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Consider an n-dimensional Ito process $$ X_t^x = x + \int_0^t\, \alpha(s)ds + \int_0^t\,\beta(s)\,dW(s) $$ driven by an n-dimensional Brownian motion; where $\alpha,\beta$ are Lipschitz functions with values in $\mathbb{R}^n$ and the set of n$\times $n symmetric positive definite matrices metrized by the Frobenius norm. Let $f:x\mapsto X_1^x$ denote its solution operator; which exists under mild conditions on $\alpha$ and $\beta$. It seems to me that, if $\mathcal{\mu}$ is a Gaussian distribution on $\mathbb{R}^n$ then so is the pushforward measure $f_{\#}\mu$, simply by virtue of the sum of "normals is normal".

Here's my (first even on MSE) question:

  1. Is this correct, and is $f_{\#}\mu$ indeed Gaussian and with non-singular covariance (if each $\beta_s$ is $n\times n$ and positive-definite)
  2. Are there more general (but "down to earth") conditions ensuring that $f_{\#}\mu$ is a Gaussian measure on $\mathbb{R}^n$ with positive-definite covariance matrix?
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  • $\begingroup$ $f$ is a map from $\mathbb R^n$ to random variables $X^x_1$ on $\Omega$ which as such is a space for which the notion of Gaussian measure is not immediately defined. Advice: take instead the infinite dimensional Wiener space $\Omega=C[0,1]$ and google what people have done decades ago about Gaussian measures on infinite dimensional spaces. Also: by the Girsanov theorem your pushforward measure $f_{\# \mu}$ should be closely related to the classical Wiener measure on $C[0,1]\,.$ $\endgroup$
    – Kurt G.
    Jan 29 at 9:51
  • $\begingroup$ $\alpha$ and $\beta$ are deterministic functions of only time right? If so then $X_t$ can be written as $$X_t = X_0 + \int_0^t \alpha(s)ds + \int_0^t \beta(s)dW_s,$$ where $X_0 \sim \mu$. If $\mu$ is a Gaussian distribution, then the above display is the sum of independent Gaussian random variables and therefore Gaussian. Furthermore, if we let $m, \Sigma$ be the mean and covariance of $\mu$ then $X_t$ has mean: $$ m + \int_0^t \alpha(s) ds$$ and covariance: $$ \Sigma + \int_0^t \beta(s)\beta(s)^T ds.$$ Thus as long as $\Sigma$ is positive definite, $f_\#\mu$ has positive definite covariance. $\endgroup$
    – Shiva
    Jan 29 at 13:39

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