4
$\begingroup$

I was going through some paid video on options. The tutor in the video asked the following question:

Person $A$ has the following portfolio at the start of April

  • Portfolio of options with vega $20,000$ expiring end of April.
  • Portfolio of options with vega $-40,000$ expiring end of May.
  • Portfolio of options with vega $15,000$ expiring end of June.

Now if the monthly implied volatility increases from $\sigma$ % to $(\sigma+1)$%, is it good for person $A$, what is his exposure.

The naive approach is to add all vega's to get $-5,000$ and say with increase in volatility he makes a loss. The tutor goes on to explain that this approach is not correct and one needs to calibrate vegas as time of expiry is different. He says one can add $20,000 + (-40,000/(\sqrt{2})) + (15,000/\sqrt{3})$.

My doubt is why is the naive approach wrong. Vega means change in options price with $1$% change in implied volatility. Doesn't vega (if obtained from pricing models like Black Scholes) itself incorporate the time to expiry factor ? Would it be wrong to say portfolio of second month changes by $-40,000*\sqrt{252}$ ( taking annualized volatility).

PS : I know I am missing something. Being a beginner please excuse me if I used any wrong terms.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ It seems like he is assuming that the shorter term volatilities change more than the longer term ones and the relatively sensitivity is proportional to $1 / \sqrt{T}$. This is not an uncommon assumption and the corresponding vegas are often referred to as "time weighted vegas". $\endgroup$
    – LocalVolatility
    Sep 19, 2018 at 9:55
  • $\begingroup$ can you give this formally as answer. Would be really helpful. Thanks. $\endgroup$
    – advocateofnone
    Sep 19, 2018 at 12:12

2 Answers 2

Reset to default
4
$\begingroup$

It seems like he is assuming that the shorter term volatilities change more than the longer term ones and the relatively sensitivity is proportional to $1 / \sqrt{T}$. Thus, this hedge is not against a parallel shift of the surface. This is not an uncommon assumption and the corresponding vegas are often referred to as "time weighted vegas".

Improve this answer
$\endgroup$
3
  • $\begingroup$ I am sorry. But what do you mean by "assuming that the shorter term volatilities change more than the longer term ones". As I mentioned isn't the time to expiry already incorporated into vega ? Really sorry I'm a beginner. Also what do you mean by "this hedge is not against a parallel shift of the surface" ? $\endgroup$
    – advocateofnone
    Sep 19, 2018 at 12:17
  • 1
    $\begingroup$ Typically, the short-term implied volatility is more volatility than the long-term implied volatility. Using the $1 / \sqrt{T}$ approach e.g. assumes that when the 1-month implied volatility increases by 1%, then the 2-month implied volatility increases by ~0.7%. You are hedged against this scenario, which is not a parallel shift. $\endgroup$
    – LocalVolatility
    Sep 19, 2018 at 12:33
  • 1
    $\begingroup$ Each maturity has a different ATM IV, you can think of it as the "term structure of IV" (just like the term structure of interest rates). A "parallel shift" means the IVs would all increase by the same amount. More commonly, in response to an event, the near term IVs increase more than the long term. $\endgroup$
    – Alex C
    Sep 19, 2018 at 21:37
2
$\begingroup$

Your tutor is calculating the increase in total variance. The black-scholes model has the variance term of sigma^2 * Time-to-expiry.

Hence, when the monthly volatility increases by 1%, the effective increase for the 3mth option is sqrt(3) * 1%, the 2mth option is sqrt(2) * 1% etc. He explicitly assumes the vega is relative to the total variance - i.e. the vega is due to an increase in the sqrt(variance = sigma^2 * T).

I personally don't think he is doing this correctly, as the industry standard of defining vega, is literally the change in price due to a change in % volatility.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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