3
$\begingroup$

Why would the value of a call option go infinity as volatility goes to infinity?

I understand how you could solve this question by taking $\sigma \rightarrow \infty$ in the solution to the black scholes equation. However, I cannot understand this on a more heuristic level. Surely as volatility goes to infinity you also have a larger chance that the option finishes out of the money (and potentially by a long way)?

$\endgroup$
5
  • $\begingroup$ The probability of finishing out-of-the money actually goes to one in the limit as the whole mass of the price distribution shifts towards zero. $\endgroup$ Apr 30 '18 at 12:22
  • $\begingroup$ @LocalVolatility exactly, so why would the price go to infinity? $\endgroup$
    – Trajan
    Apr 30 '18 at 12:44
  • $\begingroup$ Because of the asymmetry mentioned in the existing answer. Keep in mind that (when rates and dividends are zero), then the forward is equal to the spot. I.e. the average stock price is still the same as the current. A lot of mass near zero is balanced by a few extremely high values. $\endgroup$ Apr 30 '18 at 12:47
  • $\begingroup$ @LocalVolatility ok $\endgroup$
    – Trajan
    Apr 30 '18 at 16:13
  • 1
    $\begingroup$ The zero strike call has no vega, as a result of the model, and so is unchanged when vol is increased to infinity. At the same time, option prices must be decreasing, as a function of strike. So the maximum value an option can have is that of the zero strike call. $\endgroup$
    – will
    Oct 3 '20 at 9:30
3
$\begingroup$

The stock price may fall a lot with high volatility, but you can only lose the price of the option if you bought the option. So the upside gets bigger but the downside is bounded.

$\endgroup$
1
  • 2
    $\begingroup$ This does not address the question of the call option price going to infinity, which is incorrect. $\endgroup$
    – Dom
    Dec 4 '19 at 17:05
22
$\begingroup$

The value of a call option does not go to infinity as the volatility goes to infinity. It tends to the discounted value of the forward $F=S_0 e^{(r-q)T}$, which when the dividend yield is zero, corresponds to the current value of the stock price $S_0$.

Let me explain why. The value of a call option increases with volatility as the upside to the option is greater if the stock is more volatile - the downside is always floored at zero so this does not change. In the limit of the volatility tending to infinity the value of a call option tends to the stock price. This is already clear from no-arbitrage considerations - you would never pay more than the discounted forward price of the stock price for a call option.

We can see this from the Black-Scholes solution to the call option. This is given by: \begin{equation} C(S_0,T)= S_0 e^{-qT} N(d_1) - K e^{-rT} N(d_2) \end{equation} where \begin{equation} d_1= \frac{1}{\sigma \sqrt{T}} \left[\ln{\left(\frac{S_0}{K}\right)} + \left(r -q + \frac{\sigma^2}{2} \right) T \right] \end{equation} and \begin{equation} d_2= \frac{1}{\sigma \sqrt{T}} \left[\ln{\left(\frac{S_0}{K}\right)} + \left(r - q - \frac{\sigma^2}{2} \right) T \right] \end{equation} When $\sigma \rightarrow \infty$ we have $d_1 \rightarrow \infty$ and $d_2 \rightarrow -\infty$. Hence as $N(x) \rightarrow 1$ and $N(-x) \rightarrow 0$ as $x \rightarrow \infty$, we have \begin{equation} C(S_0,T) \rightarrow S_0 e^{-qT} \end{equation} Understanding why this is the case is not so clear. It relies on us realising that as $\sigma \rightarrow \infty$, the probability distribution for $S_T$ becomes spread out along the whole $S_T>0$ support but with a significant probability mass accumulating on $S_T=0$.

However no-arbitrage conditions require the expectation of $S_T$ at time $T$ must equal the forward price such that \begin{equation} \int_0^{\infty} S_T g(S_T) dS_T = S_0 e^{(r-q)T} \end{equation} where $g(S_T)$ is the probability density function for the terminal stock price $S_T$.

In the Black-Scholes formula the first term is actually the discounted expected value of the in-the-money stock price \begin{equation} e^{-rT} \int_K^{\infty} S_T g(S_T) dS_T = S_0 e^{-qT} - e^{-rT} \int_0^{K} S_T g(S_T) dS_T. \end{equation}

As $\sigma \rightarrow \infty$, the distribution becomes so smeared out that the integral of $S_T$ from $0$ to $K$ is negligible (the only significant probability mass is on $S_T=0$ which does not contribute) compared to the integral from $K$ to $\infty$. Hence the second term on the right hand side becomes negligible by comparison to the first term and so \begin{equation} e^{-rT} \int_K^{\infty} S_T g(S_T) dS_T \rightarrow S_0 e^{-qT} \end{equation} and \begin{equation} \lim_{\sigma \rightarrow \infty} C(S_0,T) \rightarrow S_0 e^{-qT}. \end{equation} We can also show that the put option price converges to the discounted strike. \begin{equation} \lim_{\sigma \rightarrow \infty} P(S_0,T) \rightarrow K e^{-rT}. \end{equation}

$\endgroup$
5
  • 1
    $\begingroup$ This is the correct answer $\endgroup$
    – Brian B
    Dec 4 '19 at 19:29
  • $\begingroup$ I would expect it to converge to the discounted value of the forward, and not the spot. $\endgroup$
    – will
    Oct 3 '20 at 9:28
  • $\begingroup$ Yes. I did it for the zero dividend case q=0. I just amended it to the more general case. $\endgroup$
    – Dom
    Oct 3 '20 at 14:06
  • $\begingroup$ @Dom just the answer I was looking for. However, can you point my in a direction of a paper where this is described? $\endgroup$ May 17 at 18:58
  • $\begingroup$ I don't have a reference. Not sure if this has been written up formally. $\endgroup$
    – Dom
    May 17 at 21:10
3
$\begingroup$

I also got a bit confused about understanding intuitively what happens in the limit with the Black-Scholes Call and Put values as the volatility goes to infinity, but after analyzing the derivation of the formula it becomes clear.

My confusion was about why the contribution of the $-K$ term disappear in the limit. My faulty reasoning was a bit like this : If $S_T$ in the limit when $\sigma \to \infty$ is supposed to be big most of the time then I felt the option values should be $e^{-rT}(S_0 \cdot e^{rT} - K)$, the discounted payoff of the forward value. And if $S_T$ would be small most of the time, the limit would be $0$. This is wrong.

As LocalVol points out in the comments above, it is important to make a distinction between the size of the underlying and the probability that it has a certain size. The price of the call is (using the risk neutral measure as usual and ignoring the initial discount factor) $$ C = E[Max[(S_T - K, 0)] = E[S_T \cdot 1_{S_T \geq K} ] - E[K \cdot 1_{S_T \geq K}], $$ where we have used indicator functions for the event that we end up in the money. Let us look at the second term first. This is just $$-E[K \cdot 1_{S_T \geq K} ] = -K \cdot \mathbb{P}(S_T \geq K).$$ That is, the value of $-K$ times the risk neutral probablility that this cash flow will occur. I will not rewrite all the math here, but we get that $$ \lim_{\sigma \to \infty} \mathbb{P}(S_T \geq K) = 0. $$ The probablity that we end up in the money goes to zero and the value of this second term goes to zero in the limit. If we were sampling the values of $S_T$, we would find that a larger proportion of paths goes down below K when $\sigma$ grows.

However, the point here is, as LocalVol points out, that although the probability of ending up in the money goes to zero, a larger and larger proportion of the expected value of $S_T$ comes from that region.
Mathematically we have $$ \begin{eqnarray*} \lim_{\sigma \to \infty} E[S_T \cdot 1_{S_T \geq K} ] &=& S_0 \cdot e^{rT} = E[S_T].\\ \lim_{\sigma \to \infty} E[S_T \cdot 1_{S_T < K} ] &=& 0. \end{eqnarray*} $$ More and more contribution of the expected value come from values of $S_T$ when $S_T \geq K$ as $\sigma$ grows, although the probability for those values to occur goes to zero.

So we have a sort of competition of limits here where the values of $S_T$ above $K$ increases faster than their probability to occur goes to zero, so to speak

$\endgroup$
0
3
$\begingroup$

Copy pasting parts of an answer I did here as it illustrates the limits of call and put option premia.

$N(d2)$ is the probability that a call option with an exercise price of $K$ is exercised in a risk-neutral world. Therefore, $(1− N(d2)$ or $N(-d2)$ is the probability that a put with the same exercise price will be exercised. Let's plot this as a function of vol.

enter image description here

How about the premium of puts and calls?

enter image description here

Even though there is supposedly zero probability of exercise for a call, I still pays a maximum that seems to be the current spot price, irrespective of the strike. There is another explanation using $N(d1)$ a bit further below in the link I added above (after the PDF of the normal distribution). For the put, however, the maximum is reached at the strike price itself. As @Will explains intuitively, a zero strike call has no vega and does not change when vol is increased to infinity.

enter image description here

At the same time, call option prices are decreasing, as a function of strike (see next chart). Thus, the maximum value a call option can have is that of the zero strike call (which is the price of the current spot when there are no interest rates and dividends). For puts, a similar argument can be made that leads to the strike being the maximum value. I link an answer/explanation where the question is flawed (delta is not the probability of exercise) but the accepted answer is correct.

enter image description here

@Jesper Tidblom's "competition of limits" is also illustrated in the answer I linked.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.