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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!

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about your last part: Do you now the result, if $\mathcal{G}\subset\mathcal{F}$ two $\sigma$-algebras, $X_1$ a $\mathcal{G}$ measurable r.v., and a r.v. $X_2$ independent of $\mathcal{G}$. For every $F:\mathbb{R}\times\mathbb{R}\to [0,\infty]$ (measurable with respect to the product $\sigma$-algebra), we have $E[F(X_1,X_2)|\mathcal{G}](\omega)=E[F(x_1,X_2)]\big|_{x_1=X_1(\omega)}=h(X_1( \omega ))$ –  user8 Jan 10 '13 at 9:05
    
yes, I know that one (at least I think I remember this theorem from somewhere). I guess I am having trouble figuring out the (maybe smaller) details in the above argument. But thanks for the help! –  s_2 Jan 10 '13 at 19:16
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2 Answers

As for the Gaussian Mixture, this is just a weighted sum of $n$ independent Gaussians distributions (each parameterized by $\Theta_i$), defining a density function of arbitrary complexity (i.e arbitrary number of moments, depending on $n$):

$P(x) = \sum_i^n w_i * g(x,\Theta_i)$, with $\sum_i^n w_i = 1$.

You should find tons of literature on that on the internet ...

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The fact that the distribution of $X=\int\limits_0^1B_1\mathrm dB_2$ conditioned on $B_1$ has density $(f_b)_{b}$ means that, for every bounded measurable function $u$, $$ E[u(X)\mid B_1]=v(B_1),\qquad v(b)=\int_\mathbb R u(x)f_b(x)\mathrm dx. $$ In the present case, for every continuous $b:[0,1]\to\mathbb R$, one can take for $f_b$ the density of the centered normal distribution with variance $s^2(b)=\int\limits_0^1b^2$, that is, calling $g$ the standard normal density, $$ f_b(\cdot)=\frac1{s(b)}g\left(\frac{\cdot}{s(b)}\right). $$ In other words, $X$ conditioned on $B_1$ is distributed like $s(B_1)Z$ where $Z$ is standard normal and independent of $B_1$. The density $f$ of the (absolute) distribution of $X$ is such that $$ f(\cdot)=\int_{C[0,1]} f_b(\cdot)\mathrm dP_{B_1}(b)=\int_0^\infty\frac1{t}g\left(\frac{\cdot}{t}\right)\mathrm d\nu(t), $$ where $\nu$ is the distribution of $s(B_1)$. Each density $x\mapsto\frac1tg\left(\frac{x}{t}\right)$ is gaussian and $\nu$ describes their mixture.

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