Given the following stochastic process

\begin{equation} x_t = \frac{u_t}{\sum_{s=1}^{t-1} u_s} \end{equation}

where $u_t \overset{i.i.d.}{\sim} \mathcal{N}(0,\sigma^2)$, $\sigma^2<\infty$, and the first observation is not available (NA) since $x_t$ is basically $\Delta R/R$ where R is a random walk. Prove or disprove that $x_t$ is memoryless. The definition of memoryless should be with respect to the Markov property, but if you know different ways to tackle the problem I would like to see different solutions

Intuitively, I would say that it is indeed memoryless, as at the numerator we have a white noise, and the denominator is still white noise given that the sum of i.i.d. white noises is still white noise. I woul like to see a rigorous proof. Thank you.

  • $\begingroup$ Can you describe your own attempt at a rigorous proof? $\endgroup$
    – mhdadk
    Commented Dec 27, 2021 at 8:31
  • 1
    $\begingroup$ I think by memoryless you mean show that $x_{t}$ is a martingale ? In that case, you want to show that $E(x_{t+1} | x_{t}) = x_{t}$. If this is correct, then I would use the term martingale because, in statistics, memoryless and martingale are not the same thing. $\endgroup$
    – mlofton
    Commented Dec 27, 2021 at 9:25
  • $\begingroup$ @mlofton but martingale is not the same as the Markov property. It requires that the distribution of the next state only depends on the previous state (but not on the history). The expectation does not need to be stationary. $\endgroup$ Commented Dec 27, 2021 at 9:28
  • $\begingroup$ @Sextus Empiricus: Right. I agree. I'm just not clear on what Kolomorogorovwannabe is trying to show ? The memoryless,property , as far as I know, is associated with the exponential. But yes, Markov property means only a function of the past. My apologies for confusion. Maybe the answers below will clarify my understanding. I'll read them now. Thanks. $\endgroup$
    – mlofton
    Commented Dec 28, 2021 at 15:57

2 Answers 2


Hint: With a bit of recursive algebra it can be shown that:

$$\frac{u_t}{u_1} = x_t \prod_{i=2}^{t-1} (1+x_i),$$

which gives the recursive form:

$$x_t = \frac{u_t}{u_1 \prod_{i=2}^{t-1} (1+x_i)}.$$

Have a go at deriving the above equations and see if this tells you anything about the "memory" of the stochastic process.

  • 1
    $\begingroup$ With your form it seems to me that $x_t$ depends on previous states so it shows some degree of memory, right? $\endgroup$ Commented Dec 27, 2021 at 14:02

For a random walk,

$$y_t = \sum_{s=1}^{t} u_s$$

you can write

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

which shows that the distribution of $y_{t+1}$ only depends on the before last step and not the entire history.

More clearly we can show it by writing out the expression for the distribution of $y_{t+1}$

$$y_{t+1} \sim N(y_t,\sigma^2)$$

and this distribution of the future state $y_{t+1}$ only depends on the present state $y_{t}$

For your case,

You can rewrite your $x_t$ in terms of this random walk. To simplify this I use a transformed/shifted variable the $z_t = x_t+1$

$$z_t = 1+x_{t} = 1 + \frac{u_t}{ \sum_{s=1}^{t-1} u_s} = \frac{\sum_{s=1}^{t-1} u_s}{ \sum_{s=1}^{t-1} u_s} + \frac{u_t}{ \sum_{s=1}^{t-1} u_s} = \frac{y_t}{y_{t-1}}$$

This $z_{t}$ is not memoryless. Intuitively, it depends on $y_t$ and also the previous step $y_{t-1}$.

We can make this more clear by deriving the distribution of $z_{t+1}$ and show that you need more than only $z_{t}$ and $u_{t+1}$. Namely, you also need to know $y_{t-1} = \sum_{s=1}^{t-1} u_t $.

$$z_{t+1} = \frac{y_{t+1}}{y_{t}} = \frac{z_ty_{t-1} + u_{t+1}}{z_ty_{t-1}} = 1 + \frac{u_{t+1}}{z_ty_{t-1}}$$

For the same $z_t$, the $y_{t-1}$ can be different. Thus the distribution of $z_{t+1}$ does not solely depend on $z_{t}$, and also on the history of $z_t$ described by $y_{t-1}$.

$$z_{t+1} \sim N\left( 1,\frac{\sigma^2}{z_t^2} \cdot \frac{1}{y_{t-1}^2} \right)$$

The same can be argued for $x_t$ as for $z_t$.

$$x_{t+1} \sim N\left( 0,\frac{\sigma^2}{(x_t+1)^2} \cdot \frac{1}{y_{t-1}^2} \right)$$

  • $\begingroup$ Side note: In this case, if you include $y_{t-1}$ as part of the state then the process is memoryless. It is like the future position of a moving particle is not only dependent on the current position but also on the present speed/momentum. $\endgroup$ Commented Dec 27, 2021 at 11:37
  • $\begingroup$ This is a wonderful answer thank you! To sum up, the conclusion is that $x_t$ shows some degree of memory because $x_t = \frac{y_t}{y_{t-1}} - 1$, so it depends on $y_t$ and $y_{t-1}$, right? It is a strange thing from my perspective because $x_t$ can be described as a ratio between random number generated from a Gaussian, so I cannot see intuitively why it depends on its history. Moreover, I'm not sure to understand your side note, what do you mean with including $y_{t-1}$ as a part of the state? $\endgroup$ Commented Dec 27, 2021 at 14:09
  • $\begingroup$ @Kolmogorovwannabe the future state $z_{t+1}$ does not just depend on the current state $z_{t}$ and some random variable $u_{t+1}$. Instead, for different histories $y_{t-1}$ you get different dependencies of the future state from the current state. $\endgroup$ Commented Dec 27, 2021 at 14:13
  • $\begingroup$ @Sextus Empiricus: A beautiful answer. My only question is what we do we say about the process ? Does it obey the Markov property ? It's definitely not memoryless and it's definitely not a martingale. Thanks. $\endgroup$
    – mlofton
    Commented Dec 28, 2021 at 16:03

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