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Does the random walk theory assume a simple symmetric random walk? In other words: does the random walk theory assume that the price rises as often as it falls? I've been looking for an answer for a while but I'm still not sure. Maybe this question is very simple, but I think that it will then be easier for other people to find an answer to this question.

Note: I‘m an undergraduate economics student.

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Random walk theory assumes that stock price can be modeled by:

$$log(S_t)=log(S_{t-1})+\epsilon_t, \epsilon_t~ iid N(0, \sigma^2).$$

In other words, stock price follows a random walk, as described above.

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  • $\begingroup$ Okay, if I see that correctly, it should mean that rising and falling prices occur with the same frequency. But if we accept a positive drift, there is a bias towards the positive classes, right? $\endgroup$
    – Rainer Niemann
    Commented Dec 11, 2021 at 21:06
  • $\begingroup$ Yes, the mean of $\epsilon$ is 0, so equal chances for growth or decline. If you model random walk with drift, i.e., $S_t=\mu+S_{t-1}+\epsilon_t$, then recursively you'll get $S_t=\mu t +S_0+\sum \epsilon_t$. If we assume that $S_0=0$, then expectaion $E(S_t)=\mu t$, which is non-constant and depends on time. If $\mu$ is postitive, then over time mean is increasing, if negative -- then decreasing. So, bias towards negative or positive classes depends on the sign of $\mu$. $\endgroup$
    – Sane
    Commented Dec 11, 2021 at 21:42
  • $\begingroup$ AFAIK, when the term "random walk" is used, they are referrring to log prices following a random walk and no drift ( so symmetric RW ) so the model would be $log(S_t) = log(S_{t-1}) + \epsilon_t $. In this manner, returns are random with zero mean. $\endgroup$
    – mark leeds
    Commented Dec 12, 2021 at 1:42
  • $\begingroup$ @markleeds Yes, it is assumed that log prices follow random walk -- edited, thanks. Aside from that, it is common to use random walk with drift. See here, equation 1: rodneywhitecenter.wharton.upenn.edu/wp-content/uploads/2014/04/… $\endgroup$
    – Sane
    Commented Dec 12, 2021 at 7:11
  • $\begingroup$ @markleeds What do the logarithmized prices mean in this context? $\endgroup$
    – Rainer Niemann
    Commented Dec 12, 2021 at 11:30
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For the stockmarket, this is often a first-pass default assumption. Add in a bit of financial spice, however, and you distribute stock returns lognormally (as opposed to normally), then strictly speaking, they cease to be symmetrical. But 99% of people could not tell the difference 99% of the time.

Turning to debt markets, this rapidly falls apart. 99% of the time, the company does not default and I get my coupon. 1% of the time, they do default and I lose 70-75% of my capital. This is clearly not symmetrical. But this can still represent a random walk!

Think of the random walk as a repetitive coin-flipping exercise. At each step, you flip a coin. But the probability of heads does not have to be 50%; and the relative payoff of heads:tails does not have to be 1:-1. Nothing in the maths of random walks requires symmetry.

Except this is often lazily assumed because it makes the entire subject much easier to think about :-)

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  • $\begingroup$ Hi @demully, thanks for your answer. Does this mean that there is no such thing as one random walk model in the context of the random walk hypothesis? $\endgroup$
    – Rainer Niemann
    Commented Dec 13, 2021 at 11:09
  • $\begingroup$ @Demully: Thanks for nice explanation but note that I was talking about equities rather than bonds. Also, you do need symmetry ( well, expectation zero which usually implies symmetry ) in the error term of the random walk because otherwise, the log price will not be a martingale. Basically, you want the expectation of the next log price to be equal to the previous log price ( maybe plus a drift ) so, if the expectation of the error term is not zero, you won't obtain that. $\endgroup$
    – mark leeds
    Commented Dec 13, 2021 at 15:56

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