2
$\begingroup$

I am trying to understand the diffrence between random walk and martingale. According to my understanding, a random walk without drift is

$$ y_{t} = y_{t-1} + u_{t} $$

where $u_{t}$ is $i.i.d.(0, \sigma_{t}^2)$ where $\text{Cov}(y_{t},y_{t-s})=0$ for $t \neq s$.

However, a martingale has just one restiction:

$$ \mathbb{E}[y_{t}|y_{t-1},y_{t-2},\dotsc] = y_{t-1} $$

and there are no restrictions on higher order moments. The distribution of $u_{t}$ is just required to satisfy $\mathbb{E}(u_{t})=0$. Heterogeneity of $u_{i}$ and/or correlation between sequences of $u_{i}$ are allowed.

Is my understanding right?

$\endgroup$
  • $\begingroup$ Random walks are discrete time processes. so trivially, a Wiener process (i.e. basically the continuous time limit of a random walk) is not technically a random walk though it is a Martingale. $\endgroup$ – Matthew Gunn May 15 '16 at 0:03
  • $\begingroup$ And the def. of a Martingale in discrete time is that the conditional expectation $E[y_t \mid y_{t-1} \ldots y_1] = y_{t-1}$. That's different than the unconditional expectation which you wrote, $E[y_t]$. $\endgroup$ – Matthew Gunn May 15 '16 at 0:05
3
$\begingroup$

For a random walk, I think $u_t\sim i.i.d.(0,\sigma^2$) (no time subscript for $\sigma^2_t$).

For a martingale,

$$ \mathbb{E}(y_{t}|y_{t-1},y_{t-2},\dotsc) = y_{t-1} $$

implies $u_t$ are non-autocorrelated, because if they were, the conditional expectation would no longer be zero:

  • $\mathbb{E}(u_t|y_{t-1},y_{t-2},\dotsc)=0$ implies $\mathbb{E}(u_t)=0$;
  • given $y_{t-1}$ and $y_{t-2}$ we have $u_{t-1}$; if it is nonzero and $\{u_t\}$ as a sequence are autocorrelated, then $\mathbb{E}(u_t|u_{t-1}) \neq 0$ and $\mathbb{E}(y_{t}|y_{t-1},y_{t-2},\dotsc) \neq y_{t-1}$, which is a contradiction.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.