Prove that a process is memoryless (simple example)

Question

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.

Answer 1

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.

Answer 2

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 $y_{t}$

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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 $z_t$.

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