Relating Discrete Proportional Dividends to Continuous Yield

# How to understand the compatibility between the discrete and continuous dividend payments


# How to understand the compatibility between the discrete and continuous dividend payments












We know that a stock has discrete dividend payings $D_i$ at $t_i$, in the pricing of a forward, we will calculate the discount cash flows $$\sum e^{-rt_i}D_i$$ then minus the sum of discount cash flows from spot time $S_0.$

And if stock has a continuous dividend paying rate $D,$ we have the forward price $$e^{-DT}S_0 - e^{-rT}K.$$ So how to understand $e^{-DT}S_0$ is same as $S_0 - \sum e^{-rt_i}D_i$ i.e the spot price minus the discounted cash flows of dividend. Or when $\Delta t = t_{i+1} - t_i\rightarrow 0，$ how to obtain the compatible result.

Suppose, we pay dividend rate $\alpha_i$(include multiplying time $\Delta t$) at time $t_i$ with $i =1,2,3$ and $S_{t_i-}$ means the price before dividend paying, we have $$S_{t_i} = (1 - \alpha_i)S_{t_i-}.$$ And we use the factor that the discounted stock price before dividend paying should be martingale. So we receive dividend $\alpha_3 S_{t_3-}$ at time $t_3,$ the expected value at time $t_2$ under risk neutral measure should be $$E[e^{-r(t_3-t_2)}\alpha_3 S_{t_3-}] = \alpha_3 S_{t_2}$$ $$= \alpha_3(1 - \alpha_2)S_{t_2-}.$$ Again to discounted to time $t_1,$ we have $$\alpha_3(1 - \alpha_2)(1 - \alpha_1)S_{t_1-}.$$ Finally, we have discounted value at time $0$ of dividend $t_3$ $$\alpha_3(1 - \alpha_2)(1 - \alpha_1)S_0$$ then dividend $t_2$ is $$\alpha_2(1 - \alpha_1)S_0,$$ then dividend $t_1$ is $$\alpha_1S_0.$$ Then the initial value minus the discounted dividend should be $$S_0 - \alpha_3(1 - \alpha_2)(1 - \alpha_1)S_0 - \alpha_2(1 - \alpha_1)S_0 - \alpha_1S_0$$ $$=(1-\alpha_3)(1 - \alpha_2)(1 - \alpha_1)S_0.$$

## Answer by Quantuple (score 7)

https://quant.stackexchange.com/a/35596

This equivalence can only be written for discrete proportional dividends. For discrete cash dividends the two spot diffusion models are too different for that relationship to be written in general form (since the div yield model guarantees strictly positive future equity prices, while using discrete cash dividends does not).

More specifically, if you have $N$ evenly spaced discrete proportional dividend payments over $[0,T[$ the div over the first period makes you move from $S_0$ to $S_0(1-q\Delta t)$ with $\Delta t=T/N$. Denote the resulting spot value by $S_{\Delta t}$. The next makes you move from $S_{\Delta t}$ to $S_{\Delta t}(1-q\Delta t)$ etc.

At the end of the day you get: $$S_{T=N\Delta t} = S_0(1-q\Delta t)^N = S_0(1-q\Delta t)^{T/\Delta t}$$ Now taking the limit as $\Delta t \to 0$, knowing that $\lim_{x \to 0} \exp(-x) = 1-x$ you heuristically get: $$\lim_{\Delta t \to 0} S_T = S_0 \exp(-q \Delta t)^{T/\Delta t} = S_0 \exp(-qT)$$

Edit

For discrete proportional dividends (payment date = ex-date), under $\Bbb{Q}$

$$ dS_t = r S_t dt + \sigma S_t dW_t^\Bbb{Q} - \sum_i D(S_{t_i^-}) \delta(t-t_i) dt $$

$$ D(S_{t_i^-})=\alpha_i S_{t_i^-}, \alpha_i \in [0,1[ $$ Applying Itô's lemma for semi-martingales with jumps and integrating from $t=0$ to $t=T$ yields \begin{align*} \ln S_T - \ln S_0 &= \left(r - \frac{\sigma^2}{2} \right) T + \sigma W_T^{\Bbb{Q}} + \sum_{i : 0 < t_i \le T} \left[ \ln(S_{t_i}) - \ln(S_{t_i^-}) \right]\\ &= \left(r - \frac{\sigma^2}{2} \right) T + \sigma W_T^{\Bbb{Q}} + \sum_{i : 0 < t_i \le T} \left[ \ln \left( \frac{S_{t_i^-}-D(S_{t_i^-})}{S_{t_i^-}} \right) \right]\\ &= \left( r - \frac{\sigma^2}{2} \right) T + \sigma W_T^{\Bbb{Q}} + \sum_{i : 0 < t_i \le T} \ln(1-\alpha_i) \end{align*} hence \begin{align} S_T = S_0 \prod_{i : 0 < t_i \le T} (1-\alpha_i) \exp \left( r T\right) \mathcal{E}[\sigma W_T^{\Bbb{Q}}] \end{align} where $\mathcal{E}[X_t]$ figures the stochastic exponential of the process $X_t$, i.e. $\mathcal{E}[X_t] = \exp(X_t - 1/2[X,X]_t)$, hence $$ F(0,T) = S_0 \prod_{i : 0 < t_i \le T} (1-\alpha_i) e^{r T} \tag{A} $$

Similarly if you assume a continuous dividend yield: $$ dS_t = (r - q) S_t dt + \sigma S_t dW_t^\Bbb{Q} $$ you would have obtained the well-known result $$ S_T = S_0 e^{(r-q)T} \mathcal{E}\left[ \sigma W_T^\Bbb{Q} \right]$$ hence $$ F(0,T) = S_0 e^{(r-q)T} \tag{B} $$