Question on Checking the stationarity of a Process

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Suppose, $Xt$ is a time series, defined by,
$Xt = cos(t+Y)$,
where $Y$ is a random variable, uniformly distributed over $[0,2pi]$.
How do I conclude that $Xt$ is a STRICT-SENSE stationary process?



What I figured out is that for stationarity, the cumulative PDF of a number of collections of random variables constituting the random process must remain unchanged when shifted in time.



But, how do I prove this mathematically or intuitively for the given time series?
I'm just starting out on learning about random processes and encountered this as an example in the Wikipedia article.






probability probability-theory statistics random-variables stationary-processes







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  • What does 'd' denote?
    – LumosMaxima
    Aug 31 at 5:47










  • ?? Equality in distribution.
    – Did
    Aug 31 at 5:49










  • How does that work? Y is uniformly distributed in [0,2pi]. Adding some other variable to it might lead Y+d to exceed the range of [0,2pi].
    – LumosMaxima
    Aug 31 at 5:53










  • Sorry, correction: $cos(Y+t)stackrel d=cos Y$.
    – Did
    Aug 31 at 5:55














up vote
-3
down vote

favorite












Suppose, $Xt$ is a time series, defined by,
$Xt = cos(t+Y)$,
where $Y$ is a random variable, uniformly distributed over $[0,2pi]$.
How do I conclude that $Xt$ is a STRICT-SENSE stationary process?



What I figured out is that for stationarity, the cumulative PDF of a number of collections of random variables constituting the random process must remain unchanged when shifted in time.



But, how do I prove this mathematically or intuitively for the given time series?
I'm just starting out on learning about random processes and encountered this as an example in the Wikipedia article.






probability probability-theory statistics random-variables stationary-processes







share|cite|improve this question























  • What does 'd' denote?
    – LumosMaxima
    Aug 31 at 5:47










  • ?? Equality in distribution.
    – Did
    Aug 31 at 5:49










  • How does that work? Y is uniformly distributed in [0,2pi]. Adding some other variable to it might lead Y+d to exceed the range of [0,2pi].
    – LumosMaxima
    Aug 31 at 5:53










  • Sorry, correction: $cos(Y+t)stackrel d=cos Y$.
    – Did
    Aug 31 at 5:55












up vote
-3
down vote

favorite









up vote
-3
down vote

favorite











Suppose, $Xt$ is a time series, defined by,
$Xt = cos(t+Y)$,
where $Y$ is a random variable, uniformly distributed over $[0,2pi]$.
How do I conclude that $Xt$ is a STRICT-SENSE stationary process?



What I figured out is that for stationarity, the cumulative PDF of a number of collections of random variables constituting the random process must remain unchanged when shifted in time.



But, how do I prove this mathematically or intuitively for the given time series?
I'm just starting out on learning about random processes and encountered this as an example in the Wikipedia article.






probability probability-theory statistics random-variables stationary-processes







share|cite|improve this question















Suppose, $Xt$ is a time series, defined by,
$Xt = cos(t+Y)$,
where $Y$ is a random variable, uniformly distributed over $[0,2pi]$.
How do I conclude that $Xt$ is a STRICT-SENSE stationary process?



What I figured out is that for stationarity, the cumulative PDF of a number of collections of random variables constituting the random process must remain unchanged when shifted in time.



But, how do I prove this mathematically or intuitively for the given time series?
I'm just starting out on learning about random processes and encountered this as an example in the Wikipedia article.






probability probability-theory statistics random-variables stationary-processes




probability probability-theory statistics random-variables stationary-processes






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 31 at 4:47

























asked Aug 31 at 4:25









LumosMaxima

378




378











  • What does 'd' denote?
    – LumosMaxima
    Aug 31 at 5:47










  • ?? Equality in distribution.
    – Did
    Aug 31 at 5:49










  • How does that work? Y is uniformly distributed in [0,2pi]. Adding some other variable to it might lead Y+d to exceed the range of [0,2pi].
    – LumosMaxima
    Aug 31 at 5:53










  • Sorry, correction: $cos(Y+t)stackrel d=cos Y$.
    – Did
    Aug 31 at 5:55
















  • What does 'd' denote?
    – LumosMaxima
    Aug 31 at 5:47










  • ?? Equality in distribution.
    – Did
    Aug 31 at 5:49










  • How does that work? Y is uniformly distributed in [0,2pi]. Adding some other variable to it might lead Y+d to exceed the range of [0,2pi].
    – LumosMaxima
    Aug 31 at 5:53










  • Sorry, correction: $cos(Y+t)stackrel d=cos Y$.
    – Did
    Aug 31 at 5:55















What does 'd' denote?
– LumosMaxima
Aug 31 at 5:47




What does 'd' denote?
– LumosMaxima
Aug 31 at 5:47












?? Equality in distribution.
– Did
Aug 31 at 5:49




?? Equality in distribution.
– Did
Aug 31 at 5:49












How does that work? Y is uniformly distributed in [0,2pi]. Adding some other variable to it might lead Y+d to exceed the range of [0,2pi].
– LumosMaxima
Aug 31 at 5:53




How does that work? Y is uniformly distributed in [0,2pi]. Adding some other variable to it might lead Y+d to exceed the range of [0,2pi].
– LumosMaxima
Aug 31 at 5:53












Sorry, correction: $cos(Y+t)stackrel d=cos Y$.
– Did
Aug 31 at 5:55




Sorry, correction: $cos(Y+t)stackrel d=cos Y$.
– Did
Aug 31 at 5:55















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