How does high IV effect a put backspread?

Question

I have a hard time understanding how high IV effects the amount of gamma obtained via a put backspread. Is it via the angle on the payoff or via the ratio one gets i.e number of OTMs one can buy? or simply just that one has bigger numbers in the credit and debit for the position?

Answer 1

In a put back-spread, you are short one put and long two puts with a lower strike.

As far as Vega (the spread's price sensitivity to IV) is concerned, it depends on where you enter the spread. Usually, you'd enter the spread where the short put is at the money or near being at the money: this ATM put will have a high Vega, whist the two puts with lower strike will have a low vega to start with. So if the underlying price stays where it is, you're actually short Vega and increasing IV should increase the value of the short put more than the value of the two OTM puts. So in this scenario, increasing IV will lower the price of the spread.

As the underlying price moves lower and your short put becomes ITM, your two long puts will gradually come closer to being ATM. Since Vega is highest for ATM options, there will come a point where Vega on the two long puts will become higher than the Vega on the short put: then, increasing IV will increase the price of the spread.

Edit: picture below shows that deep OTM or ITM options have low sensitivity to IV (so if the options are on the SPX, the sensi to VIX for ITM or OTM options is much lower than for ATM options: that should answer your question as to "why the one ATM put benefits from higher VIX more than two OTM puts"):

Answer 2

Below, Gamma is denoted by $\Gamma$, and $IV=\sigma$:

$$ \Gamma = \left(\frac{1}{S_0*\sigma*\sqrt{T}}\right)*\left(\frac{1}{\sqrt{2\pi}}e^{\frac{-d_1^2}{2}} \right) $$

The expression in the first bracket is inversely proportional to $\sigma$, so isolating just this first expression: higher $\sigma$ will trivially lower this first expression.

The second bracketed expression is a Standard Normal PDF, where:

$$d_1=\frac{ln\left(\frac{S_0}{Ke^{-rT}}\right)}{\sigma\sqrt{T}}+0.5\sigma\sqrt{T}$$.

For OTM put options, $ln \left( \frac{S_0}{Ke^{-rT}} \right)$ is positive, because $S_0>Ke^{-rT}$. So increaing $\sigma$ will make this log-term smaller, which will make $d_1$ closer to zero and therefore the PDF $\left(\frac{1}{\sqrt{2\pi}}e^{\frac{-d_1^2}{2}} \right)$ will move closer to its peak. At the same time, increasing $\sigma$ will make the expresison $0.5\sigma\sqrt{T}$ larger, moving the $d_1$ away from zero, and therefore moving the PDF away from its peak.

So in conclusion: increasing IV decreases the first bracketed expression, and it is not possible to say with certainty whether it increases or decreases the second bracketed expression: whether the second bracketed expression increases with increasing $\sigma$ will depend on:

(i) how much OTM the option is,

(ii) the current level of $\sigma$,

(iii) Time to maturity $T$.

In general, increasing IV decreases Gamma via the first bracketed expression, but there can be limited cases where deep OTM options will gain Gamma from rapid increases in IV. To get a clearer idea, the best way to go about this is to try to plot the gamma for different strikes and play around with the $\sigma$ parameter.