I have a question about the distribution of an Ito Integral. Consider the integral $$ \int_0^1 B_1(r) \mathrm{d}B_2(r), $$ where $B_1$ and $B_2$ are two independent standard Brownian motions. I am interested in the result (that I have read in a paper on time series analyis) that the above Ito Integral has the distribution of a Gaussian Mixture (first question: can anybody give me a rigorous definition of "Gaussian Mixture"). The argument that was provided is as follows:
1) Since $B_1$ and $B_2$ are independent, we can condition on $B_1$ without changing the distribution of $B_2$ (i.e., the conditional is the same as the unconditional distribution). I get that part.
2) So we can condition the above integral on $B_1$, hence treat $B_1$ as constant in the integral (why is that, or more precisely: how can I argue that rigorously?), and obtain that, conditionally on $B_1$, the integral has the distribution of a Gaussian random variable with mean zero and variance $\int_{0}^1B_1(r)^2\mathrm{d}r$. I get the second part (just the Ito integral of a deterministic function is a mean zero Gaussian random variable with the above variance - I am fine with that).
My problem is really the conditioning part (does the conditional distribution exist? why can I just treat the conditional variable as constant (intuitively this is fine - but rigorously?)); I guess my question is: given $Z:=f(X,Y)$, for two random variables $X,Y$, is the conditional distribution (when does it exist?) of $Z$ given $X=x$ in general just the distribution of $f(x,Y)$? If so, why, and who can I obtain that result generally from the definition of a conditional distribution (without further assumptions, like discreteness, etc.)?
And then again the part with the Gaussian Mixture - what exactly is that object (who do I have to interpret it?)?
Many thanks for any kind of help!
Cheers!