Determine traffic speed distribution while driving I assume that the vehicle speed of cars on a highway is normally distributed around the posted speed limit. I could verify this by sitting by the road with a radar gun and measure vehicle speed for a large number of vehicles passing by.
How could I do this while driving at exactly the speed limit?
Assuming that I have a radar gun in my car, I would measure a positive speed for cars passing me and a negative speed for cars I pass. If I then add my own speed to the observed speed and I should get the same distribution as if I were sitting by the side of the road.
But I feel that I'm missing something: if I keep to the speed limit, then during any period of time, I observe a lot of cars that are either a lot slower than me or a lot faster than me. I'll see a few cars that are a bit slower or a bit faster. However, I'll see (almost) no cars that drive the speed limit (other than the ones in my direct vicinity), because I will never be able to observe them. As a result, I will get a lot of observations in the tail ends of the distribution and almost none near the mean (almost as if sampling from the inverse distribution...?).

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*Can I do a test on my observations to assert whether they fit a Gaussian distribution in the first place?

*If yes, what operation or transformation do I need to apply to my observations to get a Gaussian distribution, if possible?

*What if I can't actually measure speed, but only classify vehicles in bins ("a bit faster/slower, a lot faster/slower, ridiculously fast/slow")?


This crossed my mind the other day while driving and I had the feeling a lot more cars were passing me than I was passing (which is probably just a bias, but nevertheless it triggered this question). I have no background in statistics other than the basics from first year university, so this may be a solved problem and I just don't know what terminology to use to find the solution.

Simplified question
Let the cars be modeled as completely randomly distributed and not interacting with each other. They flow along each other with different speeds and for every speed they are randomly and independently spaced. (Note this is not realistic but adjusting for complications, e.g. trains of cars driving at the same speed, would make the question too much broad)

One single speed
Let's focus for the moment on only one single speed of cars (later one we think about multiple speeds). The counts of these cars will be Poisson distributed and depend on the concentration of these cars and on their density in space.
For some spatial density of $\rho$ cars per km, driving with a speed difference of $v$ km/hour. You would expect on average to pass $\rho \cdot v$ cars per hour and the distribution of the count of these cars in an hour will be a Poisson distribution with a mean of $\lambda = \rho \cdot \vert v \vert$ (using the absolute value $\vert v\vert$ because we can not have negative rates and also the sign of the speed does not influence the rate).
Two speeds
So... when we single out the speed of cars with a single velocity, the distribution will be Poisson distributed.
When we look at multiple cars then we will have a combination of overlapping Poisson distributions.
Say we have two different speeds. Then the probability that the next car is being observed is of speed $v_1$ or speed $v_2$ will depend on the relative concentration of cars with these speeds and with the hitting rate of those speeds. The cars with a higher speed difference will be encountered relatively more often.
Say that the presence is $\rho_1$ and $\rho_2$ (with $f_1+f_2=1$) then the average hitting ratio's of the two will be $\rho_1 \cdot v_1$ and $\rho_2 \cdot v_2$. The relative probability of encountering a car of speed 1 or speed 2 will be $\frac{\rho_1 \cdot v_1}{\rho_2 \cdot v_2}$.
Generalized to many speeds
Using the above intuitive reasoning we can expand the relative probability of encountering a car at some speed as follows.
Let $f(v)$ be the density distribution of the probability that a car has  speed $v$ relative to us.
Let $g(v)$ be the density distribution of the probability that we encounter a car with a speed difference of $v$.
We compute $g(v)$ as
$$g(v) = \frac{f(v)\cdot \vert v \vert}{\int f(v)\cdot \vert v \vert dv} =   f(v) \cdot \frac{\vert v \vert }{\overline {\vert v \vert}}$$
If $f(v)$ is a normal distribution with $\mu_v=0$ then $g(v)$ will resemble a Chi distribution with 2 degrees of freedom. That is, it will be of the form $x e^{-x^2}$ (and you need to mirror it to account for negative velocity as well). In the case that the average relative speed is not equal to zero, then I am not sure what you are getting. It won't seem like a standard (known) distribution to me.
