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In a notes on "Option Pricing using Fourier Transform": Price of plain vanila call is given by $$ C(t, S_t) = e^{-rT}\mathbb{E}^{\mathbb{Q}}[(S_T -K)^+|\mathcal{F}_0] = e^{-rT} \int_K^{\infty} (S_T -K)\mathbb{Q}(S_T|\mathcal{F}_0) dS_T$$ It is a standard formula. In the next step the author claims to use, a change of variable from $S_T$ to $\ln S_T$ and writes $$ C(T, K) = e^{-rT} \int_{\ln K}^{\infty} (e^{\ln S_T} -e^{\ln K})\mathbb{Q}(\ln S_T|\mathcal{F}_0) d \ln S_T$$ which needs some explanation. To be precise, I think it should be $\ln S_T$ in place of $e^{\ln S_T}$ in the integral and the rest is fine.

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Note that the second expression is not based on a substitution of the first expression; it is a different view: \begin{align*} e^{-rT}E\left((S_T-K)^+\right) &= e^{-rT}E\left(\left(e^{\ln S_T}-e^{\ln K}\right)^+\right)\\ &=e^{-rT}\int_{-\infty}^{\infty}\left(e^{\ln S_T}-e^{\ln K}\right)^+Q(\ln S_T\mid \mathcal{F}_0)\, d\ln S_T\\ &=e^{-rT}\int_{\ln K}^{\infty}\left(e^{\ln S_T}-e^{\ln K}\right)Q(\ln S_T\mid \mathcal{F}_0)\, d\ln S_T. \end{align*}

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  • $\begingroup$ I am still not clear. I think, the step, $E[(S_T - K)^+] = E[(\ln S_T - \ln K)^+]$ should agree on the change of variable step in the question above. The expectations on both sides of equality are integrals, which can only be equal if we can get one from the other via some chnage of variable. $\endgroup$ – MathNovice Nov 22 '15 at 19:50
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    $\begingroup$ The identity $E((S_T-K)^+) = E((\ln S_T-\ln K)^+)$ is incorrect. What I did is to treat $\ln S_T$ as a random variable, and $Q(\ln S_T \mid \mathcal{F}_0)$ its density function. $\endgroup$ – Gordon Nov 22 '15 at 22:15
  • $\begingroup$ I am sure it is something really very basic but still: Another author does the same step as: $$ e^{-rT}\mathbb{E}^{\mathbb{Q}}[(S_T -K)^+|\mathcal{F}_0] = e^{-rT} \int_{\ln K}^{\infty} (e^{\ln S_T} - K)\mathbb{Q}(\ln S_T|\mathcal{F}_0) d \ln S_T$$. For me it has the same question; Why not transform $S_T$. @Gordon: For a beginner like I am, the first line of your answer states this identity. In addition, book by Rouah, at page 4 states that, $E^Q [\mathbb{1}_{S_T > K} ] = Q(S_T > K) = Q(\ln S_T > \ln K)$. $\endgroup$ – MathNovice Nov 24 '15 at 17:59
  • $\begingroup$ $E\left((S_T-K)^+\right) = E\left(\left(e^{\ln S_T}-e^{\ln K}\right)^+\right)$ is not the same as $E\left((S_T-K)^+\right) = E\left(\left(\ln S_T-\ln K\right)^+\right)$. $\endgroup$ – Gordon Nov 24 '15 at 18:03
  • $\begingroup$ Got it. So we are finding expectation of the r.v. $(S_T - K)^+$ using density of the r.v. $\ln S_T$ from $\ln K$ to $\infty$ and we claim that they are the same. $\endgroup$ – MathNovice Nov 24 '15 at 18:12
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I think it's ok $$ S_T = e^{\ln S_T} $$

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  • $\begingroup$ Well thats obvious, but how can we leave one of the $S_T$'s as it is and apply the change of variable on the rest? Point being $S_T - K$ remains $e^{\ln S_T} - e^{\ln K}$ but rest been replaced by $\ln S_T$. $\endgroup$ – MathNovice Nov 22 '15 at 19:45
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    $\begingroup$ Well, if you like, you could set $u=\ln(S_T)$ to make it more clear that the integral is over $u$, not $S_T$. You would get $e^u$ instead of $e^{\ln S_T}$. $\endgroup$ – Olaf Nov 22 '15 at 20:27
  • $\begingroup$ I agree that the notation is poor and I would not have written it this way. As Olaf says, replace $\ln S_T$ everywhere by $u$ and it makes much more sense. I go through all this in detail in "more mathematical finance". $\endgroup$ – Mark Joshi Nov 22 '15 at 23:17
  • $\begingroup$ Using $u = \ln (S_T)$ converts every $S_T$ into $e^u$? $\endgroup$ – MathNovice Nov 24 '15 at 18:04

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