Particle with normally distributed initial position and velocity Some time ago I asked myself, "suppose I have a particle and I know its velocity and initial position are normally distributed; what is the distribution of its position after a period of time $\Delta t$?"
I don't know how to solve this, but from an empirical investigation, I suspect the result may be 
$$x_f \sim \mathcal{N}(\mu_v \Delta t+\mu_i, \sqrt{(\sigma_v \Delta t)^2+\sigma_i^2}),
$$ 
though I have no idea why. Can someone provide an intuitive (or unintuitive) explanation for this formula?
 A: Let's consider the one-dimensional case. A body has an unknown initial position, $x_i$, at time $t_i$, described by the pdf
$$
p_i(x_i) = \text {Gauss}(x_i; \mu_i, \sigma_i).
$$
We desire the pdf for the body's position $x_f$ at a later time $t_f$, but don't know the body's velocity, $v$.  We only know that
$$
p_v(v) = \text {Gauss}(v; \mu_v, \sigma_v)
$$
and that 
$$
x_f = x_i + v (t_f -t_i) \equiv x_i + v\Delta t
$$
We may write, using marginalisation and remembering Jacobians from changes of variables of a pdf,
$$
\begin{align}
p_f(x_f) =& \int dv\, p_f(x_f \,\vert\, v)\cdot p_v(v)\\
=& \int dv\, p_i(x_f - v \Delta t) \left|\frac{dx_i}{dx_f}\right|\cdot p_v(v)\\
=& \int dv\, \text{Gauss}(x_f - v \Delta t; \mu_i, \sigma_i) \cdot \text {Gauss}(v; \mu_v, \sigma_v)\\
=& \int d(v\Delta t)\, \text{Gauss}(x_f - v \Delta t; \mu_i, \sigma_i) \cdot \text {Gauss}(v\Delta t; \mu_v \Delta t, \sigma_v \Delta t)\end{align}
$$
This is the convolution of two Gaussians. The result is well-known - when we convolute Gaussians, we sum the means and add the standard deviations in quadrature. Thus, for our case,
$$
p_f(x_f) = \text{Gauss}(x_f; \mu_i +\mu_v \Delta t, \sqrt{(\sigma_v \Delta t)^2 + \sigma_i^2})
$$
This is precisely the result you found empirically. 
