# ATM strike Heston model

I'm thinking about the heston model. price of the asset $$S^1=(S_t^1)_{t \leq T}$$ fullfills the differential equation $$dS_t^1=S_t^1(\mu dt + \sqrt{V_t} dB_t^1)$$ the stochastic volatility is given by $$V=(V_t)_{t \leq T}$$ $$dV_t= \kappa (\theta-V_t)dt+ \sigma \sqrt{V_t} d B_t^V$$

$$dB_t^V dB_t^1=d[B^V,B^1]_t= \rho dt$$

Now with an exchange of measure I get the following $$d \tilde{S_t^1}=\tilde{S_t^1} \sqrt{V_t} d W_t^V$$

$$d V_t= \kappa (\theta- V_t)dt+ \sigma \sqrt{V_t} d W_t^V$$

$$dW^1 dW^V= \rho dt$$

$$\tilde{S_t^1}=S_0 e^{X_t}$$

$$dX_t=-\frac{1}{2}V_t dt+ \sqrt{V_t} dW_t^1$$

Now I want to define that a call option is at-the-money $$K=S_0 e^x$$ with $$x$$ the log moneyness. So the option is ATM if $$K=S_0$$. I also saw the definition for ATM that $$K=F_t$$. My question now is are these two definitions for ATM the same or not?

• What is $x$? Do you want to set the ATM strike to be the forward to the option maturity? Jan 7, 2019 at 18:03
• You may assume $\mu=r$ under the risk-neutral measure. How did you make the measure change? I would suggest you edit your question so that it is easier for understanding. Jan 7, 2019 at 18:05
• Moneyness is a concept independent of the underlying model. When we speak of ATM we usually specify ATM spot, or ATM forward (ATMF). The first is simply $K=S_t$ the other is $K = F_t(T)$, where $T$ is the maturity of the option considered since ATMF is maturity dependent. $t$ is today/current time.
– user34971
Jan 8, 2019 at 4:48
• So I can not mix up these two definitions?
– P.G.
Jan 8, 2019 at 10:19
• No you can't, unless the risk neutral drift for the spot process $S_t$ is zero in which case $S_t = F_t(T) \,\, \forall T$
– user34971
Jan 10, 2019 at 0:53