The classical random walk can be described as the evolution of the position $X\_t$ of a walker for integers $t \geqslant 0$, where $X\_0 = 0$ and $X\_t = X\_{t-1} + V\_t$ for $t \geqslant 1$, where the "speed" $V\_t$ at each time step is uniformly random $V\_t \in\_{\mathrm R} \{-1,+1\}$ and independent at each time step.
It is well-known that this process yields a position which obeys a symmetric binomial distribution with mean $0$ and variance $t$, and that $t^{-1/2} X\_t$ tends to a Gaussian distribution with variance $1$.


I am interested in a variant in which the speed itself increases or decreases by random "boosts", behaving like a classical random walk, and where the position is governed by the speed as it evolves over time. That is, we have
$$\begin{aligned}
 X\_0 &= 0 & V\_0 &= 0 \\
 X\_t &= X\_{t-1} + V\_t & V\_t &= V\_{t-1} + A\_t & A\_t \in\_{\mathrm R} \{-1,+1\}.
\end{aligned}$$
**Question.** What is the probability distribution of $X\_t$ as a function of $t$?


*Remark #1.*
This process is in effect a discrete-time variant of a second-order stochastic differential equation of a form
$$\begin{aligned}
 \frac{\mathrm dx}{\mathrm dt} &= v, & \frac{\mathrm dv}{\mathrm dt} = \xi\_t
\end{aligned}$$
where $W\_t = \int\_{0}^t \mathrm d\tau \, \xi\_\tau\,$; though I do not know enough to be able to distil what probability density function one would expect for $x(T)$ from the references I found.


*Remark #2.* I originally requested an answer regarding the density function of the continuous version $x(t)$ as a fall-back. I have accepted an answer on that question for now; and that answer implicitly provides an answer for the discrete-time version in which the steps are normally distributed instead of being of fixed size. However, I will preferentially accept and reward any answer on the discrete-time version of the problem.



Walk with randomised boosts
The *position process* $x\_t$ satisfies $$
x\_t = x\_0 + t v\_0 + \int\_0^t W\_s ds \;.
$$ Because $\int\_0^t W\_s ds \sim \mathcal{N}(0,\frac{1}{3} t^3)$, a simple change of variables shows that $$
x\_t \sim \mathcal{N}( x\_0 + t v\_0, \frac{t^3}{3}) \;.
$$




---


The following relates $x\_t$ to a discrete-time weak approximation $X\_N$, which is obtained by weakly approximating the Brownian increments in the velocity process $v\_t$ by Rademacher random variables. 



> 
> **Claim.** Given $t>0$ and $N \in \mathbb{N}$, set 
>  $$
> \tag{$\star$}
> X\_n = X\_{n-1} + h V\_n \;, \quad V\_n = V\_{n-1} + \sqrt{h} A\_n \;, \quad 1 \le n \le N
> $$ with initial conditions $X\_0 = x\_0$, $V\_0 = v\_0$, and where $\{ A\_i \}\_{i=1}^N$ are independent Rademacher random variables and $h=\frac{t}{N}$. For any $t>0$, $$
> X\_{N} \overset{d}{\to} x\_t \quad \text{as $N \to \infty$} \;.
> $$ 
> 
> 
> 


*Proof.* Unraveling the recurrence relation ($\star$) yields,
$$
X\_N = x\_0 + t v\_0 + \frac{t^{3/2}}{N^{1/2}} S\_N 
$$
where $S\_N = \sum\_{i=1}^N A\_i \frac{(N-i+1)}{N}$.
Note that $S\_N$ is a sum of *independent* and *uniformly bounded* random variables each with mean zero. Moreover, the variance $s\_N^2$ of $S\_N$ goes to $\infty$ as $N \to \infty$ since 
$$
s\_N^2 = \operatorname{Var}(S\_N) = \frac{1+3 N + 2 N^2}{6 N} \;.
$$ 
By the [Lyapunov Central Limit Theorem](https://en.wikipedia.org/wiki/Central_limit_theorem#Lyapunov_CLT), $$
\frac{S\_N}{s\_n} \overset{d}{\to} \mathcal{N}(0,1) \;,
$$ from which it follows that, $$
X\_N \overset{d}{\to} \mathcal{N}(x\_0 + t v\_0, \frac{t^3}{3} ) \;.
$$


