# Analytical Solution for Heston Model

What is the rationale of equation 7 in this paper? Could you please provide a step-by-step demonstration of this equality?

• By the way - this is a horribly written paper. It makes big claims about a better analytical valuation function for European vanilla options in the Heston model and the corresponding probability density functions. However, the authors don't even present their formulas or discuss the relationship to the (vast) existing literature. A rule of thumb is - when you find a non-published, non-cited paper called "breakthrough in ###", then consume it with caution. – LocalVolatility Jan 16 '17 at 22:09
• This paper does not provide actual claimed result. It is just some advertisement from the authors to claim that they know it. No wonder the paper is not published ;) – Ezy Nov 12 '18 at 14:20

This equation is unrelated to the Heston model. It is simply the value of a European call under the a constant coefficient geometric Brownian motion, i.e. the Black and Scholes (1973) model. Here $\nu$ is the constant volatility and $\mu$ is the risk-neutral drift of the asset. For a stock you could for example have $\mu = r - q$ where $r$ is the risk-free interest rate and $q$ is the dividend yield.

You find a derivation of this formula in almost any introductory book on continuous time finance - e.g. Shreve's "Continuous Time Finance II" or Musiela and Rutkowksi's "Martingale Methods in Financial Modelling".

Note that there is one mistake in the formula: the authors forget to discount the call price. In their notation it should read

$$C_0 = \frac{1}{2} \color{red}{e^{-r \left( T - t_0 \right)}} \left( S_0 e^{\mu \left( T - t_0 \right)} \mathrm{erfc} \left( -\frac{d_+}{\sqrt{2}} \right) - K \mathrm{erfc} \left( -\frac{d_-}{\sqrt{2}}\right) \right),$$

where

$$d_\pm = \frac{1}{\sqrt{\nu \left( T - t_0 \right)}} \left( \ln \left( \frac{S_0}{K} \right) + \left( \mu \pm \frac{1}{2} \nu \right) \left( T_0 - t \right) \right)$$

Usually you find this formula expressed in terms of the cumulative normal distribution function $\mathcal{N}(x)$ instead of the complementary error function $\mathrm{erfc}(x)$. The connection between the two is

\begin{eqnarray} \frac{1}{2} \mathrm{erfc} \left( -\frac{d_\pm}{\sqrt{2}} \right) & = & \frac{1}{2} \left( 1 - \mathrm{erf} \left( -\frac{d_\pm}{\sqrt{2}} \right) \right)\\ & = & \frac{1}{2} \left( 1 + \mathrm{erf} \left( \frac{d_\pm}{\sqrt{2}} \right) \right)\\ & = & \mathcal{N} \left( d_\pm \right). \end{eqnarray}

You then get the more familiar expression

$$C_0 = e^{-r \left( T - t_0 \right)} \left( S_0 e^{\mu \left( T - t_0 \right)} \mathcal{N} \left( d_+ \right) - K \mathcal{N} \left( d_- \right) \right).$$