin{align*} p &= E\Big[\Big(\dfrac{S_{T_2}}{S_{T_1}}-1\Big)^+ \Big|\mathbb{F}_0\Big]\\ &= E\Big[E\Big[\Big(\dfrac{S_{T_2}}{S_{T_1}}-1\Big)^+ \Big|\mathbb{F}_{T_1}\Big]\Big|\mathbb{F}_0\Big]\\ &= AVG_{i=1...N}\Big[ HestonCall(S_0=1,K=1,T=(T_2-T_1),hestonparam=\{v_0=v^{i}_{k(T_1)},...\}) \Big] \end{align*} where $'...'$ means the other Heston parameters remain unchanged.
Based on my understanding, these two methods should give the same result. However, they are not.
There must be something wrong with the way I come up with the second method, but I completely got stuck here. Could anyone help to point out which part is wrong? Thanks a lot!
