2
$\begingroup$

Suppose I have observations of a stochastic process $\{X_t\}$. I.e., time series data. Say presently $X_t < u$ for a level $u$, for every $t=1,2,...,N$.

From these observations, how can I model or otherwise calculate the probability $P(X_{N + h} \geq u)$, for some $h$? On the other hand, are expected hitting times easier to work with? Such as $E[\tau | X_t < u ]$, where $\tau = \inf\{T: X_t \geq u\}$?

The only thing I can think of is to build a generative model for $\{X_t\}$, estimate its parameters from the data, then estimate the hitting times by simulating future paths of the time series, but most models assume things too many nice conditions like stationary, Gaussian noise, etc which make things tough in a non-stationary, non-Gaussian case. Is there a general literature on this type of problem?

$\endgroup$
1
  • $\begingroup$ Not sure what to do in case of non-gaussian case (I think you need to have some assumption about the distribution of noise), but if series is stationary, forecast of $X_{N+h}$ is a point forecast for $E[X_{N+h}]$ and if N is large, the forecast's asymptotic distribution is usually known (depends on estimation process). This can be used to get $P(X_{N+h} \geq u)$ using z-table. The case of non-stationary would be fairly complicated. You'd have to start with $h=1$ and take $\widetilde{u}\equiv u-X_N$ and so on. $\endgroup$
    – Dayne
    Aug 20 '19 at 7:16
1
$\begingroup$

So let me consider only the Gaussian case: Here also two possibilities - Stationary and non-Stationary.

Stationary Case Let $\{x_t\}_{t=0}^{N}$ be an instance of the stochastic process $\{X_t\}$. Since, we are assuming that $X_t$ is stationary, let's also assume that we have a good model to forecast it.

Let $\mu_h \equiv \hat{X}_{t+h}$ be the estimate of $E[X_{t+h}|\{x_t\}_{t=0}^{N}]$; and $\sigma_h^2$ be the estimate of $Var(X_{t+h}|\{x_t\}_{t=0}^{N})$ that we have from the forecasting model.

A Side note: The standard error reported by forecasting model is not the $\sigma_h^2$ we need. The reported standard error is the standard deviation of $\mu_h$ (which is an estimate of actual mean). For details see this.

Coming back to the question. Since we have estimates of the mean and variance of $X_{t+h}$, we know the approximate asymptotic distribution of $X_{t+h}$.

So, $X_{t+h} \sim AN(\mu_h,\sigma_h^2)$. So $P(X_{N+h}>u)=1-P(X_{N+h}\leq u)=1-F(u|\mu_h,\sigma_h^2)$; where F is the normal CDF with $\mu_h,\sigma_h^2$ parameters.

Non-Stationary Case This would follow the same concept but the result will be fairly ugly.

So assume that while $\{X_t\}$ is not stationary, $\{\Delta X_t\}$ is stationary.

Now, what we have from the forecasting model, instead, is the estimate of $E[\Delta X_{t+h}|\{\Delta x_t\}_{t=1}^{N}]$

See that, $X_{N+h}=\sum\limits_{j=1}^{h} (\Delta X_{N+j})+X_N$

So, $E[X_{N+h}]=\sum\limits_{j=1}^{h} (E[\Delta X_{N+j}])+x_N$ ................................(1)

$Var(X_{N+h})=Var(\sum\limits_{j=1}^{h} \Delta X_{N+j})$ $=\sum\limits_{i=1}^h \sum\limits_{j=1}^h Cov(\Delta X_{N+i},\Delta X_{N+j})$

$=\sum\limits_{i=1}^h Var(\Delta X_{N+i})+\sum\limits_{k=0}^{h-1} (k+1)\gamma_{h-k}$ ...........................(2)

where $\gamma_j \equiv Cov(\Delta X_{t},\Delta X_{t+j}) \,\,\,\,\forall t$ (this flows from stationarity of $\{\Delta X_t\}$.

Now let $\mu_{dh}$, $\sigma_{dh}^2$ and $\hat\gamma_j$ be the estimates of $E[\Delta X_{N+h}]$, $Var(\Delta X_{N+h})$ and $Cov(\Delta X_{t},\Delta X_{t+j})$, respectively, from the model.

So using (1): $\mu_{h}=x_N+\sum\limits_{j=1}^h \mu_{dj}$

Similarly from (2),

$\sigma_{h}^2=\sum\limits_{j=1}^h \sigma_{dj}+\sum\limits_{k=0}^{h-1} (k+1)\hat\gamma_{h-k}\,\,\,\,\,$(for $h<N$)

So now we have estimates of the parameters of the asymptotic distribution of $X_{N+h}$ so as in case of stationarity above, we can use:

$X_{t+h} \sim AN(\mu_h,\sigma_h^2)$

Hope there aren't any mistakes.

$\endgroup$

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.