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...?).

*

*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.
 A: I'm going to address the idealized situation where we have a stationary density of cars per unit length along the road, per unit speed, $\rho(v)$. This distribution is time and space invariant.  For concreteness I'm going to write $\rho(v) = \rho(v ; \mu, \sigma^2)$ to express the fact that the distribution has a location $\mu$ and scale $\sigma$ parameter, though nothing in what follows requires the distribution to Gaussian.
We're going to be considering frames of reference moving with speed $v_0$.  In that case we have $\rho(u=v-v_0)du = \rho(u; \mu-v_0, \sigma^2)du$ (this is just variable substitution, and, physically, is a manifestation of Galilean invariance).
The rate at which cars will pass a fixed reference point in the moving reference frame will be given by $I(u)du  = u \rho(u) du= (v-v_0)\rho(v-v_0; \mu-v_0, \sigma^2)dv$.  The reasoning behind this goes like this:  How many cars moving at (relative) speed $u$ pass a reference point $x_0$ in the (moving) reference frame between times $t$ and $t+\delta t$?  All of the cars with (relative) speed $u$ at locations between $x-u(t+\delta t)$ and $x-u t$. The density (per unit speed) is $\rho(u)*u*\delta t$, so the rate per unit time is $u\rho(u)du$
So in the end, the rate (cars/unit time) you'll see cars moving with relative speed in the (infinitesimal) interval $[u, u+du]$  when moving at speed $v_0$ relative to the road is
$I(u; v_0)du = u \rho(u; \mu-v_0, \sigma^2)du$.
A: Consider doing your experiment at least twice at two different speeds, eg, 65 mph and 70 mph. (This is in the US. Elsewhere use two speeds in kph.)
Each experiment gives you information about the number or fraction of cars traveling at the other speed, as well as twice as much information about all the other speeds. Weighting the numbers at these two speeds would be appropriate, either fifty-fifty or taking the two totals into account.
I have counted the numbers of cars traveling faster or slower than I was on the tri-state toll road around Chicago. I concluded that the median speed was 70 miles per hour when the speed limit was 55 mph for nearly the total distance. During rush hours, of course, the median can be as low as 10 mph, or less, for long stretches.
I do not know how accurate radar guns are in moving cars. It would be worth finding out if there is a bias, either due to angle of measurement or due low differences in speed. I do not think you have to take relativity due to moving traffic into account.
Happy measurement.
A: 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.
A: I think your goal of characterizing the distribution of highway speeds while on the highway yourself is doomed to fail, as you've rightly pointed out that you'll never pass anyone going the same speed as you. Suppose you're in an autonomous car with the front and back windshield blacked out, so that you can only observe cars when you pass them. It's impossible for you to tell if the highway is completely empty, or if it's full of cars moving the exact same speed as you. If you do pass or are passed by some other cars, it'll be impossible to know if those are the only other cars and very common speeds, or if they're a minority of people not driving the same speed as you. You might have some additional mileage if you're able to assume that speeds are in fact normally distributed around your own speed, as you can get a sense of a variance in speed from the cars that do pass you, but I don't think you can get very far without some means of reasoning about the number of people driving the same speed as you, which the distributional assumption allows.
A: This was a bit of a rabbit hole. None of the existing answers really answered my question, but the comments by "whuber" (pointing to this very similar post) and "Dave" pointed me in the right direction. I'll summarize here what I learned. I simulated everything to check my assumptions.

*

*Assuming that vehicle speeds are normally distributed and that distribution is constant in time and space, passing a car or me being passed can be considered a sampling from that distribution.

*The fact that I'm driving does not matter, because, as "Dave" points out, the properties of the normal distribution are invariant for the change of reference (linear operation).

*If I do have a (reliable) radar gun, I can use the Maximum Likelihood Estimator to estimate the parameters of the distribution ($\mu, \sigma$). The fact that I will observe only more much faster/slower vehicles does not really matter, as long as the observations don't skew towards one of those categories (e.g. I get passed a lot by much faster cars, but I don't pass a lot of much slower cars).

*If I don't have a radar gun, I can count the number of cars I pass and that pass me and use that as a z-score (see aforementioned post. In this way, assuming that the distribution is centered around the speed limit and if my speed is not equal to the average speed, I can estimate $\sigma$ from my observations. I tried to combine the results from counting at two different speeds (what "David Smith" suggests) and then estimate both $\mu$ and $\sigma$, but this doesn't give good results, (probably for reasons that are obvious to people more versed in statistics).

