# Does the random walk theory assume a simple symmetric random walk?

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.

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.

• 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? Commented Dec 11, 2021 at 21:06
• 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 $\mu$.
– Sane
Commented Dec 11, 2021 at 21:42
• 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. Commented Dec 12, 2021 at 1:42
• @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/…
– Sane
Commented Dec 12, 2021 at 7:11
• @markleeds What do the logarithmized prices mean in this context? Commented Dec 12, 2021 at 11:30

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 :-)

• 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? Commented Dec 13, 2021 at 11:09
• @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. Commented Dec 13, 2021 at 15:56