Let's say I've computed the price of a call using Monte Carlo with $S_0 = 100$ and $K = 80$, using $T = 0.1$ and $r = 0$ to be $\$20.00095$. This price estimate comes with a $95\%$ confidence interval of $[19.99969 , 20.00221]$. The issue, then, is trying to estimate implied vol, including a confidence interval, since the $95\%$ lower bound on price is below intrinsic and hence the inverse problem has no solution.

Even if we set the confidence interval to be $[20, 20.00221]$, the implied vol estimate would be $0.2172$ with a $95\%$ confidence interval $[0, 0.2324]$, a huge spread.

I'm already using a massive amount of stock prices to get the confidence intervals this tight, so I'd prefer not to just ramp up the number of simulations. All of the paths finish in the money for this case. Is there a variance reduction technique, perhaps, for dealing with these deep ITM issues?

  • $\begingroup$ I guess that by "massive amount of stock prices" you mean "massive amount of stock paths" hence simulations, right? $\endgroup$
    – Quantuple
    Commented Jun 8, 2016 at 14:39
  • $\begingroup$ @Quantuple Right, to the point of it is not reasonable to add more due to time constraints. $\endgroup$
    – bcf
    Commented Jun 8, 2016 at 14:43

1 Answer 1


[Short answer]

IMHO there is a fundamental problem with wanting to extract a sound implied volatility figure out of a deep ITM option's price. You should use out-of-the-money forward options (OTMF) instead: put options for strikes smaller than the forward price (left wing of the volatility surface) and call options otherwise (right wing of the volatility surface).

[Long answer]

To illustrate my point, let $V$ denote the $t$-value of a European option, which we split into 2 components $$ V = V_i + V_e $$ according the following thought experiment:

  • The intrinsic value, $V_i$, is defined as what you would get if you could exercise the option immediately at time $t$, or equivalently, what your final gain would look like should the underlying price be frozen to its current value up to the contract's expiry. $V_i$ is always positive (but can be zero).
  • The extrinsic or time value, $V_e$, is the remaining part. It accounts for the fact that the underlying price is expected to evolve and not remain frozen. $V_e$ can be positive or negative.

By construction, we have that the intrinsic value $V_i$ does not depend on the future volatility as it is something we have defined assuming the underlying remained frozen. In contrast, the time value $V_e$ does depend on future volatility, more or less strongly depending on the remaining time to maturity $\tau = T-t$ and where the current spot price $S_t$ is located with respect to the strike $K$.

By definition, the price of a strongly ITM option essentially corresponds to intrinsic value $$ V = V_i + V_e \approx V_i $$ Because $V_i$ does not depend on volatility, it is very difficult to imply a robust volatility figure from an ITM option price (the fraction $V_e$ of the option price which truly depends on volatility is very small relative to the full option price $V$)

On the contrary, strongly OTM options essentially reflect time value $$ V = V_i + V_e \approx V_e $$ which makes it easier to imply volatility (the fraction $V_e$ of the option price which truly depends on volatility is very important relative to the full option price $V$)

Hence you should prefer OTM options to ITM options when it comes to inferring implied volatilities.

  • $\begingroup$ I just noticed you wrote $V_e$ can be negative. Is this not an arbitrage opportunity for Europeans? $\endgroup$
    – bcf
    Commented Jun 13, 2016 at 14:48
  • $\begingroup$ Also, the arbitrage-consistent price for the corresponding put in my problem (via parity) is $\$0.00095$, with a confidence interval of $[-0.00031, 0.00221]$. Thus the issue of no solution arises again, even for puts. OTOH using the same paths I used above to price a put results in a put price of $\$0$, since all paths finish above $80$. What do you think of importance sampling these paths for pricing puts, then using the resulting prices for IV comps? But, then they wouldn't be consistent (via parity) with the IVs from the calls... $\endgroup$
    – bcf
    Commented Jun 13, 2016 at 15:03
  • $\begingroup$ @bcf No, as long as $V = V_i + V_e > 0$ you are fine. Compare for instance a European call vs European put without dividends. For the former both the intrinsic and extrinsic values are always positive. For the second, for some ITM strikes, time value is negative (the "price" graph is below the "payoff" graph). This is why the price of an American put is different from that of a European counterpart (when time value is negative, you are better off exercising). $\endgroup$
    – Quantuple
    Commented Jun 13, 2016 at 15:35
  • $\begingroup$ It's essentially numerical precision issues that you are facing... even if you were using the BS formula it would be impossible to retrieve the true volatility at some point due to the finite machine precision (using smart implementations it is possible to reach double precision, see jaeckel.org "Let's be rational" ). Remember that in practice there is a minimum trade price meaning that figures such that \$0.00095 will typically be floored to for instance (1 cent). Can we have an idea of what moneyness level and time to maturity are you considering? $\endgroup$
    – Quantuple
    Commented Jun 13, 2016 at 15:40
  • $\begingroup$ If you obtain a better MC estimator (using importance sampling or any other variance reduction technique), you may be better off for that particular option (by that I mean having a 95% CI which is at least in the positive region), but you'll eventually face the same problem for another option as you go more and more OTM. $\endgroup$
    – Quantuple
    Commented Jun 13, 2016 at 15:42

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