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I want to create a variable which is assumed to be the acceleration of a car. I assume it should has zero mean and normal distribution. But the acceleration cannot change rapidly. How do I make it change smoothly (slowly) overtime?

opmfan
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  • Integrate Brownian motion? – hmakholm left over Monica Jun 27 '16 at 18:32
  • @HenningMakholm: I don't know if it must be that complicated. You can imagine that when you drive a car, the speed can't be the same overtime. So the acceleration should be slowly changing around 0. I just want a variable having that property. – opmfan Jun 27 '16 at 19:05

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So basically you are looking for a distribution that changes smoothly. There are many, take the Cauchy distribution with mean $0$ and the scale parameter near $2$ to $3$

Qwerty
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$$ a_{n + 1} = a_{n} + 10^{-6}\sigma $$ where $\displaystyle{\sigma}$ take values $\displaystyle{\pm 1}$ with probability $\displaystyle{1 \over 2}$ each.

I imagine you are thinking in a computer simulation so you can play by changing the $\displaystyle{\sigma}$ prefactor which is actualy $\displaystyle{10^{-6}}$.
Felix Marin
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  • The problem with this is that $a_n$ is not stable (nor stationary). A better way is to add a multiplier $a_{n + 1} = \alpha a_{n} + \beta \sigma$ where $0< \alpha < 1$ (in terms of Autoregresive processes, this would amount to move the pole inside the unit circle) – leonbloy Jun 27 '16 at 20:30
  • @leonbloy Thanks. I didn't think in that posibility. This is somehow new for me. – Felix Marin Jun 27 '16 at 20:33
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I assume it should has zero mean and normal distribution. But the acceleration cannot change rapidly.

In signal processing, that amounts to generate a stochastic process with low frequency. The simplest way is a AR(1) (autoregresive process) of the form

$$X[n]= \alpha X[n-1] + \beta e[n]$$

where $e[n]$ is zero mean white gaussian noise, (typically generated by a pseudorandom number generator), $0<\alpha<1$ is a "memory" parameter that tunes the cut-off frequency and $\beta>0$ is a factor that determines the variance.

leonbloy
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  • This is exact what I want. Thank you very much ^.^ – opmfan Jun 28 '16 at 19:10
  • you're welcome. BTW, notice that with $\alpha=1$ this indeed amounts to "integrate brownian motion" (cf Henning Makholm comment), with $\alpha <1$ we only add a decreasing exponential window – leonbloy Jun 28 '16 at 22:21