Can discrete time series be non-stationary?












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$begingroup$


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. Example of non-stationary time series



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







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$endgroup$

















    1












    $begingroup$


    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. Example of non-stationary time series



    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







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      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. Example of non-stationary time series



      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







      share|cite|improve this question









      $endgroup$




      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. Example of non-stationary time series



      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 3 hours ago









      Linas KleizaLinas Kleiza

      294




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          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
            47 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♦
            35 mins ago











          Your Answer





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          active

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          3












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
            47 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♦
            35 mins ago
















          3












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
            47 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♦
            35 mins ago














          3












          3








          3





          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          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
            47 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♦
            35 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
            47 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♦
            35 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
          47 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
          47 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♦
          35 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♦
          35 mins ago


















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