Calibration of Heston using implied vol as $v_0$

Question

I am looking at the difference if you calibrated the heston from market data using objective function minimisation.

In scenario 1, I calibrate all the parameters from market data In scenario 2, I calculate the implied volatility using market data for strikes close to ATM and 1-2 weeks expiry, then I use that as my parameter for $v_0$ in the Heston and calibrate the rest of the parameters.

In other words, scenario 2 is calibrating with 1 less unknown parameter.

For scenario 2, when calculating the implied vol's, if I use options that expire in 1-2 weeks with strikes 1 above and below of $S_0$ (So if $S_0=4002.15$, then use strikes $K=4000$ and $K=4005$, calculate those implied vols to get an estimation of implied vol for $S_0 = 4002.15$).

Would I then only need to linearly interpolate with degree=1 because the vol surface is quite flat in that area? Or would I need to use options further away from ATM to get a better fit for the implied vol at $S_0$ (Since implied vol is curved in reality and not flat).

(I'm not calculating implied vol of ITM options, I'm just using put options with put-call parity to get implied vols for $K<S_0$)

Answer 1

Using the ATM implied vol of short term options is indeed a common practice for $v_0$ as in your Scenario 2. Linear interpolation should be enough, given that 1 week is somewhat arbitrary anyway. In Scenario 1, you will want to use this for the initial guess of the minimization.

Of course, it is not much more costly to use 3 points parabola interpolation, so if you really worry about those details, you may as well use the 3 points parabola, but it's not necessary unless the market quotes are sparse (not like your example of 4000 and 4005 but more like 3500 and 4500).

$v_0$ (and $\theta$) may also be implied from the curve of variance swap prices. See F. Guillaume, W. Schoutens (2010) "Use a reduced Heston or reduce the use of Heston?" This is also more in line with Bergomi variance curve approach.