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 ...


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.

  • $\begingroup$ Thank you very much for such comprehensive reply, but unfortunately the last lines made me confused. First of all, 7th fro below line, I guess there should be a plus instead of minus. Probably a typo, but then I just don't get your idea after "Expand it further in Δt". Don't you get $O(\Delta t^2)$ variables everywhere, which are zero?. And even before that, we still got the coefficients to substitute from the equation 8th from below, these were my problem ... the formulas simply did not cancel out. Could you please elaborate the last steps? Much appreciated for your effort $\endgroup$
    – Makina
    Mar 6 '19 at 13:32
  • $\begingroup$ @Makina, I corrected the typo and added some details. You should consider two different expansions, the one for the up case and another one for the down case. Given the asymptotic results due to CLT we get to the BS formula. Note that $\hat\mu N$ and $\hat\sigma^2N$ are equal to $\mu T$ and $\sigma^2 T$, respectively. So, I don't see any other coefficient that does not get simplified. I plug the $u,d,q$ in $O(\sqrt{\Delta t})$ coefficients to get eq. (2) in $O((\Delta t)^{3/2})$. $\endgroup$
    – FunnyBuzer
    Mar 6 '19 at 17:52

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.

  • $\begingroup$ What point are you doing the Taylor expansion around? $\endgroup$ Oct 2 '19 at 17:19

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