mutable = truefalse -- This option specifies whether the tree should be mutable or not. The default is true.

The BinomialTree(G, S, $P_u$, $P_d$, opts) calling sequence constructs a recombining binomial tree approximating a certain stochastic process, typically a GeometricBrownianMotion. The constructed tree will be based on the discretizations of the time and the state spaces given by G and S.

Assume that the time grid G consists of N points $T_1$, $T_1$, ..., $T_N$. Then the resulting binomial tree will have N levels, each level representing possible states of the discretized process at time $T_i$, $i=1..N$. The parameter S contains all possible states of the discretized process. The number of elements of S should be equal to $2N-1$, and the elements of S must be sorted in descending order.

At level $i$, $i=1..N$ the tree has $i$ nodes, $S_{i,1}$, ..., $S_{i,i}$. Each node $S_{i,j}$ has two descendants at level $i+1$, $S_{i+1,j}$ (the upper descendant), and $S_{i+1,j+1}$ (the lower descendant). The initial state of the underlying process will be equal to $S_N$. For odd $i$, the states of the underlying at the level $i$ are $S_{N-i}$, $S_{N-i+2}$, ..., $S_{N-2}$, $S_N$, $S_{N+2}$, ..., $S_{N+i-2}$, $S_{N+i}$. For even $i$, the states of the underlying at the level $i$ are $S_{N-i}$, $S_{N-i+2}$, ..., $S_{N-1}$, $S_{N+1}$, ..., $S_{N+i-2}$, $S_{N+i}$.

The transition probabilities (i.e. the probability of going from $S_{i,j}$ to $S_{i+1,j}$ and the probability of going from $S_{i,j}$ to $S_{i+1,j+1}$) are defined by $P_u$ and $P_d$. Both $P_u$ and $P_d$ can be either non-negative real constants or one-parameter operators. If $P_u$ and  $P_d$ are given in the operator form the corresponding transition probabilities at level $i$ will be calculated as $P_u\left(\mathrm{dt}\right)$ and $P_d\left(\mathrm{dt}\right)$ respectively, where $\mathrm{dt}=T_{i+1}-T_i$.

The BinomialTree(T, S, $P_u$, $P_d$, opts) calling sequence is similar except that in this case a uniform time grid with step size $\frac{T}{N}$ is used instead of G. In this case N will be deduced from the size of the state array S.

The BinomialTree(T, N, $S_0$, $S_u$, $P_u$, $S_d$, $P_d$, opts) calling sequence will construct a binomial tree based on a uniform time grid with step size $\frac{T}{N}$. Each tree node $S_{i,j}$ will have two descendants $S_{i+1,j}=S_{i,j}{S}_u$ (the upper descendant) and $S_{i+1,j+1}=S_{i,j}{S}_d$ (the lower descendant). The transition probabilities will be calculated the same way as above. By default $S_d$ is set to $\frac{1}{S_u}$ and $P_d$ is set to $1-P_u$.

The resulting data structure can be inspected using the GetUnderlying and GetProbabilities commands and can be further manipulated using the SetUnderlying and SetProbabilities commands.

$\mathrm{with}\left(\mathrm{Finance}\right)\:$

$S≔\left[7.9\,7.5\,7.1\,6.5\,5.\,3.7\,3.3\,2.95\,2.8\right]$

$S≔\left[7.9\,7.5\,7.1\,6.5\,5.\,3.7\,3.3\,2.95\,2.8\right]$

$T≔\mathrm{BinomialTree}\left(3\,S\,0.3\right)\:$

$\mathrm{TreePlot}\left(T\,\mathrm{thickness}=2\,\mathrm{axes}=\mathrm{BOXED}\,\mathrm{gridlines}=\mathrm{true}\right)$

$\mathrm{TreePlot}\left(T\,\mathrm{thickness}=2\,\mathrm{axes}=\mathrm{BOXED}\,\mathrm{gridlines}=\mathrm{true}\,\mathrm{color}=\mathrm{red}..\mathrm{blue}\,\mathrm{scale}=\mathrm{logarithmic}\right)$

$\mathrm{GetUnderlying}\left(T\,5\,1\right)$

$7.900000000$

$\mathrm{GetProbabilities}\left(T\,3\,1\right)$

$\left[0.3000000000\,0.7000000000\right]$

$\mathrm{P1}≔\mathrm{TreePlot}\left(T\,\mathrm{thickness}=2\,\mathrm{axes}=\mathrm{BOXED}\,\mathrm{gridlines}=\mathrm{true}\right)\:$

$\mathrm{SetUnderlying}\left(T\,5\,1\,8.5\right)$

$\mathrm{P2}≔\mathrm{TreePlot}\left(T\,\mathrm{thickness}=2\,\mathrm{axes}=\mathrm{BOXED}\,\mathrm{color}=\mathrm{red}\,\mathrm{gridlines}=\mathrm{true}\right)\:$

$\mathrm{plots}\left[\mathrm{display}\right]\left(\mathrm{P1}\,\mathrm{P2}\right)$

$G≔\mathrm{TimeGrid}\left(\left[0\,1.5\,2.0\,2.5\,3.0\right]\right)$

$G≔\mathbf{module}\left(\right)\mathbf{end\; module}$

$S≔\left[7.9\,7.5\,7.1\,6.5\,5.\,3.7\,3.3\,2.95\,2.8\right]$

$S≔\left[7.9\,7.5\,7.1\,6.5\,5.\,3.7\,3.3\,2.95\,2.8\right]$

$T≔\mathrm{BinomialTree}\left(G\,S\,0.3\right)\:$

$\mathrm{TreePlot}\left(T\,\mathrm{thickness}=2\,\mathrm{axes}=\mathrm{BOXED}\,\mathrm{gridlines}=\mathrm{true}\right)$

$\mathrm{TreePlot}\left(T\,\mathrm{thickness}=2\,\mathrm{axes}=\mathrm{BOXED}\,\mathrm{gridlines}=\mathrm{true}\,\mathrm{color}=\mathrm{red}..\mathrm{blue}\,\mathrm{scale}=\mathrm{logarithmic}\right)$

$\mathrm{GetUnderlying}\left(T\,5\,1\right)$

$7.900000000$

$\mathrm{GetProbabilities}\left(T\,3\,1\right)$

$\left[0.3000000000\,0.7000000000\right]$

$\mathrm{P1}≔\mathrm{TreePlot}\left(T\,\mathrm{thickness}=2\,\mathrm{axes}=\mathrm{BOXED}\,\mathrm{gridlines}=\mathrm{true}\right)\:$

$\mathrm{SetUnderlying}\left(T\,5\,1\,8.5\right)$

$\mathrm{P2}≔\mathrm{TreePlot}\left(T\,\mathrm{thickness}=2\,\mathrm{axes}=\mathrm{BOXED}\,\mathrm{color}=\mathrm{red}\,\mathrm{gridlines}=\mathrm{true}\right)\:$

$\mathrm{plots}\left[\mathrm{display}\right]\left(\mathrm{P1}\,\mathrm{P2}\right)$

$\mathrm{Su}≔1.1$

$\mathrm{Su}≔1.1$

$\mathrm{Sd}≔0.95$

$\mathrm{Sd}≔0.95$

$T≔\mathrm{BinomialTree}\left(3\,20\,100\,\mathrm{Su}\,0.3\,\mathrm{Sd}\,0.7\right)\:$

$\mathrm{TreePlot}\left(T\,\mathrm{thickness}=2\,\mathrm{axes}=\mathrm{BOXED}\,\mathrm{gridlines}=\mathrm{true}\right)$

$\mathrm{TreePlot}\left(T\,\mathrm{thickness}=2\,\mathrm{axes}=\mathrm{BOXED}\,\mathrm{gridlines}=\mathrm{true}\,\mathrm{color}=\mathrm{red}..\mathrm{blue}\,\mathrm{scale}=\mathrm{logarithmic}\right)$

$\mathrm{T2}≔\mathrm{BinomialTree}\left(3\,20\,100\,\mathrm{Su}\,0.3\,\mathrm{Sd}\,0.7\,\mathrm{mutable}=\mathrm{false}\right)\:$

$\mathrm{plots}\left[\mathrm{display}\right]\left(\mathrm{TreePlot}\left(\mathrm{T2}\,\mathrm{thickness}=2\,\mathrm{axes}=\mathrm{BOXED}\,\mathrm{gridlines}=\mathrm{true}\,\mathrm{color}=\mathrm{red}\right)\,\mathrm{TreePlot}\left(T\,\mathrm{thickness}=2\,\mathrm{axes}=\mathrm{BOXED}\,\mathrm{gridlines}=\mathrm{true}\,\mathrm{color}=\mathrm{blue}\right)\right)$

$\mathrm{T3}≔\mathrm{BinomialTree}\left(3\,20\,100\,\mathrm{Su}\,0.3\,\mathrm{mutable}=\mathrm{false}\right)\:$

$\mathrm{plots}\left[\mathrm{display}\right]\left(\mathrm{TreePlot}\left(\mathrm{T3}\,\mathrm{thickness}=2\,\mathrm{axes}=\mathrm{BOXED}\,\mathrm{gridlines}=\mathrm{true}\,\mathrm{color}=\mathrm{red}\right)\,\mathrm{TreePlot}\left(T\,\mathrm{thickness}=2\,\mathrm{axes}=\mathrm{BOXED}\,\mathrm{gridlines}=\mathrm{true}\,\mathrm{color}=\mathrm{blue}\right)\right)$

$\mathrm{GetUnderlying}\left(T\,5\,1\right)$

$146.4100000$

$\mathrm{GetProbabilities}\left(T\,3\,1\right)$

$\left[0.3000000000\,0.7000000000\right]$

$\mathrm{P1}≔\mathrm{TreePlot}\left(T\,\mathrm{thickness}=2\,\mathrm{axes}=\mathrm{BOXED}\,\mathrm{gridlines}=\mathrm{true}\right)\:$

$$\begin{gathered} \mathbf{for}i\mathbf{to}20\mathbf{do} \\ \mathrm{SetUnderlying}\left(T\,i\,i\,100{0.9}^{i-1}\right) \\ \mathbf{end}\mathbf{do} \end{gathered}$$

$\mathrm{P2}≔\mathrm{TreePlot}\left(T\,\mathrm{thickness}=2\,\mathrm{axes}=\mathrm{BOXED}\,\mathrm{color}=\mathrm{red}\,\mathrm{gridlines}=\mathrm{true}\right)\:$

$\mathrm{plots}\left[\mathrm{display}\right]\left(\mathrm{P1}\,\mathrm{P2}\right)$

Brigo, D., Mercurio, F., Interest Rate Models: Theory and Practice, New York: Springer-Verlag, 2001.

Glasserman, P., Monte Carlo Methods in Financial Engineering, New York: Springer-Verlag, 2004.

Hull, J., Options, Futures, and Other Derivatives, 5th. edition. Upper Saddle River, New Jersey: Prentice Hall, 2003.

The Finance[BinomialTree] command was introduced in Maple 15.

For more information on Maple 15 changes, see Updates in Maple 15.