Speed in m/s is normally distributed, but same data expressed as "Time for 10 meters" is not I am trying to understand why the same data can be normally distributed if expressed in one way, but not normally distributed if expressed in another way.
I have a variable that is "time taken to walk 10 meters" (in seconds). This data is not normally distributed (Shapiro-Wilk: W = 0.632, df = 108, Sig. <0.001, +see "histogram 1" below).
I expressed this same variable as "speed" (in meters per second). I computed it by dividing 10 meters by the time taken to complete the distance, (ex. 14 sec to complete 10 meters becomes 10/14 = 0.71 m/s).
When I now check "speed" for normality, it is very much normally distributed (Shapiro-Wilk: W = 0.984, df = 108, Sig. = 0.234, +see "histogram 2" below).
Am I doing something wrong, or is there a logical explanation for this? While adding the tags, stackexchange mentioned "inverse Gaussian distribution" - is this what is happening here?


 A: Yes, this is an instance of inverse Gaussian. It has been observed that there is an inverse relationship between the cumulant generating function of the time to cover a unit distance and the cumulant generating function of the distance covered in a unit time. Because the distance covered in a unit time (in this case, walking speed) is approximately normal, then the time to cover a unit distance (which is roughly the first-hitting time of 1-dimensional Brownian particle) is by definition approximately inverse Gaussian.
A: The image below illustrates intuitively why the transformed variable has a different distribution:
I have drawn two parallel lines.

*

*On the lowest line I have plotted evenly spaced points at $0.1, 0.2, ..., 1.1, 1.2$ which represent the velocity $v$.

*On the upper line I have draw points according to the formula $t=0.1/v$ (note I reversed the axis it has 1.2 on the left and 0 on the right)

I have drawn lines connecting the different points. You can see that the evenly spaced points $v$ are not transforming into evenly spaced points $t$ but instead the points are more dense in the low values than in the high values.
This squeezing will happen also to the density distribution. The distribution of times $t$ will not be just the same as the distribution of $v$ with a transformed location. Instead you also get a factor that is based on how much the space gets stretched out or squeezed in.

*

*For instance: The region $0.1 < v < 0.2$ gets spread out over a region $0.5 < t <1$ which is a region with a larger size. So the same probability to fall into a specific region gets spread out over a region with larger size.


*Another example: The region $0.4 < v < 0.5$ gets squeezed into a region $0.2 < t <0.25$ which is a region with a smaller size. So the same probability to fall into a specific region gets compressed into a region with smaller size.
In the image below these two corresponding regions $0.4 < v < 0.5$ and $0.2 < t <0.25$ and the area under the density curves are colored, the two different colored areas have the same area size.
So as the distribution for the times $g(t)$ you do not just take the distribution of the velocity $f(v)$ where you transform the variable $v=0.1/t$ (which actually already make the distribution look different than the normal curve, see the green curve in the image), but you also take into account the spreading/compressing of the probability mass over larger/smaller regions.

note: I have taken $t=0.1/v$ instead of $t = 100/v$ because this makes the two scales the same and makes the comparison of the two densities equivalent (when you squeeze an image then this will influence the density).

See more about transformations:
https://en.wikipedia.org/wiki/Random_variable#Functions_of_random_variables
The inverse of a normal distributed variable is more generally:
$$t = a/v \quad \text{with} \quad f_V(v) = \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{1}{2}\frac{(v-\mu)^2}{\sigma^2}}$$
then
$$g_T(t) = \frac{1}{\sqrt{2 \pi \sigma^2}} \frac{a}{t^2} e^{-\frac{1}{2}\frac{(a/t-\mu)^2}{\sigma^2}}$$
you can find more about it by looking for the search term 'reciprocal normal distribution' https://math.stackexchange.com/search?q=reciprocal+normal+distribution
It is not the same as 'inverse Gaussian distribution', which relates to the waiting time in relation to Brownian motion with drift (which can be described by a Gaussian curve).
