# Pushforward of Gaussian Measure by Solution Operator to SDE

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?
• $f$ 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 $C[0,1]\,.$ Jan 29 at 9:51
• $\alpha$ 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. Jan 29 at 13:39