# Value of Call Option as Volatility goes to Infinity

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)?

• 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. – LocalVolatility Apr 30 '18 at 12:22
• @LocalVolatility exactly, so why would the price go to infinity? – Permian Apr 30 '18 at 12:44
• 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. – LocalVolatility Apr 30 '18 at 12:47
• @LocalVolatility ok – Permian Apr 30 '18 at 16:13

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.

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

The value of a call option DOES NOT go to infinity as the volatility goes to infinity. It tends 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. We can see this from the Black-Scholes solution to the call option. This is given by: $$$$C(S_0,T)= S_0 N(d_1) - K e^{-rT} N(d_2)$$$$ where $$$$d_1= \frac{1}{\sigma \sqrt{T}} \left[\ln{\left(\frac{S_0}{K}\right)} + \left(r + \frac{\sigma^2}{2} \right) T \right]$$$$ and $$$$d_2= \frac{1}{\sigma \sqrt{T}} \left[\ln{\left(\frac{S_0}{K}\right)} + \left(r - \frac{\sigma^2}{2} \right) T \right]$$$$ When $$\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$$, we have $$$$C(S_0,T) \rightarrow S_0$$$$ Understanding why this is the case is not so clear. It relies on us realising that as $$\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 $$$$\int_0^{\infty} S_T g(S_T) dS_T = S_0 e^{rT}$$$$ where $$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 $$$$e^{-rT} \int_K^{\infty} S_T g(S_T) dS_T = S_0 - e^{-rT} \int_0^{K} S_T g(S_T) dS_T.$$$$

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 $$$$e^{-rT} \int_K^{\infty} S_T g(S_T) dS_T \rightarrow S_0$$$$ and $$$$C(S_0,T) \rightarrow S_0.$$$$

• This is the correct answer – Brian B Dec 4 '19 at 19:29