Suppose X is a time series that follows the process:

$X[t]=\rho X[t-1] + \varepsilon[t]$

where $\varepsilon[t] \sim \mathcal{N}(0.0, \sigma)$, $\varepsilon[t]$ is IID and $X[0] \sim \mathcal{N}(0.0, \sigma)$,.

My interest is computing a conditional expectation using mathematical:

Allowing for notational sloppiness, I want to focus on the subset of $X[t]$ that satisfy:

$$ \tilde{X}[t]\ < Q( X[t] < Q(0.1, X[t]) $$ Where $Q(0.1, X[t])$ is the $0.1$ quantile observation of $X[t]$.

Then find the conditional expectation of $\tilde{X}[t]$:

$$\mathbb{E}[\ \tilde{X}[t] \ |\ ( \tilde{X}[t-3] < Q(0.5, \tilde{X}[t-3]))] $$

In words, I want to find the subset of the process that falls below the 0.1 quantile observation. Then find the mean of the subset that satisfies the condition before and also had a realization three periods ago that is lower than the median of all observations from three periods for this particular subset of realizations.

  • $\begingroup$ Define AR1..... $\endgroup$ Jan 10 at 2:56
  • $\begingroup$ For reference: Autoregressive Model wiki page. $\endgroup$
    – Syed
    Jan 10 at 3:29
  • $\begingroup$ Should we assume that the $\epsilon_i$'s are all independent (which is implied by the title but not made explicit in the question) and that $X_0\sim N(0,\sigma)$? $\endgroup$
    – JimB
    Jan 10 at 4:29
  • $\begingroup$ Yep, I added some more information to clarify that. $\endgroup$ Jan 10 at 4:38
  • 1
    $\begingroup$ I want to emphasize that $X_0 \sim N(0,\sigma/\sqrt{1-\rho^2})$ (which is the value in the link given by @Syed) which means that $E(X_t | X_{t-3} < 0)$ is $-\frac{\sqrt{\frac{2}{\pi }} \rho ^3 \sigma }{\sqrt{1-\rho ^2}}$. (I haven't got a closed-form for when $1/10=0.1$ is used.) $\endgroup$
    – JimB
    Jan 10 at 21:55

1 Answer 1


If in Mathematica terms by $Q(0.05, X_t)$ you mean

x = InverseCDF[NormalDistribution[0, σ/Sqrt[1 - ρ^2]], 1/20]

then because the marginal distributions of the $X_i$ are all normal with mean zero and variance $\sigma^2/(1-\rho^2)$, $Q(0.05, X_t)$ is always less than $Q(0.10, X_{t-3})$, then mean of $X_t$ conditional on $X_t < x$ is found with

dist = TruncatedDistribution[{-∞, InverseCDF[NormalDistribution[0, σ/Sqrt[1 - ρ^2]], 1/20]},
  NormalDistribution[0, σ/Sqrt[1 - ρ^2]]]

Conditional mean

But I suspect this is not the conditioning you want.

  • $\begingroup$ You are right Jim. Thinking about it some more, I am after something that is definitely not like this. What I want to compute is a conditional expectation. I first want to focus on $X_t$ less than $Q(0.1,X_t)$ . So let's define these as: $$\tilde{X_t} < Q(0.1, X_t)$$ Then the object I am interested in can be summarised in sloppy notation as: $$\mathbb{E}_t[ \tilde{X_t} \ | \ \tilde{X}_{t-3} \ < \ Q(0.5, \tilde{X}_{t-3}) ] $$ $\endgroup$ Jan 10 at 14:45
  • $\begingroup$ Sounds good. Editing your question to reflect that would be helpful to others. In short, $X_t$ and $X_{t-3}$ will have a bivariate normal distribution so that conditional expectation will be easy to obtain. $\endgroup$
    – JimB
    Jan 10 at 15:49

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