Intuition for Brownian motion time-inversion formula Background: I am studying a course in Stochastic Financial Models, and have been introduced to Brownian motions (BMs) and Gaussian processes.
The BM time-inversion formula states that if $W_t$ is a BM, then $$B_t := tW_{1/t}$$ is also a BM.
I know how to prove the result, through Gaussian processes, similar to the answer here.
However, the proof is unenlightening and does not offer any intuition. In fact, my course lecturer does not offer any intuition either.
Question: Please provide some intuition why the formula “should be” true.
Example of my intuitions for another formula:

*

*“Scaling” formula: if $W_t$ is a BM, then $cW_{t/c^2}$ is also a BM.

*Intuition: picture time on the $x$ axis and the motion on the $y$ axis. BMs “don’t go that far up or down”, i.e. their standard deviation is “only” sd$(W_t)=\sqrt t$, so to compensate for the factor of $c$ stretch in the $y$ axis, we have to stretch by a factor of $c^2$ in the $x$ axis.

 A: The idea is that we're essentially turning the $t$-axis inside out around the point $t=1$, i.e. we're taking $B_t$ for $1 \le t < \infty$ and compressing it into the interval $[0,1]$, and taking $B_t$ for $0 < t \le 1$ and stretching it out to the interval $[1,\infty)$.  This is sort of like the reason that $\sin(1/x)$ is discontinuous at $x=0$: we're taking all of the infinitely many oscillations between $-1$ and $1$ that $\sin(x)$ experiences on $[1,\infty)$ and compressing them down into $[0,1]$.
When we take all the variation in $B_t$ on $[1,\infty)$ and compress it into $[0,1]$, we need to rescale the size of the fluctuations to account for the fact that they're happening on a much smaller time scale.  The amount that we need to rescale by turns out to be a factor of $t$, which is probably easiest to find by making the variance correct, but makes sense: For small $t$, we need to compress $B_{1/t}$ by a lot because it came from farther along originally in the Brownian motion path and hence had more time to fluctuate.
Similarly, when we take the fluctuations in $B_t$ on $[0,1]$ and stretch them out to $[1,\infty)$ we need to rescale to make them larger.  The intuition for the factor of $t$ is essentially the same: For large $t$, we need to stretch $B_{1/t}$ by a lot because it originally came from close to $0$ and hence didn't have time to fluctuate much, so we need to amplify the fluctuations it did have by more.
A: This explanation is intuitive but very far from stringent. I think one issue with building an intuition for time inversion scaling based on previously obtained intuition for the regular scaling formula $c W_{t/c^2}$ is that the regular scaling formula is a kind of "uniform global scaling" whereas the time inversion scaling can be seen as more of a "local scaling". Assume that we have built an intuition for how $c W_{t/c^2}$ is BM. I wanted to show how one can "build a bridge" to time inversion scaling.
Consider the time interval $t_1-t_0$, where we will later set $t_0=0$. If we now take a large step back so we can see a big chunk, the interval $t-t_0$, of the random walk for which the interval $t_1-t_0$ is relatively tiny, $t_1\ll t$, then we can consider the BM to be fairly constant (in some sense) in the $t_1-t_0$ interval compared to the much larger $t-t_0$ interval.
Let's now use the regular scaling formula on the little interval, with $c=t_1-t_0$:
$$c W_{t/c^2}=(t_1-t_0) W_{t/(t_1-t_0)^2}$$
Here we need  $c=t_1-t_0$ and $W_{t_1-t_0}$  to be "small" compared to the expected variation of $W_t$.
Set $t_0=0$ to get:
$$c W_{t/c^2}=t_1 W_{t/t_1^2}$$
If we let $t_1$ grow and catch up with $t$ (or $t$ shrink down to $t_1$) we get
$$t_1 W_{t/t_1^2} \rightarrow t W_{t/t^2}=t W_{1/t}$$
and this provides a link between the regular scaling formula and the time inversion formula. Apologies for the utter and complete lack of stringency.
Several things are problematic here. For example, what do we mean with the BM to be relatively (fairly) constant? How can we be allowed to let $t_1$ approach $t$? However, I suspect making it stringent would be a larger undertaking and not give the wanted intuition. The key, I think, is to realize that any subinterval $t_1-t_0$ will look like a regularly scaled BM when compared to a much larger interval. For stringent proofs, see:
Prove the time inversion formula is brownian motion
or
http://www.math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Stolarski.pdf
It is very difficult to get a nice computer plot that shows time inversion. This can be understood, for example, as you needing an extremely large number of terms in a, say, Fourier series or Karhunen–Loève representation (https://en.wikipedia.org/wiki/Wiener_process). And if you try this in the obvious naïve way by just summing terms you will inevitably get a BM that is highly resolved close to $t=0$ but poorly resolved for larger value of $t$. In a way you get penalized for not being able to represent this as a true infinite series.
A: This property and all similar ones stem from a simple fact: BM is the sum of uncorrelated normal random variables. It does not make any difference if the time indices are reversed, inverted, or shuffled, as long as the sigma algebra is not compromised. The correlation depends on the intervals that overlap, the corresponding time index of which could show up as minimum, maximum, or some complicated combination of indices.
