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In a case such as the Cox-Ingersoll-Ross where $$ \mathrm{d}{R\left(t\right)}=\left(\alpha-\beta R\left(t\right)\right)\mathrm{d}{t}+\sigma\sqrt{R\left(t\right)}\mathrm{d}{W\left(t\right)}, $$ is it wrong to do the following: \begin{align*} \mathrm{d}\left(e^{\beta t}R\left(t\right)\right)&=\mathrm{d}\left(e^{\beta t}\right)R\left(t\right)+\mathrm{d}\left(R\left(t\right)\right)e^{\beta t} \\ &= \beta e^{\beta t}R\left(t\right)\mathrm{d}t+e^{\beta t}\mathrm{d}{R\left(t\right)}. \end{align*}

Edit : add a $\mathrm{d}t$ in $\mathrm{d}\left(e^{\beta t}\right)=\beta e^{\beta t}\mathrm{d}t$

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Yes it is correct since $t\to e^{\beta t}$ has finite variation, you have no quadratic term in the Ito's lemma.

More precisely, if $A$ is a finite variation process and $X$ a semi-martingale you have : $$d(AX)_t =A_t dX_t + X_t dA_t$$

If now $X$ and $Y$ are two semi-martingales you have : $$d(XY)_t =X_t dY_t + Y_t dX_t + d\left\langle X, Y\right\rangle_t$$ where $\left\langle X , Y \right\rangle$ is the crochet between $X$ and $Y$.

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  • $\begingroup$ Thank you very much for this, this is exactly the reasoning that i had but wasn't sure of it! $\endgroup$
    – wilsnunn
    Apr 20, 2016 at 13:37

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