2

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.

qp212223
  • 1,626

1 Answers1

2

With a continuous dividend yield $\delta$, the end-of-period expected value of the stock price under the risk-neutral probability measure should be the forward price

$$\tag{*}\mathbb{E}(S) = F= S_0e^{(r-\delta)\Delta t},$$

and then it easily follows that

$$p = \frac{ S_0e^{(r-\delta)\Delta t}-S_d}{S_u-S_d} = \frac{ e^{(r-\delta)\Delta t}-d}{u-d}= \ldots $$

We can show that (*) is the correct forward price by an arbitrage argument. If $F > S_0e^{(r-\delta)\Delta t}$ then we could borrow $S_0e^{-\delta \Delta t}$ in cash, buy $e^{-\delta \Delta t}$ shares of stock and enter into a forward contract to sell the stock at the end of the period for $F$. At the end of the period with dividend reinvestment the number of shares grows to $1$. Delivering the one share of stock we receive the delivery price $F$ and repay our borrowings with interest totaling $S_0e^{(r- \delta) \Delta t}$ for a risk free gain of

$$F - S_0e^{(r-\delta)\Delta t} > 0$$

This is impossible in the absence of abritrage opportunity. By a similar argument we can show that it is impossible to have $F < S_0e^{(r-\delta)\Delta t}$.

RRL
  • 90,707