Ito's Lemma in the multi-dimensional case

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Suppose that $S_1$ and $S_2$ are two stochastic process defined by : $$ dS_i(t) = S_i(t)(rdt + \sigma_idB_i^t)$$ for $i = 1,2$

How to find the stochastic differential equation verified by : $$S_3(t) = \frac{S_1(t) + S_2(t)}{2}$$

Same question for : $$S_4(t) = \sqrt{(S_1(t)S_2(t)}$$

For example, let's consider $S_4$:

For $S_4$, if we use $\phi(x,y) = \sqrt{xy}$, we have : $$\phi'_x(x,y) = \frac{\sqrt{y}}{2\sqrt{x}}$$ $$\phi'_y(x,y) = \frac{\sqrt{x}}{2\sqrt{y}}$$ and : $$\phi''_{xx}(x,y) = \frac{-\sqrt{y}}{4}x^{-3/2}$$

Am i right ?

So the first part of Ito's formula should give us :

$$dS_4(t) = \frac{\sqrt{S_2}}{2\sqrt{S_1}}dS_1 + \frac{\sqrt{S_1}}{2\sqrt{S_2}}dS_2 + ...$$

I stop right there because i know it's already false... (i have the result and it has nothing to do with that).

It doesn't seem to be the right way to apply the Ito's formula...

Even for the second part, will it be something like :

$$\frac{1}{2}* \frac{-1}{4}\sqrt{S_2}S_1^{-3/2}d<S_1,S_1> + ... $$

Could you help me ?

Thanks a lot.

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1) For $S_3(t) = \frac{S_1(t) + S_2(t)}{2}$, there is no much you could do except,

$$ dS_3(t) = S_3(t)rdt +\frac12\left[ S_1(t)\sigma_1dB_1^t+S_2(t) \sigma_2dB_2^t\right]$$

2) For $S_4(t) = \sqrt{(S_1(t)S_2(t)}$, you may proceed as follows,

$$dS_4^2(t)=S_2(t)dS_1(t)+S_1(t)dS_2(t)+dS_1(t)dS_2(t)$$

$$= S_4^2(t)[(2r+\rho\sigma_1\sigma_2)dt+\sigma_1dB_1^t+\sigma_2dB_2^t]$$

where $\rho dt=<dW_1^tdW_2^t>$. Rewrite $dS_4^2(t)=2S_4(t)dS_4(t) + [dS_4(t)]^2$,

$$dS_4(t)= \frac12S_4(t)[(2r+\rho\sigma_1\sigma_2)dt+\sigma_1dB_1^t+\sigma_2dB_2^t] -\frac1{2S_t(t)}[dS_4(t)]^2$$ $$= S_4(t)\left[\left(r-\frac18\sigma_1^2-\frac18\sigma_2^2+\frac14\rho\sigma_1\sigma_2\right)dt+\frac12\sigma_1dB_1^t+\frac12\sigma_2dB_2^t\right]$$