Prelude
A valuation tree introduces a lattice structure for derivatives valuation. To not overburden notation, let us define the tree by $K$, the fixed number of nodes exiting each state with corresponding probabilities $p_1,p_2,\ldots,p_K$ and relative jump sizes $J_1,J_2,\ldots,J_K$. Forthermore, $N$ is the number of time steps in the tree. With a time-to-maturity of $T$, the length of each time step is $\Delta t=T/N$. We assume a time-invariant tree, i.e. all parameters are fixed.
With the choice of $K$ ($K=2$ binomial, $K=3$ trinomial...) we introduce $2K-1$ degrees of freedom to our model: $K-1$ jump probabilities (they need to sum to one) and $K$ jump sizes.
Under the risk neutral measure, the tree must induce the risk neutral expectation at each time step:
$$S_t\sum p_{k}J_k=F(t+\Delta t)$$
I.e. the (risk neutral) expectation of the asset price at the next step is the forward price. Thus, we have $2K-2$ degrees of freedom in any tree.
We finally fix the tree's parameters by adding model assumptions: We could add computational features, e.g. assuming a recombining tree, reducing the space complexity from $O(N^K)$ to $O(N^1)$; or we might impose a certain distributional assumptions and try to match its moments.
Specific example: Binomial Tree
Given the prelude, in a binomial tree we are left with the two free asset jump parameters $J_1\equiv U$,$J_2\equiv D$.
A canonical binomial tree introduces the assumption that the tree has to be recombining, i.e. $UD=1$, leaving us with one free parameter, $U$. The classical Cox-Ross-Rubinstein 1979 binomial tree postulates that $U$ be chosen such that the distribution of $S_{t+\Delta t}$ converges to a lognormal distribution with risk neutral drift and variance $\sigma^2 \Delta t$ as $\Delta t \to 0$. Thus:
- $\mathbb{E_Q}\left(S_{t+\Delta t}\right)=p_{\mathbb{Q}}S_tU+(1-p_{\mathbb{Q}})S_t1/U\stackrel{!}{=}F_{t+\Delta t}=S_te^{(r-y)\Delta}$
- $\mathbb{V_Q}\left(S_{t+\Delta t}\right)\stackrel{!}{=}F^2\left(e^{\sigma^2\Delta t}-1\right)$ or, more simply, $\mathbb{E_Q}\left(S_{t+\Delta t}^2\right)\Rightarrow p_{\mathbb{Q}}U^2+(1-p_{\mathbb{Q}})\frac{1}{U^2}\stackrel{!}{=}e^{2(r-y)\Delta}e^{\sigma^2\Delta t}$
At closer inspection, we see that there is only one dof ($U$). You could solve for this nonlinear system of equations via some root search, or you do it the old school way: After solving for the risk neutral probability,
$$
\begin{align}
p_\mathbb{Q}S_{t+\Delta t}^u+\left(1-p_\mathbb{Q}\right)S_{t+\Delta t}^d\stackrel{!}{=}F_{t+\Delta t}&=S_te^{(r-y)\Delta t}\\
\Leftrightarrow p_\mathbb{Q}U+\left(1-p_\mathbb{Q}\right)D&=e^{(r-y)\Delta t}\\
\Rightarrow p_{\mathbb{Q}}=\frac{e^{(r-y)\Delta t}-D}{U-D}
\end{align}
$$
we need to find a factor $U$ such that condition 2 holds for $\Delta t \to 0$. Let's break it out:
$$
\begin{align}
p_\mathbb{Q}U^2+(1-p_\mathbb{Q})D^2&=F^2e^{\sigma^2\Delta t}\\
\frac{F-D}{U-D}(U+D)(U-D)+D^2&=F^2e^{\sigma^2\Delta t}\\
FU+FU^{-1}-1&=F^2e^{\sigma^2\Delta t}\\
U+U^{-1}&=F^{-1}+Fe^{\sigma^2\Delta t}\\
\end{align}
$$
At this point, let's take $\Delta t \to 0$ and linearize all terms around $X=e^x\approx 1+ x$, and linearise $U\approx 1 + u + \frac{1}{2}u^2$, as well as $U^{-1}\approx 1-u+\frac{1}{2}u^2$ for some yet unknown $u$:
$$
\begin{align}
(1+u+\frac{1}{2}u^2)+(1-u+\frac{1}{2}u^2)&=(1-f)+(1+f)(1+\sigma^2\Delta t)\\
\Leftrightarrow 2+u^2&=2+\sigma^2\Delta t + f\sigma^2\Delta t
\end{align}
$$
Now as $\Delta t \to 0$, all higher order terms vanish and we are left with
$$
2+u^2=2+\sigma^2\Delta t \Rightarrow u=\sigma \sqrt{\Delta t}
$$
As stated in the prelude, there exist multiple ways to solve this. You could even define your binomial tree differently and start by postulating condintions on the log moments, i.e. that $p log(U) + (1-p)log(D)=\mu$ and such...
HTH?
