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I am reading an article and came across this expression and would appreciate some explanation.

"We have a function $$u(F)=2\Big[\frac{F-L}{L}-ln\frac{F}{L}\Big]$$

Since $u''(F)>0$ a Taylor series expansion with second-order remainder of $u(F_n)$ about $F_n=L$ implies $$u(F_n)=\int_0^L\frac{2}{K^2}(K-F_n)^+dK+\int_L^\infty\frac{2}{K^2}(F_n-K)^+dK$$"

Now, with what I know I would have the Taylor series to be $$\frac{2}{L^2}(L-F_n)$$ I do not see how the integrals play a part in the series expansion. If someone could shed some light over this it would be extremely helpful. Thanks!

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  • $\begingroup$ the function $u(F)$ has a discontinuous second derivative at $F=L$, and a first derivative that vanishes identically for $F<L$, hence the more involved expressions, see risk.net/data/Pay_per_view/risk/technical/2004/… $\endgroup$ May 17, 2014 at 9:25
  • $\begingroup$ Thank you Carlo! I am actually reading the exact same article only I do not see the need for an indefinite integral to describe the series expansion. $\endgroup$
    – Ptouma
    May 17, 2014 at 9:30

1 Answer 1

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you can find this worked out in Appendix 1 of Towards a Theory of Volatility Trading

the general identity (for $L>0$) is

$$u(F)=u(L)+u'(L)[(F-L)^+-(L-F)^+]$$ $$\qquad+\int_0^L u''(K)(K-F)^+\,dK+\int_L^\infty u''(K)(F-K)^+ dK,$$

where $(K-F)^+$ means $K-F$ if $F<K$ and zero if $F>K$.

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