# p.d.f. of a position variable from stochastic velocity p.d.f.

I have a stochastic process, $v(t)$, that represents a velocity, and has a known probability distribution function $f(x,t)$ which is time-varying. I am interested to acquire a probability distribution (as a function of time) for a stochastic process $p(t)$ that is the integral wrt time of the velocity variable. How can this be done? Does it require stochastic integration? (I am not trying to integrate wrt a random process, just wrt time). Can I compute the sum of the velocity variable at multiple times (by convolution) and then take a limit of ∆t$\to$0? Thanks

• Is $v(t)$ continuous? If not, this raises some interesting difficulties, doesn't it? – Brian Tung Apr 20 '15 at 17:05
• My $v(t)$ is indeed continuous. – Daniel Naftalovich Apr 20 '15 at 23:56
• So then, can you articulate exactly how $v(t)$ is selected for a given $t$, with all of $v(t'), t' < t$ as history? It can't be selected arbitrarily from the PDF $f(x, t)$, for otherwise, it would not be continuous in general. – Brian Tung Apr 21 '15 at 0:15
• Sorry I meant that the pdf of $v(t)$ is continuous – Daniel Naftalovich Apr 21 '15 at 4:13
• So, just so I'm clear, the object will have an uncountably infinite number of different velocities within any positive interval of time, no matter how small, yes? – Brian Tung Apr 21 '15 at 4:33

We assume that the PDF of $v(t)$, namely $f(x, t)$, always has a finite first moment (that is, the mean).

Let us consider the behavior of the position from $t = 0$ to $t = \Delta t$, where $\Delta t$ is chosen small enough that the variation in the first moment is as small as we like. (That we can do this is established by the continuity of $f(x, t)$.)

As a first approximation, we might choose a single velocity $v(0)$, from the distribution $f(x, 0)$. Then the position at $t = \Delta t$ is given by $p(\Delta t) = p(0)+v(0)\Delta t$.

Next, let us shrink the time window to $\Delta t/2$. We choose a velocity $v(0)$ according to $f(x, 0)$ that will apply for $t \in [0, \Delta t/2)$, and then a second velocity $v(\Delta t/2)$ according to $f(x, \Delta t/2)$ that will apply for $t \in [\Delta t/2, \Delta t)$. Then the position at $t = \Delta t$ is given by $p(\Delta t) = p(0)+[v(0)+v(\Delta t/2)]\Delta t/2$. In other words, the average speed over the entire interval of time from $0$ to $\Delta t$ is the average of the two chosen speeds.

Now, imagine further subdividing the interval into $N$ slices, each having its own attendant speed. No matter how we divide the interval, the average speed over the interval will be the average of $N$ values, each chosen from the PDF $f(x, \cdot)$, with the time parameter somewhere in the interval $[0, \Delta t)$. Because we stipulated that the PDF's first moments differ by an arbitrarily small amount anywhere over that interval, we can apply the central limit theorem, and in the limit as $N \to \infty$, the average velocity is a fixed value, arbitrarily close to the first moment of $f(x, 0)$.

We can apply the above reasoning to the behavior at any time $t$, not just at $t = 0$. Therefore, in short, the velocity at any time $t$ is deterministic, equal to the first moment of $f(x, t)$. If we denote this first moment by $\overline{v}(t)$, then the position at time $t$ is equal to $p(t) = p(0) + m(t)$, with $m(t)$ being the amount of motion between time $0$ and time $t$, given by

$$m(t) = \int_{t'=0}^t \overline{v}(t') \, dt'$$

If $p(0)$ is a random variable with PDF $g(x, 0)$, then the corresponding PDF at time $t$ is given by

$$g(x, t) = g(x-m(t), 0)$$

That is, $g(x, t)$ is $g(x, 0)$, translated rightward by $m(t)$.