Pricing Calls by Integrating the Log-Price Density

Summary

The document explains why a call-option pricing integral can be written in terms of the log of the terminal asset price. The key point is that the second expression is not obtained by merely substituting a logarithm into the original integral over price. Instead, it rewrites the expected payoff using the distribution of log-price: the payoff becomes the difference between exponentiated log-price and strike, and the density is that of the log-price variable.

Since a call pays only when the terminal price exceeds the strike, the integral over log-price begins at the log of the strike. This gives the same expected discounted payoff in a different variable representation. The answer clarifies the role of the exponential transformation, but does not show the Jacobian relationship needed to convert a density in price directly into one in log-price; the distinction between reformulating the expectation and performing a change of variables is essential.

Key ideas

• A call payoff expressed using log-price remains a function of exponentiated log-price.
• The corresponding integral uses the probability density of log-price rather than the price density.
• The integration threshold is the logarithm of the strike because the payoff is positive above that level.
• Rewriting an expectation in terms of a transformed random variable differs from substituting into an integral without adjusting its density.

Tags

# Fourier Transform


# Fourier Transform












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.

## Answer by Gordon (score 1, accepted)

https://quant.stackexchange.com/a/21882

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*}

## Answer by Mark Joshi (score 1)

https://quant.stackexchange.com/a/21879

I think it's ok $$ S_T = e^{\ln S_T} $$

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.