Deriving Heston Call Probabilities Under Stock and Bond Numeraires

# Confused about term disappearing in the heston model after going from stock to bond numeraire


# Confused about term disappearing in the heston model after going from stock to bond numeraire












Where does $\varphi(-i)$ go in the heston model? I know that we end up with (following the steps in [1][1] ) \begin{align*} \mathbb{Q}\big[\{S_T\geq K\}\big] &= \frac{1}{2}+\frac{1}{\pi}\int_0^\infty \Re\left(\frac{e^{-i\ln(K)u}\varphi(u)}{iu}\right)\mathrm{d}u, \\ \mathbb{S}\big[\{S_T\geq K\}\big] &= \frac{1}{2}+\frac{1}{\pi}\int_0^\infty \Re\left(\frac{e^{-i\ln(K)u}\varphi(u-i)}{iu\varphi(-i)}\right)\mathrm{d}u, \end{align*} With $\varphi$ the characteristic function under the bond numeraire.

However, most formulas I see online are of the form \begin{equation} C(t,S_t) = S_t \Big( \frac{1}{2} + \frac{1}{\pi}\int^{\infty}_{0} \Re ( \frac{ \varphi{(u-i)}}{iu } e^{-iu \ln K} ) du \Big) - Ke^{-r(T-t)} \Big( \frac{1}{2} + \frac{1}{\pi}\int^{\infty}_{0} \Re \Big( \frac{ \varphi{(u)}}{iu}e^{-iu \ln K} \Big) du \Big) \end{equation} And one can show that this is the correct form by doing a Monte Carlo simulation. Still, I don't understand what cancels out with $\varphi(-i) \;=\; \mathbb{E}^{Q}\bigl[e^{i(-\,i)\,X_{T}}\bigr] \;=\; \mathbb{E}^{Q}\bigl[e^{X_{T}}\bigr] \;=\; \mathbb{E}^{Q}\bigl[S_{T}\bigr] \;=\; S_{t}\,e^{\,r\,(T - t)}.$ Shouldn't the final formula be something along the lines of \begin{equation} \Big(S_t \frac{1}{2} + e^{-r(T-t)}\frac{1}{\pi}\int^{\infty}_{0} \Re ( \frac{ \varphi{(u-i)}}{iu } e^{-iu \ln K} ) du \Big) \end{equation} [1]: Deriving the solution for European call option in the Heston Model

## Answer by Kevin (score 2, accepted)

https://quant.stackexchange.com/a/83601

Geman, El-Karoui, and Rochet (1995) show that call option prices are given by \begin{align*} C = S_0e^{-qT}\mathbb{S}\big[\{S_T\geq K\}\big] - Ke^{-rT} \mathbb{Q}\big[\{S_T\geq K\}\big]. \end{align*} where $\mathbb{S}$ and $\mathbb{Q}$ are risk-neutral probability measures using the reinvested stock and risk-free bank account as numeraire, respectively. Let $k=\ln(K)$. By the inversion formula from Gil-Pelaez (1951) inversion formula, we obtain \begin{align*} \mathbb{S}\big[\{S_T\geq K\}\big] &= \mathbb{S}\big[\{\ln(S_T)\geq \ln(K)\}\big] \\ &= 1- F^{\mathbb{S}}_{\ln(S_T)}(k) \\ &= \frac{1}{2}+\frac{1}{\pi}\int_0^\infty \text{Re}\left(\frac{e^{-iku}\varphi_{\ln(S_T)}^{\mathbb{S}}(u)}{iu}\right)\mathrm{d}u, \end{align*} where $\varphi_{\ln(S_T)}^{\mathbb{S}}$ is the characteristic function of $\ln(S_T)$ under $\mathbb{S}$. Using the Radon-Nikodym derivative, we get \begin{align*} \varphi_{\ln(S_T)}^{\mathbb{S}}(u) &= \mathbb{E}^{\mathbb{S}}\left[e^{iu\ln(S_T)}\right] \\ &= \mathbb{E}^{\mathbb{Q}}\left[e^{iu\ln(S_T)}\frac{e^{\ln(S_T)}}{\mathbb{E}^{\mathbb{Q}}[e^{\ln(S_T)}]}\right] \\ &=\frac{\mathbb{E}^{\mathbb{Q}}\left[e^{(iu+1)\ln(S_T)}\right]}{\mathbb{E}^{\mathbb{Q}}\left[e^{\ln(S_T)}\right]}\\ &=\frac{\mathbb{E}^{\mathbb{Q}}\left[e^{i(u-i)\ln(S_T)}\right]}{\mathbb{E}^{\mathbb{Q}}\left[e^{i(-i)\ln(S_T)}\right]} \\ &= \frac{\varphi_{\ln(S_T)}^\mathbb{Q}(u-i)}{\varphi_{\ln(S_T)}^\mathbb{Q}(-i)}. \end{align*} Similarly, \begin{align*} \mathbb{Q}\big[\{S_T\geq K\}\big] &= \mathbb{Q}\big[\{\ln(S_T)\geq \ln(K)\}\big] \\ &= 1- F^{\mathbb{Q}}_{\ln(S_T)}(k) \\ &= \frac{1}{2}+\frac{1}{\pi}\int_0^\infty \text{Re}\left(\frac{e^{-iku}\varphi_{\ln(S_T)}^{\mathbb{Q}}(u)}{iu}\right)\mathrm{d}u. \end{align*}

Consequently, putting everything together, and denoting the risk-neutral characteristic function of the terminal log asset price by $\varphi$ (as opposed to $\varphi^\mathbb{Q}_{\ln(S_T)}$), we arrive at \begin{align*} \mathbb{S}\big[\{S_T\geq K\}\big] &= \frac{1}{2}+\frac{1}{\pi}\int_0^\infty \text{Re}\left(\frac{e^{-iku}\varphi(u-i)}{iu\varphi(-i)}\right)\mathrm{d}u, \\ \mathbb{Q}\big[\{S_T\geq K\}\big] &= \frac{1}{2}+\frac{1}{\pi}\int_0^\infty \text{Re}\left(\frac{e^{-iku}\varphi(u)}{iu}\right)\mathrm{d}u. \end{align*}