# Calibration of $\rho$ in the heston model

When calibrating the Heston model, the gradient of the price of the call/cost function wrt $$\rho$$ (correlation between $$S$$ and $$V$$), is a lot less than the other parameters like $$v_0$$ and $$\bar{v}$$. As a result, any algorithm that uses the gradient, will over-adjust $$\rho$$ straight to the boundary because it requires a larger change in $$\rho$$ to reduce the cost function compared to the other parameters. Even if you use box constraints like a sigmoid function, it will take rho to -1 or 0, if your initial guess was over or under.

If you start with a really small damping/scaling factor, it won't matter as over time, $$\rho$$ will still want to go to the boundary. I thought of making $$\rho's$$ steps arbitrarily smaller than the others, but that sort of defeats the purpose of using gradient based optimisation.

It seems like a fundamental issue - if the gradient of 1 parameter is far less than the others, it will always be over-adjusting the smaller gradient parameter. How do people normally overcome this and getting a reasonable $$\rho$$ and not $$\rho =\{-1,0\}$$

In other words, you could choose any value of $$\rho$$ and still find a local minima. How do people normally deal with this?

• I have a question. The parameters you give initially is a good candidate to be the objective solution? As I told you in a similar question you asked some time ago. You are basically asking how to prevent the gradient method from adjusting the update of the objective function for $\rho = \{0 , -1\}$ which (unfortunately) in Heston's model is the one that guarantees the largest negative variation of descent. You do realize that what I just wrote is a contradiction right? My advice try generating from a uniform (0,1) as many sets of $\theta^{(0)}$ to pass to the algorithm and see what happens Commented Nov 14, 2023 at 13:35
• Ok, I'll try that. But yes, the only way that I can prevent $\rho$ from instantly going to 0 or -1 is setting my damping factor to like 1e-9, but then $\rho$ will inevitably go to -1 or 0 anyway. I realise that we have many local minimums and realistically, $\rho$ could be considered futile as you can always make small adjustments to v parameters instead. Im more so asking a general idea of how to tackle this problem because out of all the heston calibration papers I've, not 1 has discussed this phenomena - gradient wrt 1 param being a lot less than the rest, resulting it going to the boundary Commented Nov 14, 2023 at 23:33
• Out of curiosity, what algorithm do you use to minimize and what programming language did you use? Commented Nov 15, 2023 at 19:40
• @SimoPape python + levenberg-marquardt algorithm. I'm not using scipy because their LM algorithm doesn't have bounds, and am using the cosine expansion method for the heston Commented Nov 15, 2023 at 23:59
• If you have MatLab I invite you to look at my GitHub repository where I calibrate Heston (and also Bates) using the COS method and the Levenberg-Marquardt algorithm (See cap4). Maybe it can give you a source of inspiration github.com/SSPerotti/MasterThesis_FFTCOS Commented Nov 17, 2023 at 14:04