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How would a rise in implied volatility on a short-term option affect the implied volatility of another short-term option with the same strike, but with slightly-longer expiry?

Assuming that the short-term volatility rise is due to an event completely contained in the time of the short-term option.

Here is my stab at this:

Let $s$ and $\ell$ be the DTE of the short-term and long-term options.

Let $V_s$ be the short term option implied volatility now, and $V_\ell$ is the longer-term option volatility.

Since variances are additive, we should be able to calculate $V_{\ell-s}$, the implied volatility between the expiry of the short-term option to the expiry of the longer-term option right? $$V_{\ell-s}^2 = V_{\ell}^2 - \frac{s}{\ell}V_s^2$$

Then if $V_s$ rises to $V_{s*}$, then $V_{\ell}$ should rise to $V_{\ell*}$ where

$$ (V_{\ell*})^2 = \frac{s}{\ell}V_{s*}^2 + V_{\ell-s}^2 $$

Giving us:

$$ (V_{\ell*})^2 = \frac{s}{\ell}\left(V_{s*}^2 - V_s^2\right)+ V_{\ell}^2 $$

From a more practical standpoint, I'm curious how this translates to the value of options? I've been reading about how calendar spreads are not always long-vega strategies?

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    $\begingroup$ Without doing any maths, the implied vol for the slightly longer-dated option will cover the period of the shorter-dated option. If the shorter-dated option starts experiencing an increase in implied vol, the market is pricing in some stressed event (could be dividends if we're talking a stock option, or a central bank rate decision if we're talking an FX or Rates option). This stress event will affect the underlying in a way that will also affect the value of the longer-dated option. So you can definitely expect the IV to also increase for the longer-dated option. $\endgroup$ – Jan Stuller Jan 11 at 10:30
  • $\begingroup$ @JanStuller Obvious. But the question is how much? $\endgroup$ – Winston Du Jan 12 at 16:57
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Short answer: I believe the quantification of the impact of an increase in the shorter-dated vol on the longer-dated vol depends on what type of vol function we select to model the Volatility term-structure of the options.

Long answer (focusing on piece-wise volatility term-structure function):

Variances of log-returns in the GBM model are proportional to the time increment because of the scaling property of Brownian motion (i.e. an assumption inherent in the model):

$$ln\left(\frac{ S_{t_{i+1}}}{S_{t_{i}}}\right)=(\mu+0.5\sigma^2)(t_{i+1}-t_i)+\sigma W(t_{i+1}-t_i)\overset{d}{=}(\mu+0.5\sigma^2)(t_{i+1}-t_i)+\sigma \sqrt{t_{i+1}-t_i}Z$$

In the GBM model, each option with expiry $T_i$ would have its own BS volatility $\sigma_{T_i}$. To compute how an increase in a shorter-dated $\sigma_{T_i}$ affects a change in some longer-dated $\sigma_{T_i}$, we could map the BS vols onto some time-dependent vol function $\tilde{\sigma}(t)$:

$$\sigma_{T_i}T_i=\int_{h=0}^{h=T_i}\tilde{\sigma}(h)dh$$

If we assume piece-wise constant $\tilde{\sigma}(t)$, we will get:

$$\sigma_{T_i}T_i=\sum_{j=1}^{i}\tilde{\sigma}_j(T_{j}-T_{j-1})$$

We then get:

  • $\sigma_{T_1}^2T_1=\tilde{\sigma}^2_1T_1$
  • $\sigma^2_{T_2}T_2-\tilde{\sigma}^2_1T_1=\tilde{\sigma}^2_2(T_2-T_1)$
  • $\sigma^2_{T_n}T_n-\sum_{j=1}^{j=n-1}\tilde{\sigma}_j^2(T_j-T_{j-1})=\tilde{\sigma}^2_{T_n}(T_n-T_{n-1})$

Focusing on $T_2$ and $T_1$ for simplicity, supposing $\sigma_{T_1}$ goes up (so that $\tilde{\sigma}_1$ also goes up), but the forward piece-wise volatility $\tilde{\sigma}_2$ between $T_2$ and $T_1$ doesn't change, the new value of the BS volatility (that I denote $\sigma_{T_2}^*$) for the option expiring at $T_2$ would be:

$$(\sigma_{T_2}^*)^2T_2=\left(\tilde{\sigma}_{1}+\delta_{\tilde{\sigma}_1}\right)^2 T_1+\tilde{\sigma}_2^2(T_2-T_1)$$

The difference between $\sigma_{T_2}^*$ and the old value $\sigma_{T_2}$ is then simply:

$$\delta_{\sigma_{T_2}}=\sqrt{(\sigma_{T_2}^*)^2-\sigma_{T_2}^2}=\sqrt{\frac{T_1}{T_2}(2\tilde{\sigma}_1\delta_{\tilde{\sigma}_1}+\delta_{\tilde{\sigma}_1}^2)}$$

To compute the impact on the value of the option expiring at $T_2$, we could plug the quantity $\delta_{\sigma_{T_2}}$ into the option Vega.

If we assume different than piece-wise constant volatility function, the quantity $\delta_{\sigma_{T_2}}$ would be different.

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  • $\begingroup$ Can you clarify all of the variables you're using? The notation is confusing. For example, what is $\tilde{\sigma}_1$ supposed to be? I thought you defined it as a function. $\endgroup$ – Winston Du Jan 15 at 8:13
  • $\begingroup$ @WinstonDu: $\tilde{\sigma}_1$ is the piece-wise constant volatility between $t_0$ and $T_1$. $\tilde{\sigma}_2$ is the piece-wise constant volatility between $T_1$ and $T_2$, and so on. Together, across all indexes $j$, the function $$\sum_{j=1}^{i}\tilde{\sigma}_j(T_{j}-T_{j-1})$$ is the piece-wise volatility function $\tilde{\sigma}(t)$. I could have probably introduced a new variable for the summation, but I didn't want to have too many variable. Hope that's a bit clearer. $\endgroup$ – Jan Stuller Jan 15 at 8:19

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