6
$\begingroup$

We consider an stock price $S$ following a normal model: $dS_t = \sigma dW_t$

We can write this as $\frac{dS_t}{S_t}=\frac{\sigma}{S_t}dW_t$

Hence we can see that $S$ follows a "log-normal" diffusion with a local volatility function $c(S)=\frac{\sigma}{S}$ which is downward sloping.

My question is: can we deduce that the log normal smile implied by this model will be downward sloping as well ? That is to say, if we have a local volatility function which is decreasing as a function of $S$, will the lognormal implied vol be decreasing as a function of the strike ?

Thanks !

$\endgroup$
  • $\begingroup$ Indeed $S_t\sim N(S_0\,,\,\sigma^2 t)$ and has Normal distribution. $\endgroup$ – user16651 Jul 20 '16 at 17:16
  • $\begingroup$ @BehrouzMaleki what does that imply regarding the log normal smile ? $\endgroup$ – Dark Jul 20 '16 at 17:26
  • $\begingroup$ Please note $S_t=S_0+\sigma W_t$ thus $S_t $ is a Brownian motion $\endgroup$ – user16651 Jul 20 '16 at 17:50
3
$\begingroup$

Since $S_T = S_0 + \sigma W_T$, \begin{align*} C &:= E\left((S_T-K)^+ \right)\\ &= E\left((S_0+\sigma W_T-K)^+ \right)\\ &=\int_{\frac{K-S_0}{\sigma \sqrt{T}}}^{\infty}(S_0+\sigma\sqrt{T} x-K) \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx\\ &=(S_0-K)\Phi\left(\frac{S_0-K}{\sigma \sqrt{T}}\right)+\frac{\sigma\sqrt{T}}{\sqrt{2\pi}}e^{-\frac{(S_0-K)^2}{2\sigma^2 T}}, \end{align*} where $\Phi$ is the cumulative distribution function of a standard normal random variable. Then, \begin{align*} \frac{dC}{d K} &= -\Phi\left(\frac{S_0-K}{\sigma \sqrt{T}}\right) <0. \end{align*} On the other hand, let $\sigma_I(K)$ be the log-normal implied volatility, that is, \begin{align*} C = C(K, \sigma_I(K)). \end{align*} Then \begin{align*} \frac{dC}{d K} &=\frac{\partial C}{\partial K} + \frac{\partial C}{\partial \sigma_I}\frac{\partial \sigma_I}{\partial K}. \end{align*} Here, \begin{align*} \frac{\partial C}{\partial K} = - \Phi(d_2), \end{align*} where \begin{align*} d_2 = \frac{\ln\frac{S_0}{K} - \frac{1}{2}\sigma_I^2 T}{\sigma_I \sqrt{T}}. \end{align*} Since \begin{align*} \lim_{K\rightarrow \infty}\frac{S_0-K}{\ln \frac{S_0}{K}} = \infty, \end{align*} we can expect that, for $K$ sufficiently large, \begin{align*} d_2 > \frac{S_0-K}{\sigma \sqrt{T}}. \end{align*} That is, \begin{align*} \frac{\partial C}{\partial \sigma_I}\frac{\partial \sigma_I}{\partial K} &= \Phi(d_2) - \Phi\left(\frac{S_0-K}{\sigma \sqrt{T}}\right) > 0. \end{align*} Then, \begin{align*} \frac{\partial \sigma_I}{\partial K} > 0, \end{align*} and the implied volatility is a increasing function of such strike levels. In conclusion, the implied volatility does not have to be a decreasing function of the strike.

$\endgroup$
  • $\begingroup$ Thanks. But don't we always have $\frac{\partial C}{\partial K} \leq 0$ regardless of the model using fhe fact that $C = E (S_T-K)^+$ ? $\endgroup$ – Dark Jul 20 '16 at 18:22
  • $\begingroup$ @Dark: I revised. $\endgroup$ – Gordon Jul 20 '16 at 20:37
  • $\begingroup$ Thank you. So we can expect a smile instead of a skew ? $\endgroup$ – Dark Jul 21 '16 at 8:37
  • $\begingroup$ @Dark:yes.smile is possible. $\endgroup$ – Gordon Jul 21 '16 at 10:39
  • $\begingroup$ I can't see anything wrong with Gordon's argument but it is counter intuitive. The market uses the approximate expression lognirmal vol(k)= lognirmal vol(atm)* sqrt(atm/k) for this problem, which is decreasing. $\endgroup$ – dm63 Jul 21 '16 at 11:02
3
$\begingroup$

The implied Black-Scholes skew will be downward sloping in the limit on both the left and the right. (I believe @Gordon's derivation claiming upward slope may have a sign error somewhere).

normal model implied vols

Left Side

For the left side it is sufficient to note that the lognormal model has no density below zero while the normal model has strictly positive density in that region. Thus, implied Black Scholes vols will approach $\infty$ as strikes $K \rightarrow 0$.

Right Side

For the right side, the simplest observation to make is that the lognormal has a much "Fatter" right tail than the normal model. This leads us to conclude that the relatively skinny tail of the normal model will give rise to smaller and smaller implied vols as we increase the strike.

Let's make this a bit more precise. It is easier to do the math if you consider pricing (and implying vols) of digital options rather than of vanilla options.

Without loss of generality we may take $F=T=1$ where $F$ is the expectation of $S_T$. The CDFs are

$$ CDF_{LN}(K) = \frac12 \left[1 + \operatorname{erf}\left(\frac{\ln K-1}{\sqrt{2}\sigma_{LN}}\right)\right] $$

and

$$ CDF_{N}(K) = \frac12\left[1 + \operatorname{erf}\left( \frac{K-1}{\sigma_N\sqrt{2}}\right)\right] $$

so the price of a digital call in the lognormal model is

$$ C_{LN}(K) = \frac12 \left[1 - \operatorname{erf}\left(\frac{\ln K-1}{\sqrt{2}\sigma_{LN}}\right)\right] $$

whereas in the normal model it is

$$ C_{N}(K) = \frac12\left[1 - \operatorname{erf}\left( \frac{K-1}{\sigma_N\sqrt{2}}\right)\right]. $$

The error function is strictly increasing, and $K$ dominates $\log(K)$ as $K \rightarrow \infty$. Thus regardless of the base volatilities $\sigma_{LN}, \sigma_{N}$ from which option prices are derived, we will eventually have

$$ C_{LN}(K) \gg C_{N}(K) $$

That is to say, a lognormal model would "expect" to see much higher option prices than the normal model is giving it. The arguments to the error function are $\log(K)$ versus $K$, leading to a negative slope in option price ratio and therefore a negative slope in implied vol.

Thus a negative slope in implied skew appears on the right side for prices derived from the normal model.

R Code

K = seq(0.5, 5, by=0.01)
F = T = sigma_N = 1
r = 0
d1 = (F-K)/(sigma_N*sqrt(T))
C = (F-K)*pnorm(d1) + (sigma_N*sqrt(T))/sqrt(2*pi)*exp(-d1^2/2)
vol_curve = implied_volatilities(C, CALL, F, K, r, 1)
plot(K, vol_curve,
     xlab='Strike', ylab="Imp_Vol",
     main="Implied Black-Scholes Vols Of A Normal Model")
$\endgroup$
  • $\begingroup$ It appears that this is a different topic. That is, what you have considered appears to be whether the lognormal vol is greater than the normal vol, while the original question asks whether the lognormal vol is a decreasing function of the strike. If you consider the derivative with respect to the strike, you might have a different conclusion. $\endgroup$ – Gordon Jul 21 '16 at 16:08
  • $\begingroup$ Thanks for the comment @Gordon. I have edited for clarity and double checked the phenomenon, adding a plot as well. I am pretty certain that your mathematics for showing the implied volatility slope is positive has a mistake somewhere. $\endgroup$ – Brian B Jul 21 '16 at 20:53
  • 1
    $\begingroup$ Thanks. If you can point out the specific error, if any, it will be more helpful $\endgroup$ – Gordon Jul 21 '16 at 22:33
1
$\begingroup$

Although it's a bit different story, there are VERY accurate approximation formulas for the implied volatility under normal model (so-called basis point volatility). Using them, you can obtain the implied vol directly without numerical root-finding like Newton's method.

This is my paper https://ssrn.com/abstract=990747 and an improvement https://ssrn.com/abstract=2420757 .

Below is some discussion on blogs:

https://www.clarusft.com/analytic-implied-basis-point-volatility/ https://www.clarusft.com/bachelier-model-fast-accurate-implied-volatility/

$\endgroup$
0
$\begingroup$

I agree with Gordon's deduction if the stock price is distributed that way under risk neutral measure. With sufficiently large K it should be monotone, but for other cases, there could be different cases. I think it can be helpful if you create a list of options with different strikes, time to maturity and spot prices to observe the multivariate relationship between implied vol and S0,K,T-t, based on the pricing formula.

$\endgroup$

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.