Dividend paying stock's risk-neutral probability proof Question: Consider the one-period binomial model with a stock that pays continuous dividend $\delta$. I want to show that the risk-neutral probability is given by $$p=\frac{\exp((r-\delta)\Delta t)-d}{u-d}$$.
Hint: The value of stock at time $\Delta t$ is: $S_0uexp(-\delta\Delta t)$ for stock moving up and $S_0dexp(\delta\Delta t)$ for stock moving down.
Approach: Here is how I started the proof.
First I form portfolio at t=0. $\Delta S-O_{t=0}$ to find $\Delta$ which is;
$$\Delta S_0uexp(\delta\Delta t)-O_{up}=\Delta S_0dexp(\delta\Delta t)-O_{down}$$ where $$\Delta=\frac{O_{up}-O_{down}}{S_0exp(\delta\Delta t)(u-d)}$$
after finding this $\Delta$ I'm sure we are supposed to substitute it into the initial portfolio but I'm confused as to how am I supposed to get rid of the O's and how to derive p from this. The professor gave me this much hint.
Any help will be greatly appreciated! Thank you!!
 A: Sometimes, its more beneficial to work via first principles, than plugging into an equation.
Let $X$ denote a claim -- in this case $X$ denotes the payoff on the stock held at time $t$. Lets simplify the notation, so that without loss of generality let $t=0$ and $t=1$, and suppose the stock has a value $u$ or $d$ at $t=1$.
What we want is the value of the claim $X$ at time $t=0$, under the risk neutral measure. Denote this measure by $p$. Thus we want to find $p$, given that we know the possible prices of the stock. That is if $X_{1}(u),X_{1}(d)$ represent the claim at time t=1, we have that.
\begin{equation}
V(X) = pX_{1}(u) + (1-p)X_{1}(d) 
\end{equation}
Now $V(X) = S_{0}exp(r\Delta t)$, since the stock price at $t=0$ is known. Thus we have,
\begin{equation}
S_{0}exp(r\Delta t) = pX_{1}(u) + (1-p)X_{1}(d) 
\end{equation}
$X_{1}(u)$ is the claim at t=1, when the market is in state $u$. That is use the hint given to calculate $X_{1}(u)$ and $X_{1}(d)$, plug it into the equation above and solve for $p$. Hope this helps, let me know if you need more details.
