If possible, I would like to proof that:E [ dW1 / dW2 ] = 0,given that W1 and W2 are two uncorrelated Wiener processes. Is this possible in any way?

You have written something that doesn't make sense dW_1/dW_2 is not even definable Sorry I simply don' t get your point, may be you could elaborate on your problem so we could understand where this question comes from ?

This issue popped up when I was trying to prove something from the Heston model. The derivation, I have to admit, is mathematically not fullproof (ok, this is perhaps an understatement ).I realize that it is a somewhat weird ratio to define, dW1 / dW2...However, I was hoping there was some mathematical interpretation (perhaps with Malliavin calculus or something) that I can use so that I can show that the expectation of it is indeed zero..

I can't reply for Malliavain Calculus but to me it doesn't make sense what you can define thoug his the ratio of 2 stochastic integral if the denominator is different of 0.

I agree with TheBridge, this ratio doesn't seem to have any immediate meaning (at least not to me). However, if you think of dW1 and dW2 as dW1 = N1 sqrt(dt) and dW2 = N2 sqrt(dt) where N1,N2 are independent standard Gaussian random variables, then dW1/dW2 = N1/N2 = C and it is well-known that C has a standard Cauchy distribution : P(x < C < x+dx) = dx / Pi * (1+x^2) ; so C is symmetrical (C = -C in distribution) but it's not integrable, i.e. its expectation is undefined.Be careful, you could be tempted to think that even if the expectation is undefined you may "act as if it was 0" since the distribution is symmetrical. For example if (Cn) is a sequence of iid Cauchy variables then the sample mean (C1+...+Cn)/n does NOT converge to 0, not even in some weak sense (in fact the sample mean has the same distribution as C1, this is stability of index 1).

Thanks both for your advice! (Especially moltabile for the extension to Cauchy)I have found another derivation of what I was trying to prove, so no need for any undefined ratio any more.Regards