I'm trying to derive the following boundary conditions for heston's stochastic volatility model.

This is p. 289 of Shreve's Stochastic calculus for finance

\begin{align} c(T, s, v) &=(s-K)^{+} \text {for all } s \geq 0, v \geq 0 \\ c(t, 0, v) &=0 \text { for all } 0 \leq t \leq T, v \geq 0 \\ c(t, s, 0) &=\left(s-e^{-r(T-t)} K\right)^{+} \text {for all } 0 \leq t \leq T, s \geq 0 \\ \lim _{s \rightarrow \infty} \frac{c(t, s, v)}{s-K} &=1 \text { for all } 0 \leq t \leq T, v \geq 0 \\ \lim _{v \rightarrow \infty} c(t, s, v) &=s \text { for all } 0 \leq t \leq T, s \geq 0 \end{align}


A bit of background for the heston's stochastic volatility model

$$ d S(t)=r S(t) d t+\sqrt{V(t)} S(t) d \widetilde{W}_{1}(t) $$

$$ d V(t)=(a-b V(t)) d t+\sigma \sqrt{V(t)} d \widetilde{W}_{2}(t) $$

There is a function c that satisfies

$$ c(t, S(t), V(t))=\widetilde{\mathbb{E}}\left[e^{-r(T-t)}(S(T)-K)^{+} \mid \mathcal{F}(t)\right], \quad 0 \leq t \leq T $$

subject to

$$ c_{t}+r s c_{s}+(a-b v) c_{v}+\frac{1}{2} s^{2} v c_{s s}+\rho \sigma s v c_{s v}+\frac{1}{2} \sigma^{2} v c_{v v}=r c $$

I eventually show that

$c(t, s, v)=s \mathbb{E}^{t, x, v} \mathbb{I}_{\{X(T) \geq \log K\}}-e^{-r(T-t)} K \mathbb{E}^{t, x, v} \mathbb{I}_{\{X(T) \geq \log K\}}$


I'm trying to derive the boundary conditions

$c(T, s, v) =(s-K)^{+} \text {for all } s \geq 0, v \geq 0$ comes from evaluating at $t = T$

$c(t, 0, v) =0 \text { for all } 0 \leq t \leq T, v \geq 0$ Heuristically it's 0 as the strike price will always be greater than the stock price. If I evaluate the PDE at $s= 0$, I get $c_t + (a-bv)c_v + \frac{1}{2}\sigma^2vc_{vv} = rc$, which looks like the black-scholes pde, not sure where I can go after this though.

$\lim _{s \rightarrow \infty} \frac{c(t, s, v)}{s-K} =1 \text { for all } 0 \leq t \leq T, v \geq 0 $, I understand that as the stock price is approaches infinity, it will almost surely finish above the strike price. Taking the limit solves this.

$\lim _{v \rightarrow \infty} c(t, s, v) =s \text { for all } 0 \leq t \leq T, s \geq 0$ confuses me completely.

Thank you.


1 Answer 1


You can't really derive or prove boundary conditions. You impose them and try to economically motivate them.

Let's consider a European-style call option and go through the boundary conditions step by step.


When the underlying asset's value is zero, then the option to buy this asset is worthless. Thus, $$C(t,S=0,v)=0.$$


As the underlying raises in value, we will certainly exercise the option and purchase the stock. The option thus converges to its payoff, i.e. $$\lim_{S\to\infty}\frac{C(t,S,v)}{S-K}=1.$$ Note that another common choice is to impose that the option's delta equals one, $$\lim_{S\to\infty}\frac{\partial}{\partial S}C(t,S,v)=1.$$


That's probably the easiest one because it is simply, as you say, the option's payoff, $$C(t=T,S,v)=\max\{S-K,0\}.$$


As variance (uncertainty) increases, extreme events become more likely. As the option's payoff is bounded from below by zero, it does not suffer too much from bad states of nature and option prices typically increase with variance (have a positive vega). Option prices can't grow to infinity though because their value is bounded to be at most $S$ (by no-arbitrage). Thus, we set $$\lim_{v\to\infty}C(t,S,v)=S.$$


As argued before, options tend to be monotone in uncertainty and decrease in value as variances decrease. A lower bound for the option price is given by no-arbitrage arguments, $$C(t,S,v=0)=\max\{S-Ke^{-r(T-t)},0\}.$$

Some Final Notes

  • I'm not fully convinced by the $v=0$ condition. If you take the final closed-form solution for the call option price (using the characteristic function of $\ln(S_T)$) and plug $v=0$ into it, I don't think you get $\max\{S-Ke^{-r(T-t)},0\}$, which is essentially a deterministic payoff. Even if $v_t=0$, the variance will converge back to its long-term mean. Thus, in models with stochastic volatility, $v=0$ does not mean all risk is gone. It holds for the Black-Scholes model though. Having said this, $\max\{S-Ke^{-r(T-t)},0\}$ is certainly a lower bound for the option price. But so is $-\frac{\pi}{12}$ and I wouldn't use this one either...
  • If you consider a dividend paying stock, you need to adjust some of these conditions. For example, an upper bound for the option value is then $S_te^{-q(T-t)}$ instead of just $S$, which we used for the $v\to\infty$ condition.
  • Heston (1993, RFS) uses slightly different conditions. He specifies a condition for delta as $S\to\infty$ (as mentioned above) and also uses a different condition for $v=0$ and just imposes that the solution satisfies an inhomogeneous first-order PDE. That's particularly neat when you solve the PDE numerically using finite differences.

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  • $\begingroup$ I think the $v\to 0 $ condition is okay-*ish*. As soon as $v$ hits zero, the underlying grows deterministically at $r$ under the RN measure. Thus, the moneyness of the option is defined at this exact point in time. Your statement regarding the upwards drift through $\kappa(\theta- v_t)dt$ is of course still correct. That's really a bit odd.Interesting... $\endgroup$ Commented Dec 9, 2020 at 11:24
  • $\begingroup$ @Kermittfrog I fully agree with you in a log-normal Black Scholes setting where $\sigma=0$ indeed implies that the stock price is deterministic. But in the Heston model, even if $v_t=0$, then $v_{t+\Delta t}>0$. So, the stock price is not deterministic but remains random because the variance will increase again (mean reversion) $\endgroup$
    – Kevin
    Commented Dec 9, 2020 at 11:26
  • $\begingroup$ @Kermittfrog I mean it's certainly a valid lower bound but option prices in the Heston model won't converge to it. They converge to the no-arbitrage bound $S_t$ at $v_t\to\infty$ but they don't approach the $(S_t-Ke^{-r(T-t)})^+$ bound as $v_t\to0$. So, it's valid and you'd probably get correct prices if you use this condition in a finite difference scheme, but it's not quite as nice/optimal as Heston's original suggestion... $\endgroup$
    – Kevin
    Commented Dec 9, 2020 at 11:29
  • 1
    $\begingroup$ I agree with you. So effectively one could try to set up a numerical 2D PDE solver and impose various restriction for $v\to 0$ and see what effect they may have on the option price... Thank Heston and Carr/Madan for the Fourier Transform methods... $\endgroup$ Commented Dec 9, 2020 at 11:33
  • $\begingroup$ @Kermittfrog I don't think we necessarily need to solve the $v=0$ PDE directly. If you approximate it by finite differences, you can solve it alongside with the interior grid points in your actual 3D Heston grid. I think Rouah's book on the Heston model describes how to do that. But yeah, I fully agree with you. :) I take Fourier methods any day over finite differences! $\endgroup$
    – Kevin
    Commented Dec 9, 2020 at 11:37

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