# How calculate expectation and variation of stochastic integral Based on Heston model?

I was calculated Heston volatility model. But I think it is wrong.

$$dS_t = \mu dt + \sqrt V_t dW_t^s$$
$$dV_t = k(\theta - V_t)dt + \sigma \sqrt V_t dW_t^v$$.
$$dW^s_t dW^v_t = \rho dt$$

take integral to calculate stock price process,

$$\int_0^t dS_s = \int_0^t\mu dt + \int_0^t \sqrt V_s dW_s^s$$ then, $$S_t-S_0 = \mu t + \int_0^t \sqrt V_s dW_s^s$$

## My derivation

We have to calculate the second term $$\int_0^t \sqrt V_s dW_s^s$$. since, volatility process know as CIR process can express blow.

$$V_t=e^{-\kappa t} V_0+\theta\left(1-e^{-\kappa t}\right)+\sigma e^{-\kappa t} \int_0^t e^{\kappa s} \sqrt{V_s} d W_s^v$$

thus take a root and multiple $$dW_t^s$$ then,
$$\sqrt V_t dW_t^s = \sqrt {(e^{-\kappa t} V_0+\theta\left(1-e^{-\kappa t}\right)+\sigma e^{-\kappa t} \int_0^t e^{\kappa s} \sqrt{V_s} d W_s^v)} dW_t^s$$.

$$= \sqrt {(e^{-\kappa t} V_0 (dW^s_t)^2+\theta\left(1-e^{-\kappa t}\right)(dW^s_t)^2+\sigma e^{-\kappa t} \int_0^t e^{\kappa s} \sqrt{V_s} d W_s^v(dW^s_t)^2)}$$

since, $$(dW_t)^2 = dt , dW_t^s dW_t^v = \rho dt \text{ and } dt dW_t^s\rho = 0$$

$$\sqrt V_t dW_t^s = \sqrt {(e^{-\kappa t} V_0+\theta\left(1-e^{-\kappa t}\right)} dW_t^s$$.

therefore,

$$\int_0^t \sqrt V_s dW_s^s = (\sqrt {(e^{-\kappa t} V_0+\theta\left(1-e^{-\kappa t}\right))} \int_0^t dW_t^s \sim \mathcal{N}(0,e^{-\kappa t} V_0+\theta\left(1-e^{-\kappa t}\right))$$

Please let me know where the wrong process is.
thank you.

Using $$s$$ for labelling $$W$$ and as an integrand caused ambiguity. Here I have relabelled $$W^s \rightarrow W^1; \qquad W^v \rightarrow W^2$$ $$dW_s^2 (dW_t^1)^2 = \underbrace{dW_s^2 dW_t^1}_{\neq \rho \ d t \ !} dW_t^1$$