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

Question

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

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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.

Answer 1

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