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is the cos method used to calculate prices of options other than the European call? Or is this method only used for calibration? Is it possible to evaluate the barrier and lookback options? I am asking from the point of view of practice

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    $\begingroup$ This follow-up paper by Fang and Oosterlee (2009) discusses the pricing of American options and discrete barrier options. This paper by Zhang and Oosterlee (2013) discusses the ASCOS method (COS method for Asian options) which is generalised to early exercise features here. I guess you can apply similar ideas to lookback options. By the way, Fang Fang and Bowen Zhang were Oosterlee's PhD students. $\endgroup$
    – Kevin
    Commented Aug 30, 2020 at 13:39
  • $\begingroup$ Thanks, I am going to read it. Do you recommend any book for learning theory about Fourier series and fourier transform? $\endgroup$
    – Math122
    Commented Aug 30, 2020 at 15:12
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    $\begingroup$ You don't need a book on Fourier transforms or Fourier series. The summary from Schmelzle includes all the properties of characteristic functions you need to understand what's going on. But if you want to, Hirsa (2013) covers the plain COS method and it's very accessible. He also addresses early exercise and barrier options but no Asian or lookback options. It's a great pick if you're into numerics. $\endgroup$
    – Kevin
    Commented Aug 30, 2020 at 17:14
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    $\begingroup$ Okay, well that's a well-known fact and proven in most standard (functional) analysis books. It doesn't hold for all functions. You best start with $L^2$ and show that $\sin$ and $\cos$ form a basis of this space (I guess it's more effort to do it for all $L^1$ functions) Then, you immediately get the Fourier series. But anyways, you won't struggle to find this result in any introductory book on higher analysis. $\endgroup$
    – Kevin
    Commented Aug 31, 2020 at 17:01
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    $\begingroup$ @KeSchn I checked only cumulants for Kou and Merton model and I found 2 errors: In Kou model $c_2$ should be equal $t\left(\sigma^2+2\frac{\lambda p}{\eta_1^2}-2\frac{\lambda(1-p)}{\eta_2^2}\right)$ and in Merton model $c_4$ should be equal $t\lambda(\mu^4+6\delta^2\mu^2+3\delta^4)$ (I use different notation). $\endgroup$
    – Math122
    Commented Sep 2, 2020 at 20:48

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