Deriving the Black–Scholes PDE from a CRR Binomial Tree

# Approximation of CRR as Black Scholes PDE


# Approximation of CRR as Black Scholes PDE












I have a formula for intermediate european option price calculated at, say, m-th possible tree value.

$S_n^{(m)}$ is a price at node after going up $n$ times and down $n - m$ times $V(S_n^{(m)}, t + n\Delta t) = e^{-r \Delta t} [p V(u S_n^{(m)}, t + (n+1)\Delta t + (1 - p) V(u S_n^{(m)}, t + (n+1)\Delta t]$

$p = \frac{e^{(r - D) \Delta t} - d}{u - d}$ - risk-neutral probability, $r$ - risk free rate, $D $ continuous dividend yield

$u = e^{\sigma \sqrt{\Delta t}}$, $d = e^{-\sigma \sqrt{\Delta t}}$, $\sigma$ is volatility of stock price

I need to figure out using taylor approximations that under limit for $\Delta t -> 0$ this formula becomes BS PDE.

Ok so first of all, I used the identity $e^x = 1 + x$ and applied it everywhere possible. Then I wrote taylor expansion for both functions of the RHS but at this point my results did not really match.

Any value beyond and equal to $O(\Delta t^2)$ I made equal to zero.

So, for instance, for the term $\frac{1}{2}\frac{d^2V}{dS^2}$ I ended with the following:

... + $\frac{1}{2}\frac{d^2V}{dS^2}S^2(p(u - 1)^2 + (1-p)(d - 1)^2)*e^{-r\Delta t}$

Unfortunatelly after substituting all the variables after rewriting them with taylor approximation, the resulting expression after all the cancellations did not turn into the $\frac{1}{2}\frac{d^2V}{dS^2}S^2\sigma^2$

I wrote taylor expansion centered around $(S_n^{(m)}, t + \Delta t)$, respectively.

Could someone let me know where I made the mistake? I did not bother continuing with other parts of the formula mainly due to getting the first one wrong ...

## Answer by FunnyBuzer (score 5, accepted)

https://quant.stackexchange.com/a/44453

Assuming continuously compounded returns for a multi-period model with $N$ being the number of periods:

\begin{cases} &\log u \quad \text{with probability q}\\ &\log d \quad \text{with probability 1-q} \end{cases} given the stock price at maturity $$\log\left(\frac{S_T}{S_0}\right)=i\log u+(N−i)\log d=i\log\left(\frac{u}{d}\right)+N\log d$$ where $i$ is a binomial r.v. under the risk-neutral measure $\mathbb Q$. $$\lim_{N\rightarrow\infty}\log\left(\frac{S_T}{S_0}\right)\sim\mathcal{N}(\mu T,\sigma^2T)$$ with \begin{align} \mu T=&\mathbb{E}[i]\log(u/d) +N\log d \\ \sigma^2T=&\text{Var}[i][\log(u/d)]^2 \end{align} By CLT, the distribution of the sum of continuously compounded returns converges to a normal distribution. Since, by assumption $u=1/d$, we have the equations: $$u=e^{\sqrt{\Delta t}}, \quad d=e^{-\sqrt{\Delta T}}, \quad q=\frac{1}{2}+\frac{\mu}{2\sigma}\Delta t, \quad \Delta t=\frac{t}{N}$$ Let us re-write the binomial distribution: $\text{Bin}\left(\left\lfloor\frac{\log(K/S_0d^N)}{\log(u/d)}\right\rfloor+1,N,q\right)$ \begin{align} 1-\text{Bin}\left(\left\lfloor\frac{\log(K/S_0d^N)}{\log(u/d)}\right\rfloor+1,N,q\right) &=\mathbb{Q}\left(i\leq \left\lfloor\frac{\log(K/S_0d^N)}{\log(u/d)}\right\rfloor+1-1\right) \\ &=\mathbb{Q}\bigg(\underbrace{\bigg(\frac{i-Nq}{\sqrt{Nq(1-q)}}\bigg)}_{=\frac{\log(S_T/S_0)-N\log d-Nq\log(u/d)}{\log(u/d)\sqrt{Nq(1-q)}}}\leq\underbrace{\frac{\left\lfloor\frac{\log(K/S_0d^N)}{\log(u/d)}\right\rfloor-Nq}{\sqrt{Nq(1-q)}}}_{=:\mathcal{A}}\bigg) \end{align} $$\left\lfloor\frac{\log(K/S_0d^N)}{\log(u/d)}\right\rfloor=\frac{\log(K/S_0d^N)}{\log(u/d)}-\varepsilon, \qquad \varepsilon\in[0,1) \\ \Rightarrow\mathcal{A}=\frac{\log(K/S_0)+N(\log d+q\log(u/d))-\varepsilon\log(u/d)}{\log(u/d)q(1-q)\sqrt N}$$

Now, substitute $$\hat\mu=q(\log d+\log(u/d)),\quad \hat\sigma=q(1-q)(\log(u/d))^2,\quad q=\frac{e^{r\Delta t}-e^{-\sigma\sqrt{\Delta t}}}{e^{\sigma\sqrt{\Delta t}}-e^{-\sigma\sqrt{\Delta t}}} \\ \text{in } f(\Delta t)=\frac{\log(K/S_0)-\hat\mu N-\varepsilon\log(u/d)}{\hat\sigma\sqrt N}$$ and compute its Taylor expansion: $$f(\Delta t)=\frac{\log(K/S_0)}{\sigma\sqrt{N\Delta t}}-\frac{2\varepsilon\sigma}{\sqrt N \sigma}+\frac{1}{2}\frac{\sigma^2N\Delta t-rN\Delta t}{\sigma N\Delta t}+O((\Delta t)^3)$$ As $h\rightarrow 0$, $N\rightarrow \infty$ and $N\Delta t=T$, $$f(\Delta t)\rightarrow \frac{\log(K/S_0)-\left(r-\frac{1}{2}\sigma^2\right)T}{\sigma\sqrt{T}} \\ \Rightarrow \text{Bin}\left(\left\lfloor\frac{\log(K/S_0d^N)}{\log(u/d)}\right\rfloor+1,N,q\right)\rightarrow \mathbf\Phi\left(\frac{\log(K/S_0)-\left(r-\frac{1}{2}\sigma^2\right)T}{\sigma\sqrt{T}}\right)$$ As a result, we have the convergence towards the Black-Scholes model: $$\text{Call}_0=S_0\mathbf\Phi(d_1)-Ke^{-rT}\mathbf\Phi(d_2)$$

To derive the PDE, note in the Binomial model it holds $$e^{r\Delta t}\text{Call}(S,t)=q\text{Call}_u(S_u,t+\Delta t)+(1-q)\text{Call}_d(S_d,t+\Delta t)\tag1$$

Plug in the value of $q=\frac{e^{r\Delta t}-e^{-\sigma\sqrt{\Delta t}}}{e^{\sigma\sqrt{\Delta t}}-e^{-\sigma\sqrt{\Delta t}}}$:

$$e^{r\Delta t}\text{Call}(S,t)=\frac{e^{r\Delta t}-e^{-\sigma\sqrt{\Delta t}}}{e^{\sigma\sqrt{\Delta t}}-e^{-\sigma\sqrt{\Delta t}}}\text{Call}(Se^{\sigma\sqrt{\Delta t}},t+\Delta t)+\frac{e^{\sigma\sqrt{\Delta t}}-e^{r\Delta t}}{e^{\sigma\sqrt{\Delta t}}-e^{-\sigma\sqrt{\Delta t}}}\text{Call}(Se^{-\sigma\sqrt{\Delta t}},t+\Delta t)$$

Expand $\text{Call}(Se^{\sigma\sqrt{\Delta t}},t+\Delta t)$ about $(S,t)$:

$$\text{Call}(Se^{\sigma\sqrt{\Delta t}},t+\Delta t)=\text{Call}(S,t)+(e^{\sigma\Delta t}-1)\frac{\partial\text{Call}}{\partial S}+\frac{1}{2}(e^{\sigma\Delta t}-1)^2S^2\frac{\partial^2\text{Call}}{\partial S^2}+\Delta t\left(\frac{\partial\text{Call}}{\partial t}+(e^{\sigma\Delta t}-1)S\frac{\partial^2\text{Call}}{\partial S\partial t}+\frac{1}{2}(e^{\sigma\Delta t}-1)^2S^2\frac{\partial^3\text{Call}}{\partial S^2\partial t}\right)+O((S)^3)$$ Then, perform the same expansion for $\text{Call}(Se^{-\sigma\sqrt{\Delta t}},t+\Delta t)$ about $(S,t)$, and re-write eq. (1)

$$e^{r\Delta t}\text{Call}(S,t)=\frac{e^{r\Delta t}-e^{-\sigma\sqrt{\Delta t}}}{e^{\sigma\sqrt{\Delta t}}-e^{-\sigma\sqrt{\Delta t}}}\text{Call}(Se^{\sigma\sqrt{\Delta t}},t+\Delta t)-\frac{e^{\sigma\sqrt{\Delta t}}-e^{r\Delta t}}{e^{\sigma\sqrt{\Delta t}}-e^{-\sigma\sqrt{\Delta t}}}\text{Call}(Se^{-\sigma\sqrt{\Delta t}},t+\Delta t)\tag2$$

Expand eq. (2) in $\Delta t$ to get

$$0=\left(r\text{Call}(S,t)+\frac{\partial\text{Call}}{\partial t}+rS\frac{\partial\text{Call}}{\partial S}+\frac{1}{2}rS^2\frac{\partial^2\text{Call}}{\partial S^2}\right)\Delta t+O((\Delta t)^{3/2})$$

As $\Delta t\rightarrow 0$ we have the Black-Scholes PDE.

## Answer by ZRH (score 2)

https://quant.stackexchange.com/a/44458

I think stuff does not cancel out because there was a mistake in the initial formula, which should instead read:

$exp(r \delta t) V(S,t)=pV(uS,t+\delta t)+(1-p)V(dS,t+\delta t)$

Developing the function V to second order Taylor terms:

$exp(r \delta t) V(S,t)=p \big[V(S,t)+\frac{\partial V}{\partial S}(uS-S)+\frac{1}{2}\frac{\partial^2 V}{\partial S^2}(uS-S)^2+\frac{\partial V}{\partial t}\delta t\big]+(1-p)\big[V(S,t)+\frac{\partial V}{\partial S}(dS-S)+\frac{1}{2}\frac{\partial^2 V}{\partial S^2}(dS-S)^2+\frac{\partial V}{\partial t}\delta t\big]$

Now minding that $uS-S\approx \sigma \sqrt{\delta t}$ and $dS-S\approx -\sigma \sqrt{\delta t}$, the above equation becomes:

$exp(r \delta t)V=p \big[V+\frac{\partial V}{\partial S}S \sigma \sqrt{\delta t}+\frac{1}{2}\frac{\partial^2 V}{\partial S^2}S^2 \sigma^2 \delta t +\frac{\partial V}{\partial t}\delta t\big]+(1-p)\big[V-\frac{\partial V}{\partial S}S \sigma \sqrt{\delta t}+\frac{1}{2}\frac{\partial^2 V}{\partial S^2}S^2 \sigma^2 \delta t+\frac{\partial V}{\partial t}\delta t\big]$

Applying first order Taylor to the expression for p:

$p=\frac{1+r \delta t-(1-\sigma \sqrt{\delta t})}{1+\sigma \sqrt{\delta t}-(1-\sigma \sqrt{\delta t})}=\frac{1}{2}+\frac{r}{2\sigma}\sqrt{\delta t}$

Plugging this into the equation and working it all out:

$exp(r \delta t)V-V=\frac{\partial V}{\partial t}\delta t+\frac{S^2 \sigma^2 }{2}\frac{\partial^2 V}{\partial S^2}\delta t+rS\frac{\partial V}{\partial S}\delta t$

and using on the LHS the first order Taylor of the discount factor:

$rV\delta t=\frac{\partial V}{\partial t}\delta t+\frac{S^2 \sigma^2 }{2}\frac{\partial^2 V}{\partial S^2}\delta t+rS\frac{\partial V}{\partial S}\delta t$

$0=\frac{\partial V}{\partial t}+\frac{S^2 \sigma^2 }{2}\frac{\partial^2 V}{\partial S^2}+rS\frac{\partial V}{\partial S}-rV$

Which is precisely the BS PDE.