这是一份liverpool利物浦大学ECON308的成功案例

$$
F_{j}=S_{i-1, j} \exp (r \Delta t)=K \exp (r \Delta t)
$$
The up jump transition probability, see Equation $10.114$, is
$$
p_{j}=\frac{F_{j}-S_{i, j}}{S_{i, j+1}-S_{i, j}}
$$
which results in a down jump probability of
$$
1-p_{j}=q=1-\frac{F_{j}-S_{i, j}}{S_{i, j+1}-S_{i, j}}=\frac{S_{i, j+1}-F_{j}}{S_{i, j+1}-S_{i, j}}
$$
Multiplying top and bottom by $S_{i, j}$ we obtain
$$
q=\frac{S_{i, j+1}-F_{j}}{S_{i, j+1}-S_{i, j}}=\frac{\left(S_{i, j+1}-F_{j}\right) S_{i, j}}{\left(S_{i, j+1}-S_{i, j}\right) S_{i, j}}=\frac{S_{i, j+1} S_{i, j}-F_{j} S_{i, j}}{S_{i, j+1} S_{i, j}-S_{i, j}^{2}}
$$
We choose to centre at the spot $S_{i, j+1} S_{i, j}=K^{2}$ and we have
$$
q=\frac{S_{i, j+1} S_{i, j}-F_{j} S_{i, j}}{K^{2}-S_{i, j}^{2}}=\frac{S_{i, j+1} S_{i, j}-F_{j} S_{i, j}}{\left(K-S_{i, j}\right)\left(K+S_{i, j}\right)}=\frac{K^{2}-F_{j} S_{i, j}}{\left(K-S_{i, j}\right)\left(K+S_{i, j}\right)}
$$

ECON308 COURSE NOTES ：

Finite-difference approximations for these derivatives can be obtained by considering a Taylor expansion about the point $f_{i, j}$. We proceed as follows:
$$
\begin{aligned}
&f_{i, j+1}=f_{i, j}+f_{i, j}^{\prime} \Delta S+\frac{1}{2} f_{i, j}^{\prime \prime}(\Delta S)^{2} \
&f_{i, j-1}=f_{i, j}-f_{i, j}^{\prime} \Delta S+\frac{1}{2} f_{i, j}^{\prime \prime}(\Delta S)^{2}
\end{aligned}
$$
Subtracting Equations $10.145$ and $10.146$ we obtain:
$$
f_{i, j+1}-f_{i, j-1}=2 f_{i, j}^{\prime} \Delta S
$$
and so
$$
f_{i, j}^{\prime}=\frac{f_{i, j+1}-f_{i, j-1}}{2 \Delta S}
$$