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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.
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    $\begingroup$ Nice question.. This is a sort of BM runned backwards.. I would not know how to give much intuition... $\endgroup$
    – Tom
    Jan 28, 2021 at 21:52
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    $\begingroup$ Time reversal (say, $W_1 - W_{1-t})$ is very intuitively also a BM. However, time inversion is very different. $\endgroup$ Jan 28, 2021 at 22:54
  • $\begingroup$ I agree it s not the same thing... but I thought that the concept was somehow similar... you re running backward, though the time transformation is non linear and you need to transform space as well.. $\endgroup$
    – Tom
    Jan 29, 2021 at 9:18
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    $\begingroup$ Thank you very much Benjamin. I'd like to inform you that I have upvoted your question, not least for the quick response but for the excellent way it has been written. $\endgroup$ Feb 2, 2021 at 11:57
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    $\begingroup$ @BenjaminWang The answer by Jap covers almost everything I wanted to say. You can take a look, thanks. $\endgroup$ Feb 3, 2021 at 5:50

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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.

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    $\begingroup$ Yes, I think this is great graphical intuition that you have provided. Inch perfect. +1. The short point is essentially that Gaussian processes depend only on the mean and variance, so any chopping and changing of the time scale is compensated by adjusting the variance and mean rightly. I highly appreciate both your answer and Jap's answer above, both are frankly better than I'd have given to this question. Benjamin will have a hard time on hand with the bounty, I think. $\endgroup$ Feb 4, 2021 at 16:28
  • $\begingroup$ I have decided to accept this answer and give the bounty to another answer, by a different user, Jap88. Future visitors, please take a look at both of these. $\endgroup$ Feb 5, 2021 at 17:32
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    $\begingroup$ Thanks @BenjaminWang $\endgroup$
    – Jap88
    Feb 5, 2021 at 21:15
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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.

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  • $\begingroup$ That comparison of "local" to "global" and the resulting limit argument illustrates what I want to see perfectly, +1. One thing you could add to your answer, is that you could look through Benjamin's notes, where he's got further information on characterizations of the BM, like the Cameron-Martin formula, and you can try to provide intuition from various sources as to why this is true. I mean, graphical intuition is one thing, but intuition via other characterizations is also important since they are meant for this purpose. $\endgroup$ Feb 3, 2021 at 5:50
  • $\begingroup$ Thank you for the intuition. I guess we are not going to get a perfectly satisfactory intuition from such "segments", because we are trying to discretize something that moves continuously. This might be as good as we can get, under the current characterizations of BM. $\endgroup$ Feb 3, 2021 at 14:02
  • $\begingroup$ @TeresaLisbon yea we also covered the reflection principle, Cameron-Martin, etc. I even used "time inversion" to prove another result, in Q4 in the example sheet. (That $\sup_{u\ge 1}W_u/u$ has the same distribution as $|W_1|$) I'm not entirely sure how these would be helpful. I've also looked up the Karhumen-Loeve representation (pointed out by Jap88), but you can't practically plug in $t\rightarrow 1/t$ into that Fourier sine series, to see the time-inversion result, can you? Thank you. $\endgroup$ Feb 3, 2021 at 14:07
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    $\begingroup$ @BenjaminWang Everything you've said is correct. What I wanted to say was that the Brownian motion is uniquely characterized in many ways : the most obvious is the mean and covariance properties, but there's Levy's characterization and also the characterization via the Cameron-Martin formula and preservation of harmonics (which is where the whole idea of involution comes in). Basically, while the other scaling properties admit "graphical" intuition, alternate intuition for time inversion comes from a more abstract source. $\endgroup$ Feb 3, 2021 at 14:17
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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.

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  • $\begingroup$ What you are saying is that $f(t)W_{g(t)}$ is also a BM, for any invertible $g(t)$, where $f(t)$ is such that we preserve unit standard deviation at all times, so $f(t) = \langle W_{g(t)}^2\rangle^{-1/2}$ $\endgroup$
    – a06e
    Jan 31, 2021 at 17:59
  • $\begingroup$ Replying to Arash (answerer), can you clarify “BM is the sum of uncorrelated normal random variables” please? Perhaps you used a different definition / characterisation? I learned that BM is 1) a.s. continuous, 2) starts at zero, 3) $W_t-W_s$ independent of $W_u$ for $0\le u\le s$, 4) $W_t-W_s \sim N(0,t-s)$. Also, I don’t really understand the last sentence in your answer either. My apologies if these are basic questions, but this is my first course in random processes, although I have taken a few courses in stats and a couple in ODEs/PDEs. Thank you so much. $\endgroup$ Jan 31, 2021 at 23:28
  • $\begingroup$ @becko I understand from your comment the following: you can (sort of) do anything to a BM’s time index, as long as it’s injective and you rescale the result to make the s.d. correct. Is there a deeper intuition though? This is not at all obvious if I think about BM as the continuous limit of a random walk (this was the motivation given in lectures). Is there an alternative way to think about this that I’m missing? Thank you. If it won’t fit in the comments, feel free to elaborate in an answer. $\endgroup$ Jan 31, 2021 at 23:38

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