Question on Strict Sense Stationarity - concluding about strict sense stationarity from WSS

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite












How do I ensure whether a WSS stationary process is Strictly Stationary or not?



The rules for checking for WSS stationary involve checking that the mean is not a function of time, and the autocorrelation is a function of the shift between the two instances of the random process.



But, from my understanding, for Strict Sense Stationarity, all higher ordered moments must be independent of t as well.



Now, manually computing all higher order moments is laborious, so is there any way to conclude about SSS from observation of the random process itself?






linear-algebra probability probability-theory random-variables







share|cite|improve this question





















  • You cannot determine strict stationarity using moments. The only general method is to use the definition. In other words , you should know how to compute all finite dimensional distributions.
    – Kavi Rama Murthy
    Sep 1 at 12:12










  • Can I find an example anywhere where they have used this general method?
    – LumosMaxima
    Sep 1 at 12:14










  • Yes. An i.i.d. seqeunce $X_n$ is strictly stationary because the distribution of $X_k,X_k+1,...,X_k+n$ is same as that of $X_1,X_2,...,X_1+n$ for any $k$.
    – Kavi Rama Murthy
    Sep 1 at 12:19











  • Thanks. In this note, can you tell me why the random process cos(t+Y) where Y is uniformly distributed in [0,1] is stationary? I found this as an example in many places, but can't comprehend the exact mathematical steps to conclude so.
    – LumosMaxima
    Sep 1 at 12:29










  • $cos (t+Y)$ is not stationary if $Y$ is uniform on $[0,1]$. It is stationary if $Y$ is uniform on $[0,2pi]$.
    – Kavi Rama Murthy
    Sep 1 at 23:29














up vote
0
down vote

favorite












How do I ensure whether a WSS stationary process is Strictly Stationary or not?



The rules for checking for WSS stationary involve checking that the mean is not a function of time, and the autocorrelation is a function of the shift between the two instances of the random process.



But, from my understanding, for Strict Sense Stationarity, all higher ordered moments must be independent of t as well.



Now, manually computing all higher order moments is laborious, so is there any way to conclude about SSS from observation of the random process itself?






linear-algebra probability probability-theory random-variables







share|cite|improve this question





















  • You cannot determine strict stationarity using moments. The only general method is to use the definition. In other words , you should know how to compute all finite dimensional distributions.
    – Kavi Rama Murthy
    Sep 1 at 12:12










  • Can I find an example anywhere where they have used this general method?
    – LumosMaxima
    Sep 1 at 12:14










  • Yes. An i.i.d. seqeunce $X_n$ is strictly stationary because the distribution of $X_k,X_k+1,...,X_k+n$ is same as that of $X_1,X_2,...,X_1+n$ for any $k$.
    – Kavi Rama Murthy
    Sep 1 at 12:19











  • Thanks. In this note, can you tell me why the random process cos(t+Y) where Y is uniformly distributed in [0,1] is stationary? I found this as an example in many places, but can't comprehend the exact mathematical steps to conclude so.
    – LumosMaxima
    Sep 1 at 12:29










  • $cos (t+Y)$ is not stationary if $Y$ is uniform on $[0,1]$. It is stationary if $Y$ is uniform on $[0,2pi]$.
    – Kavi Rama Murthy
    Sep 1 at 23:29












up vote
0
down vote

favorite









up vote
0
down vote

favorite











How do I ensure whether a WSS stationary process is Strictly Stationary or not?



The rules for checking for WSS stationary involve checking that the mean is not a function of time, and the autocorrelation is a function of the shift between the two instances of the random process.



But, from my understanding, for Strict Sense Stationarity, all higher ordered moments must be independent of t as well.



Now, manually computing all higher order moments is laborious, so is there any way to conclude about SSS from observation of the random process itself?






linear-algebra probability probability-theory random-variables







share|cite|improve this question













How do I ensure whether a WSS stationary process is Strictly Stationary or not?



The rules for checking for WSS stationary involve checking that the mean is not a function of time, and the autocorrelation is a function of the shift between the two instances of the random process.



But, from my understanding, for Strict Sense Stationarity, all higher ordered moments must be independent of t as well.



Now, manually computing all higher order moments is laborious, so is there any way to conclude about SSS from observation of the random process itself?






linear-algebra probability probability-theory random-variables




linear-algebra probability probability-theory random-variables






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 1 at 6:34









LumosMaxima

378




378











  • You cannot determine strict stationarity using moments. The only general method is to use the definition. In other words , you should know how to compute all finite dimensional distributions.
    – Kavi Rama Murthy
    Sep 1 at 12:12










  • Can I find an example anywhere where they have used this general method?
    – LumosMaxima
    Sep 1 at 12:14










  • Yes. An i.i.d. seqeunce $X_n$ is strictly stationary because the distribution of $X_k,X_k+1,...,X_k+n$ is same as that of $X_1,X_2,...,X_1+n$ for any $k$.
    – Kavi Rama Murthy
    Sep 1 at 12:19











  • Thanks. In this note, can you tell me why the random process cos(t+Y) where Y is uniformly distributed in [0,1] is stationary? I found this as an example in many places, but can't comprehend the exact mathematical steps to conclude so.
    – LumosMaxima
    Sep 1 at 12:29










  • $cos (t+Y)$ is not stationary if $Y$ is uniform on $[0,1]$. It is stationary if $Y$ is uniform on $[0,2pi]$.
    – Kavi Rama Murthy
    Sep 1 at 23:29
















  • You cannot determine strict stationarity using moments. The only general method is to use the definition. In other words , you should know how to compute all finite dimensional distributions.
    – Kavi Rama Murthy
    Sep 1 at 12:12










  • Can I find an example anywhere where they have used this general method?
    – LumosMaxima
    Sep 1 at 12:14










  • Yes. An i.i.d. seqeunce $X_n$ is strictly stationary because the distribution of $X_k,X_k+1,...,X_k+n$ is same as that of $X_1,X_2,...,X_1+n$ for any $k$.
    – Kavi Rama Murthy
    Sep 1 at 12:19











  • Thanks. In this note, can you tell me why the random process cos(t+Y) where Y is uniformly distributed in [0,1] is stationary? I found this as an example in many places, but can't comprehend the exact mathematical steps to conclude so.
    – LumosMaxima
    Sep 1 at 12:29










  • $cos (t+Y)$ is not stationary if $Y$ is uniform on $[0,1]$. It is stationary if $Y$ is uniform on $[0,2pi]$.
    – Kavi Rama Murthy
    Sep 1 at 23:29















You cannot determine strict stationarity using moments. The only general method is to use the definition. In other words , you should know how to compute all finite dimensional distributions.
– Kavi Rama Murthy
Sep 1 at 12:12




You cannot determine strict stationarity using moments. The only general method is to use the definition. In other words , you should know how to compute all finite dimensional distributions.
– Kavi Rama Murthy
Sep 1 at 12:12












Can I find an example anywhere where they have used this general method?
– LumosMaxima
Sep 1 at 12:14




Can I find an example anywhere where they have used this general method?
– LumosMaxima
Sep 1 at 12:14












Yes. An i.i.d. seqeunce $X_n$ is strictly stationary because the distribution of $X_k,X_k+1,...,X_k+n$ is same as that of $X_1,X_2,...,X_1+n$ for any $k$.
– Kavi Rama Murthy
Sep 1 at 12:19





Yes. An i.i.d. seqeunce $X_n$ is strictly stationary because the distribution of $X_k,X_k+1,...,X_k+n$ is same as that of $X_1,X_2,...,X_1+n$ for any $k$.
– Kavi Rama Murthy
Sep 1 at 12:19













Thanks. In this note, can you tell me why the random process cos(t+Y) where Y is uniformly distributed in [0,1] is stationary? I found this as an example in many places, but can't comprehend the exact mathematical steps to conclude so.
– LumosMaxima
Sep 1 at 12:29




Thanks. In this note, can you tell me why the random process cos(t+Y) where Y is uniformly distributed in [0,1] is stationary? I found this as an example in many places, but can't comprehend the exact mathematical steps to conclude so.
– LumosMaxima
Sep 1 at 12:29












$cos (t+Y)$ is not stationary if $Y$ is uniform on $[0,1]$. It is stationary if $Y$ is uniform on $[0,2pi]$.
– Kavi Rama Murthy
Sep 1 at 23:29




$cos (t+Y)$ is not stationary if $Y$ is uniform on $[0,1]$. It is stationary if $Y$ is uniform on $[0,2pi]$.
– Kavi Rama Murthy
Sep 1 at 23:29















active

oldest

votes











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: "69"
;
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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2901431%2fquestion-on-strict-sense-stationarity-concluding-about-strict-sense-stationari%23new-answer', 'question_page');

);

Post as a guest






















active

oldest

votes













active

oldest

votes









active

oldest

votes






active

oldest

votes















 

draft saved


draft discarded















































 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2901431%2fquestion-on-strict-sense-stationarity-concluding-about-strict-sense-stationari%23new-answer', 'question_page');

);

Post as a guest
































































-

這個網誌中的熱門文章

How to combine Bézier curves to a surface?

圖片
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite My aim is to smooth the terrain in a video game. Therefore I contrived an algorithm that makes use of Bézier curves of different orders. But this algorithm is defined in a two dimensional space for now. To shift it into the third dimension I need to somehow combine the Bézier curves. Given two curves in each two dimensions, how can I combine them to build a surface? I thought about something like a curve of a curve. Or maybe I could multiply them somehow. How can I combine them to cause the wanted behavior? Is there a common approach? In the image you can see the input, on the left, and the output, on the right, of the algorithm that works in a two dimensional space. For the real application there is a three dimensional input. The algorithm relays only on the adjacent tiles. Given these it will define the control points of the mentioned Bézier cur...
閱讀完整內容

Mutual Information Always Non-negative

圖片
Clash Royale CLAN TAG #URR8PPP up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 '12 at 15:38 Cameron Buie 83.6k 7 71 154 asked Jun 17 '12 at 15:30 Omri 359 3 13 3 Convexity of the function $tmapsto tlog t$. – Did Jun 17 '12 at 15:33 In addition, the convexity properties require the coefficients in the linear combination sum 1. Then, as p(x,y) is a probability distribution, it fullfits such condition. – Francisco Javier Delgado Ceped Aug 10 at 18:44 add a comment  |  up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 ...
閱讀完整內容

Why am i infinitely getting the same tweet with the Twitter Search API?

圖片
up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia...
閱讀完整內容