I'm attempting to derive the Turnbull and Wakeman (1991) asian option pricing algorithm.
I am currently stuck at deriving the cumulants.
First, we let:
$$A(t)=\frac{S[T-n+1]+S[T-n+1]R_{(T-n+1)+1}+S[T-n+1]R_{(T-n+1)+1}...S[T]}{n}$$ $$=\frac{[S(0)R_{i}][L_{i+1}]}{n}$$
where $i=T-n+1$, $L_{T+1}=L_{i+1}=1$, $L_{T-n+1+j}=L_{i+j}=1+R_{i+j}L_{i+j+1}$, and $j\in[1,...,n-1]$.
We know that:
$$ [S(0)R_{i}][L_{i+1}]>K$$ $$Y\equiv R_{i}L_{i+1}>\frac{nK}{S(0)}$$
and that the kth moment of Y is:
$$E(Y^{k})=E(R_{i}^{k})E(L_{i+1}^{k})$$
We know that $E(R_{i}^{k})$ is log-normal distributed, so:
$$E[R_{i}^{k}]=e^{(\mu-\frac{1}{2}\sigma^{2}h)k+\frac{1}{2}\sigma^{2}hk^{2}}$$
where $h=(T-t)$.
We can derive that:
$$E[L_{i+1}]=1+E[R_{i+1}]E[L_{i+2}]$$ $$E[L_{i+1}^{2}]=1+E[R_{i+1}^{2}]E[L_{i+2}^{2}]+2E[R_{i+1}]E[L_{i+2}]$$ $$E[L_{i+1}^{3}]=1+E[R_{i+1}^{3}]E[L_{i+2}^{3}]+3[R_{i+1}^{2}]E[L_{i+2}^{2}]+3[R_{i+1}]E[L_{i+2}]$$ $$E[L_{i+1}^{4}]=1+E[R_{i+1}^{4}]E[L_{i+2}^{4}]+4E[R_{i+1}^{3}]E[L_{i+2}^{3}]+6[R_{i+1}^{2}]E[L_{i+2}^{2}]+4[R_{i+1}]E[L_{i+2}]$$
What should I do next?