# Risk neutral probability for stock with continuous dividend

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.

• 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. Commented Dec 3, 2020 at 6:52
• 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. Commented Dec 3, 2020 at 21:15

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:

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...

HTH?