Setting: binomial tree with one step over time $\Delta t$. I'm trying to derive the risk neutral probability for a stock which pays a continuous dividend, say $\delta$. i.e. probability $p$ such that $$e^{r \Delta t} S_0 = S_u p + S_d(1-p)$$

where $S_u, S_d$ are the values of the stock in the up and down states respectively. This immediately gives $$p = \frac{S_0 e^{r \Delta t} - S_d}{S_u - S_d}$$

Now if we assume $S$ has volatility $\sigma$, we should be getting $S_d = S_0 e^{-\sigma \sqrt{\Delta t} - \delta \Delta t}$ and $S_u = S_0 e^{\sigma \sqrt{\Delta t} - \delta \Delta t}$ so that $$p = \frac{e^{r \Delta t} - e^{- \sigma \sqrt{\Delta t} - \delta \Delta t} }{ e^{ \sigma \sqrt{\Delta t} - \delta \Delta t} - e^{- \sigma \sqrt{\Delta t} - \delta \Delta t}} = \frac{e^{(r+ \delta)\Delta t} - e^{- \sigma \sqrt{\Delta t}} }{ e^{ \sigma \sqrt{\Delta t}} - e^{- \sigma \sqrt{\Delta t}}}$$

but this is wrong because the formula that's given in my course's lecture notes on this is $$ p = \frac{e^{(r- \delta)\Delta t} - e^{- \sigma \sqrt{\Delta t}} }{ e^{ \sigma \sqrt{\Delta t}} - e^{- \sigma \sqrt{\Delta t}}}$$

(the only difference is the $r-\delta$ in the numerator instead of the $r+ \delta$). I don't understand why my assumptions on the values for $S_u$ and $S_d$ are wrong. Any help would be massively appreciated.

MY POTENTIAL EXPLANATION: perhaps the value of $S_u$ should be $S_0 e^{\sigma \sqrt{\Delta t} + \delta \Delta t}$ (and similarly with $S_d$) because we work with the payoff of owning one unit of the stock, so if we increase with upward factor $e^{\sigma \sqrt{\Delta t}}$ we GAIN the value of the dividend, not lose it.

  • 3
    $\begingroup$ Your first equation should read $e^{(r-\delta)\Delta t}$ on the LHS. All else follows then. The expected price must equal the forward price. And the forward includes any dividend expectation. $\endgroup$ Commented Dec 3, 2020 at 6:52
  • $\begingroup$ This is for the same reason I give in my potential explanation, right? The payoff of owning one share at time $0$ is $S_0\exp(\pm \sigma \sqrt{\Delta t} + \delta \Delta t)$ after $\Delta t$ because of the dividend. $\endgroup$
    – qp212223
    Commented Dec 3, 2020 at 21:15

1 Answer 1



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:

  1. $\mathbb{E_Q}\left(S_{t+\Delta t}\right)=p_{\mathbb{Q}}S_tU+(1-p_{\mathbb{Q}})S_t1/U\stackrel{!}{=}F_{t+\Delta t}=S_te^{(r-y)\Delta}$
  2. $\mathbb{V_Q}\left(S_{t+\Delta t}\right)\stackrel{!}{=}F^2\left(e^{\sigma^2\Delta t}-1\right)$ 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...



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