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Consider a standard binomial tree. Let $u = e^{(r - \delta)h + \sigma\sqrt{h}}$ and $d = e^{(r - \delta)h - \sigma\sqrt{h}},$ where $\delta$ is the continuously compounded dividend yield, $h$ is the length of one period in a binomial model, and $\sigma$ is volatility.

I am told in my textbook that the risk-neutral probability $p*$ is given by:

$$p^* = \frac{e^{(r - \delta)h} - d}{u - d} = \frac{1}{1 + e^{\sigma\sqrt{h}}}.$$

I tried deriving the second equality as follows:

$\begin{align}\frac{e^{(r - \delta)h} - d}{u - d} &= \frac{e^{(r - \delta)h} - e^{(r - \delta)h - \sigma\sqrt{h}}}{e^{(r - \delta)h + \sigma\sqrt{h}} - e^{(r - \delta)h - \sigma\sqrt{h}}}\\ &= \frac{e^{(r - \delta)h}(1 - e^{-\sigma\sqrt{h}})}{e^{(r - \delta)h}(e^{\sigma\sqrt{h}} - e^{-\sigma\sqrt{h}})}\\ &= \frac{1 - e^{-\sigma\sqrt{h}}}{e^{\sigma\sqrt{h}} - e^{-\sigma\sqrt{h}}}\end{align}.$

Now at this point I am stuck and I'm unsure if it's either something algebraic I am not seeing, or if there is some property of forward trees that we can use to reach the conclusion.

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Here's one algebraic way to derive it: $$ \frac{(1 - e^{-\sigma\sqrt{h}})(1 + e^{\sigma\sqrt{h}})}{(e^{\sigma\sqrt{h}} - e^{-\sigma\sqrt{h}})(1 + e^{\sigma\sqrt{h}})} = \frac{1 - e^{-\sigma\sqrt{h}} + e^{\sigma\sqrt{h}} - 1}{(e^{\sigma\sqrt{h}} - e^{-\sigma\sqrt{h}})(1 + e^{\sigma\sqrt{h}})} = \frac{e^{\sigma\sqrt{h}} - e^{-\sigma\sqrt{h}}}{(e^{\sigma\sqrt{h}} - e^{-\sigma\sqrt{h}})(1 + e^{\sigma\sqrt{h}})} = \frac{1}{1 + e^{\sigma\sqrt{h}}} $$

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