Can discrete time series be non-stationary?
$begingroup$
We all know about time series that grow over time, but it seems like we only ever see continuous values such as the plot shown below.
But is there such thing as a time series that might start out with balanced mix of class 0 and class 1, but eventually turn into mostly class 1's?
time-series stationarity
$endgroup$
add a comment |
$begingroup$
We all know about time series that grow over time, but it seems like we only ever see continuous values such as the plot shown below.
But is there such thing as a time series that might start out with balanced mix of class 0 and class 1, but eventually turn into mostly class 1's?
time-series stationarity
$endgroup$
add a comment |
$begingroup$
We all know about time series that grow over time, but it seems like we only ever see continuous values such as the plot shown below.
But is there such thing as a time series that might start out with balanced mix of class 0 and class 1, but eventually turn into mostly class 1's?
time-series stationarity
$endgroup$
We all know about time series that grow over time, but it seems like we only ever see continuous values such as the plot shown below.
But is there such thing as a time series that might start out with balanced mix of class 0 and class 1, but eventually turn into mostly class 1's?
time-series stationarity
time-series stationarity
asked 3 hours ago
Linas KleizaLinas Kleiza
294
294
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Per the title, certainly discrete series can be nonstationary.
The total count of events in a Poisson process is one example - let $N_t$ be the number of events by time $t$ and consider the time series formed by looking at $N_t$ at integer times: $N_1$, $N_2$, $N_3$,...
We have that $E(N_t)=lambda t$, $text{Var}(N_t)=lambda t$; so the above time series is not even weakly stationary.
But is there such thing as a time series that might start out with balanced mix of class 0 and class 1, but eventually turn into mostly class 1's?
Sure, you could readily construct a binary time series process that was non-stationary.
Let's make an example. Let $Y_t$ be the random binary value at time $t$ ($t=1,2,...$). Let $S_t$ bet the number of $1$'s observed up to time $t$; i.e. $S_0=0,S_t=sum_{i=1}^t Y_i,,: t = 1,2,...$.
Now make $p_{t+1}=P(Y_{t+1}=1|S_t)=frac{1}{1+exp(-alpha S_t)}$ for some $alpha>0$.
So $p_1=0.5$. It will remain at $0.5$ until you observe a $1$, at which point $p$ increases. e.g. if $alpha=0.3$, then after the first $1$ is observed, $p$ increases to about $0.574$ whereas if $alpha=0.1$, $p$ only increases to about $0.525$.
(There are many other ways to make nonstationary binary series)
Here's an example using the above scheme with $alpha=0.005$, for the first 1000 observations. You can see that it starts out with a similar density of 0 and 1 values but the 1-strip rapidly starts to get heavier and the 0-strip gets lighter and lighter, until near the end the 0's have become quite rare.
The conditional probability of a 1 for the last observation ($y_{1000}$) was about 0.979.
$endgroup$
$begingroup$
Thanks for the example. But for continuous non-stationary time series you can take the differences to make it stationary. Is there an analogous procedure for non-stationary discrete time series?
$endgroup$
– Linas Kleiza
52 mins ago
1
$begingroup$
Not in general, no; it would work with the Poisson example I raised, but cannot work for the binary one. (For that matter, it's not always the case that differencing results in stationarity with continuous series. For some very particular sets of assumptions differencing produces stationarity, but not all)
$endgroup$
– Glen_b♦
39 mins ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "65"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f389230%2fcan-discrete-time-series-be-non-stationary%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Per the title, certainly discrete series can be nonstationary.
The total count of events in a Poisson process is one example - let $N_t$ be the number of events by time $t$ and consider the time series formed by looking at $N_t$ at integer times: $N_1$, $N_2$, $N_3$,...
We have that $E(N_t)=lambda t$, $text{Var}(N_t)=lambda t$; so the above time series is not even weakly stationary.
But is there such thing as a time series that might start out with balanced mix of class 0 and class 1, but eventually turn into mostly class 1's?
Sure, you could readily construct a binary time series process that was non-stationary.
Let's make an example. Let $Y_t$ be the random binary value at time $t$ ($t=1,2,...$). Let $S_t$ bet the number of $1$'s observed up to time $t$; i.e. $S_0=0,S_t=sum_{i=1}^t Y_i,,: t = 1,2,...$.
Now make $p_{t+1}=P(Y_{t+1}=1|S_t)=frac{1}{1+exp(-alpha S_t)}$ for some $alpha>0$.
So $p_1=0.5$. It will remain at $0.5$ until you observe a $1$, at which point $p$ increases. e.g. if $alpha=0.3$, then after the first $1$ is observed, $p$ increases to about $0.574$ whereas if $alpha=0.1$, $p$ only increases to about $0.525$.
(There are many other ways to make nonstationary binary series)
Here's an example using the above scheme with $alpha=0.005$, for the first 1000 observations. You can see that it starts out with a similar density of 0 and 1 values but the 1-strip rapidly starts to get heavier and the 0-strip gets lighter and lighter, until near the end the 0's have become quite rare.
The conditional probability of a 1 for the last observation ($y_{1000}$) was about 0.979.
$endgroup$
$begingroup$
Thanks for the example. But for continuous non-stationary time series you can take the differences to make it stationary. Is there an analogous procedure for non-stationary discrete time series?
$endgroup$
– Linas Kleiza
52 mins ago
1
$begingroup$
Not in general, no; it would work with the Poisson example I raised, but cannot work for the binary one. (For that matter, it's not always the case that differencing results in stationarity with continuous series. For some very particular sets of assumptions differencing produces stationarity, but not all)
$endgroup$
– Glen_b♦
39 mins ago
add a comment |
$begingroup$
Per the title, certainly discrete series can be nonstationary.
The total count of events in a Poisson process is one example - let $N_t$ be the number of events by time $t$ and consider the time series formed by looking at $N_t$ at integer times: $N_1$, $N_2$, $N_3$,...
We have that $E(N_t)=lambda t$, $text{Var}(N_t)=lambda t$; so the above time series is not even weakly stationary.
But is there such thing as a time series that might start out with balanced mix of class 0 and class 1, but eventually turn into mostly class 1's?
Sure, you could readily construct a binary time series process that was non-stationary.
Let's make an example. Let $Y_t$ be the random binary value at time $t$ ($t=1,2,...$). Let $S_t$ bet the number of $1$'s observed up to time $t$; i.e. $S_0=0,S_t=sum_{i=1}^t Y_i,,: t = 1,2,...$.
Now make $p_{t+1}=P(Y_{t+1}=1|S_t)=frac{1}{1+exp(-alpha S_t)}$ for some $alpha>0$.
So $p_1=0.5$. It will remain at $0.5$ until you observe a $1$, at which point $p$ increases. e.g. if $alpha=0.3$, then after the first $1$ is observed, $p$ increases to about $0.574$ whereas if $alpha=0.1$, $p$ only increases to about $0.525$.
(There are many other ways to make nonstationary binary series)
Here's an example using the above scheme with $alpha=0.005$, for the first 1000 observations. You can see that it starts out with a similar density of 0 and 1 values but the 1-strip rapidly starts to get heavier and the 0-strip gets lighter and lighter, until near the end the 0's have become quite rare.
The conditional probability of a 1 for the last observation ($y_{1000}$) was about 0.979.
$endgroup$
$begingroup$
Thanks for the example. But for continuous non-stationary time series you can take the differences to make it stationary. Is there an analogous procedure for non-stationary discrete time series?
$endgroup$
– Linas Kleiza
52 mins ago
1
$begingroup$
Not in general, no; it would work with the Poisson example I raised, but cannot work for the binary one. (For that matter, it's not always the case that differencing results in stationarity with continuous series. For some very particular sets of assumptions differencing produces stationarity, but not all)
$endgroup$
– Glen_b♦
39 mins ago
add a comment |
$begingroup$
Per the title, certainly discrete series can be nonstationary.
The total count of events in a Poisson process is one example - let $N_t$ be the number of events by time $t$ and consider the time series formed by looking at $N_t$ at integer times: $N_1$, $N_2$, $N_3$,...
We have that $E(N_t)=lambda t$, $text{Var}(N_t)=lambda t$; so the above time series is not even weakly stationary.
But is there such thing as a time series that might start out with balanced mix of class 0 and class 1, but eventually turn into mostly class 1's?
Sure, you could readily construct a binary time series process that was non-stationary.
Let's make an example. Let $Y_t$ be the random binary value at time $t$ ($t=1,2,...$). Let $S_t$ bet the number of $1$'s observed up to time $t$; i.e. $S_0=0,S_t=sum_{i=1}^t Y_i,,: t = 1,2,...$.
Now make $p_{t+1}=P(Y_{t+1}=1|S_t)=frac{1}{1+exp(-alpha S_t)}$ for some $alpha>0$.
So $p_1=0.5$. It will remain at $0.5$ until you observe a $1$, at which point $p$ increases. e.g. if $alpha=0.3$, then after the first $1$ is observed, $p$ increases to about $0.574$ whereas if $alpha=0.1$, $p$ only increases to about $0.525$.
(There are many other ways to make nonstationary binary series)
Here's an example using the above scheme with $alpha=0.005$, for the first 1000 observations. You can see that it starts out with a similar density of 0 and 1 values but the 1-strip rapidly starts to get heavier and the 0-strip gets lighter and lighter, until near the end the 0's have become quite rare.
The conditional probability of a 1 for the last observation ($y_{1000}$) was about 0.979.
$endgroup$
Per the title, certainly discrete series can be nonstationary.
The total count of events in a Poisson process is one example - let $N_t$ be the number of events by time $t$ and consider the time series formed by looking at $N_t$ at integer times: $N_1$, $N_2$, $N_3$,...
We have that $E(N_t)=lambda t$, $text{Var}(N_t)=lambda t$; so the above time series is not even weakly stationary.
But is there such thing as a time series that might start out with balanced mix of class 0 and class 1, but eventually turn into mostly class 1's?
Sure, you could readily construct a binary time series process that was non-stationary.
Let's make an example. Let $Y_t$ be the random binary value at time $t$ ($t=1,2,...$). Let $S_t$ bet the number of $1$'s observed up to time $t$; i.e. $S_0=0,S_t=sum_{i=1}^t Y_i,,: t = 1,2,...$.
Now make $p_{t+1}=P(Y_{t+1}=1|S_t)=frac{1}{1+exp(-alpha S_t)}$ for some $alpha>0$.
So $p_1=0.5$. It will remain at $0.5$ until you observe a $1$, at which point $p$ increases. e.g. if $alpha=0.3$, then after the first $1$ is observed, $p$ increases to about $0.574$ whereas if $alpha=0.1$, $p$ only increases to about $0.525$.
(There are many other ways to make nonstationary binary series)
Here's an example using the above scheme with $alpha=0.005$, for the first 1000 observations. You can see that it starts out with a similar density of 0 and 1 values but the 1-strip rapidly starts to get heavier and the 0-strip gets lighter and lighter, until near the end the 0's have become quite rare.
The conditional probability of a 1 for the last observation ($y_{1000}$) was about 0.979.
edited 2 hours ago
answered 3 hours ago
Glen_b♦Glen_b
210k22401743
210k22401743
$begingroup$
Thanks for the example. But for continuous non-stationary time series you can take the differences to make it stationary. Is there an analogous procedure for non-stationary discrete time series?
$endgroup$
– Linas Kleiza
52 mins ago
1
$begingroup$
Not in general, no; it would work with the Poisson example I raised, but cannot work for the binary one. (For that matter, it's not always the case that differencing results in stationarity with continuous series. For some very particular sets of assumptions differencing produces stationarity, but not all)
$endgroup$
– Glen_b♦
39 mins ago
add a comment |
$begingroup$
Thanks for the example. But for continuous non-stationary time series you can take the differences to make it stationary. Is there an analogous procedure for non-stationary discrete time series?
$endgroup$
– Linas Kleiza
52 mins ago
1
$begingroup$
Not in general, no; it would work with the Poisson example I raised, but cannot work for the binary one. (For that matter, it's not always the case that differencing results in stationarity with continuous series. For some very particular sets of assumptions differencing produces stationarity, but not all)
$endgroup$
– Glen_b♦
39 mins ago
$begingroup$
Thanks for the example. But for continuous non-stationary time series you can take the differences to make it stationary. Is there an analogous procedure for non-stationary discrete time series?
$endgroup$
– Linas Kleiza
52 mins ago
$begingroup$
Thanks for the example. But for continuous non-stationary time series you can take the differences to make it stationary. Is there an analogous procedure for non-stationary discrete time series?
$endgroup$
– Linas Kleiza
52 mins ago
1
1
$begingroup$
Not in general, no; it would work with the Poisson example I raised, but cannot work for the binary one. (For that matter, it's not always the case that differencing results in stationarity with continuous series. For some very particular sets of assumptions differencing produces stationarity, but not all)
$endgroup$
– Glen_b♦
39 mins ago
$begingroup$
Not in general, no; it would work with the Poisson example I raised, but cannot work for the binary one. (For that matter, it's not always the case that differencing results in stationarity with continuous series. For some very particular sets of assumptions differencing produces stationarity, but not all)
$endgroup$
– Glen_b♦
39 mins ago
add a comment |
Thanks for contributing an answer to Cross Validated!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f389230%2fcan-discrete-time-series-be-non-stationary%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown