# 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?

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 $$0.1,0.2,...,1.1,1.20.1, 0.2, ..., 1.1, 1.2$$ which represent the velocity $$vv$$.
• On the upper line I have draw points according to the formula $$t=0.1/vt=0.1/v$$ (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 $$vv$$ are not transforming into evenly spaced points $$tt$$ 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 $$tt$$ will not be just the same as the distribution of $$vv$$ 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 $$0.1 gets spread out over a region $$0.5 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 $$0.4 gets squeezed into a region $$0.2 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 $$0.4 and $$0.2 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 $$g(t)g(t)$$ you do not just take the distribution of the velocity $$f(v)f(v)$$ where you transform the variable $$v=0.1/tv=0.1/t$$ (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 $$t=0.1/vt=0.1/v$$ instead of $$t=100/vt = 100/v$$ 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).

https://en.wikipedia.org/wiki/Random_variable#Functions_of_random_variables

The inverse of a normal distributed variable is more generally:

$$t=a/vwithfV(v)=1√2πσ2e−12(v−μ)2σ2t = a/v \quad \text{with} \quad f_V(v) = \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{1}{2}\frac{(v-\mu)^2}{\sigma^2}}$$

then

$$gT(t)=1√2πσ2at2e−12(a/t−μ)2σ2g_T(t) = \frac{1}{\sqrt{2 \pi \sigma^2}} \frac{a}{t^2} e^{-\frac{1}{2}\frac{(a/t-\mu)^2}{\sigma^2}}$$

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).