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Hi I have some question in below payoff function derived from Taylor expansion to be used for pricing Variance Swap:[$]f(S_T) = f(S_*) + f'(S_*)(S_T-S_*) + \int_{0}^{S_*} f''(K)(K-S_T)^{+}dK + \int_{S_*}^{\infty} f''(K)(S_T-K)^{+}dK[$]In the proof:[$]f(S_T) [$][$]= \int_{0}^{S_*} f(K)\Delta(S_T-K)dK + \int_{S_*}^{\infty} f(K)\Delta(S_T-K)dK[$][$]= f(K)1_{S_T<K}\Big|_{0}^{S_*} - \int_{0}^{S_*} f'(K)1_{S_T<K}dK + f(K)1_{S_T\geq K}\Big|_{S_*}^{\infty} - \int_{S_*}^{\infty} f'(K)1_{S_T \geq K}dK [$][$] = ...[$]for [$]f(K)1_{S_T\geq K}\Big|_{S_*}^{\infty}[$], I just don't understand why its equal to [$] +f(S_*)1_{S_T \geq S_*}[$] instead of [$] -f(S_*)1_{S_T \geq S_*}[$]Any one have idea on that? Thanks, Remarks: [$]\Delta(S_T-K)[$] is the Dirac function

agree -- in the line above the line with the ..., the signs (+,-,+,-) should be (+,-,-,+)

But it was just simply integration by parts. Why its (+,-,-,+)?

Because different choices are being made for v in the \int u dv parts formula

Another proof of the same result can be found in http://feb.kuleuven.be/drc/AFI/research ... i-1377.pdf I hope it helps you. kind regards,Daniel

When I first looked at the formula I checked it for the call option payoff [$]f ( S_T )\,\, =\,\, max\,\{ S_T \,\,-\,\,K\, ,\,\,0\,\}[$]. The integration is quite simple and also it is quite an appropriate illustrative example.

Yes, another good one is [$]f(S_T) = \log \frac{S_T}{F_{0,T}}[$], where [$]F_{0,T}[$] is the forward price of the stock for sale at time T.When the stock is the SPX, the formula is the basis for the VIX (eqn (1)) whichis, in turn, an approximate fair strike (in vol pts) for the variance swap of the thread title.

Sorry my bad my calculus background is no good, would you mind to tell me why below is incorrect by negative sign? I have never tried integration by part with indicator function. Thanks, [$]\int_{S_*}^{\infty} f(K)\Delta(S_T-K)dK[$][$]= \int_{S_*}^{\infty} f(K)d(1_{S_T \geq K})[$]

But I saw his proof for the Lemma seems skipped some step which I don't understood...He said: [$]u(x)[$][$]= u(a) + xu'(x) - au'(a) + \int_{x}^{a}Ku''(K)dK[$]By Considering the cases x < a and x > a[$]= u(a) + u'(a)(x-a) + \int_{inf I}^{a}u''(K)(K-x)_{+}dK + \int_{a}^{sup I}u''(K)(x-K)_{+}dK[$]I don't know how he did with the term [$]xu'(x)[$] as it disappeared in the next line...

QuoteOriginally posted by: yktsuiSorry my bad my calculus background is no good, would you mind to tell me why below is incorrect by negative sign? I have never tried integration by part with indicator function. Thanks, [$]\int_{S_*}^{\infty} f(K)\Delta(S_T-K)dK[$][$]= \int_{S_*}^{\infty} f(K)d(1_{S_T \geq K})[$]Fix a value for [$]S_T[$], say 2. Now plot [$]1_{S_T \geq K}[$] vs. [$]K[$]. See?

QuoteOriginally posted by: yktsuiBut I saw his proof for the Lemma seems skipped some step which I don't understood...He said: [$]u(x)[$][$]= u(a) + xu'(x) - au'(a) + \int_{x}^{a}Ku''(K)dK[$]By Considering the cases x < a and x > a[$]= u(a) + u'(a)(x-a) + \int_{inf I}^{a}u''(K)(K-x)_{+}dK + \int_{a}^{sup I}u''(K)(x-K)_{+}dK[$]I don't know how he did with the term [$]xu'(x)[$] as it disappeared in the next line...Proof: Assume now that [$]x<a.[$] Then we have that [$]\left( x-K\right) _{+}=0[$] if [$]K>a[$]and therefore [$]\int_{a}^{\sup I}u^{\prime\prime}\left( K\right) \left(x-K\right) _{+}[$]d[$]K=0[$] Furthermore, we have that[$]\begin{eqnarray*}\int_{\inf I}^{a}u^{\prime\prime}\left( K\right) \left( K-x\right)_{+}\text{d}K & =&\int_{\inf I}^{x}u^{\prime\prime}\left( K\right)\underset{=0}{\underbrace{\left( K-x\right) _{+}}}\text{d}K+\int_{x}^{a}u^{\prime\prime}\left( K\right) \left( K-x\right) _{+}\text{d}K\\ & =&\int_{x}^{a}u^{\prime\prime}\left( K\right) \left( K-x\right)\text{d}K\\& =&\int_{x}^{a}u^{\prime\prime}\left( K\right) K\text{d}K-x\int_{x}^{a}u^{\prime\prime}\left( K\right) \text{d}K\\& =&\int_{x}^{a}u^{\prime\prime}\left( K\right) K\text{d}K-xu^{\prime}(a)+xu^{\prime}(x).\end{eqnarray*}[$]We can then write[$]\begin{eqnarray*}& u\left( a\right) +u^{\prime}\left( a\right) \left( x-a\right)+\int_{\inf I}^{a}u^{\prime\prime}\left( K\right) \left( K-x\right)_{+}\text{d}K+\int_{a}^{\sup I}u^{\prime\prime}\left( K\right) \left(x-K\right) _{+}\text{d}K\\& =u\left( a\right) +u^{\prime}\left( a\right) \left( x-a\right)+\int_{x}^{a}u^{\prime\prime}\left( K\right) K\text{d}K-xu^{\prime}(a)+xu^{\prime}(x).\\& =u\left( a\right) +u^{\prime}\left( a\right) x-u^{\prime}\left(a\right) a+\int_{x}^{a}u^{\prime\prime}\left( K\right) K\text{d}K-xu^{\prime}(a)+xu^{\prime}(x)\\& =u\left( a\right) +xu^{\prime}(x)-u^{\prime}\left( a\right) a+\int_{x}^{a}u^{\prime\prime}\left( K\right) K\text{d}K\\& =u(x).\end{eqnarray*}[$]A similar procedure works for [$]x\geq a[$]