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random experiment with two different functions on unit interval



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Random variable and probability space notionsRandom Walk with EdgesFinding functions where the increase over a random interval is Poisson distributedNumber of days until dayCan an observed event in fact be of zero probability?Unit random processmodels of coins and uniform distributionHow to get the number of successes given $n$ trials , probability $P$ and a random variable $X$Absorbing Markov chain in a computer. Is “almost every” turned into always convergence in computer executions?Stopped random walk is not uniformly integrable










0












$begingroup$


Let $X=[0,1]$, and functions $f(x)=x$, $g(x)=2x$ mod $1$, and the probability of chosing $f,g$: $mu(f)=mu(g)=frac12$. Now if $x$ is the starting point, then what will be a general expression of state after $n$ iteration?



One of my friends says it will be $X_n=(eta_1+eta_2+dots+eta_n)x$ mod $1$, where $eta_i$ are i.i.d taking values $0,1$but I don't understand his explanations.



This is some random experiment I understand but how the process will go on, I can not visualize. Each step of the random experiment one of the above function will play their role and with probability fifty-fifty, right? So If I consider a diagram according to the experiment or iteration, I will get some curve also? Thanks for describing me mathematically.










share|cite|improve this question











$endgroup$
















    0












    $begingroup$


    Let $X=[0,1]$, and functions $f(x)=x$, $g(x)=2x$ mod $1$, and the probability of chosing $f,g$: $mu(f)=mu(g)=frac12$. Now if $x$ is the starting point, then what will be a general expression of state after $n$ iteration?



    One of my friends says it will be $X_n=(eta_1+eta_2+dots+eta_n)x$ mod $1$, where $eta_i$ are i.i.d taking values $0,1$but I don't understand his explanations.



    This is some random experiment I understand but how the process will go on, I can not visualize. Each step of the random experiment one of the above function will play their role and with probability fifty-fifty, right? So If I consider a diagram according to the experiment or iteration, I will get some curve also? Thanks for describing me mathematically.










    share|cite|improve this question











    $endgroup$














      0












      0








      0


      0



      $begingroup$


      Let $X=[0,1]$, and functions $f(x)=x$, $g(x)=2x$ mod $1$, and the probability of chosing $f,g$: $mu(f)=mu(g)=frac12$. Now if $x$ is the starting point, then what will be a general expression of state after $n$ iteration?



      One of my friends says it will be $X_n=(eta_1+eta_2+dots+eta_n)x$ mod $1$, where $eta_i$ are i.i.d taking values $0,1$but I don't understand his explanations.



      This is some random experiment I understand but how the process will go on, I can not visualize. Each step of the random experiment one of the above function will play their role and with probability fifty-fifty, right? So If I consider a diagram according to the experiment or iteration, I will get some curve also? Thanks for describing me mathematically.










      share|cite|improve this question











      $endgroup$




      Let $X=[0,1]$, and functions $f(x)=x$, $g(x)=2x$ mod $1$, and the probability of chosing $f,g$: $mu(f)=mu(g)=frac12$. Now if $x$ is the starting point, then what will be a general expression of state after $n$ iteration?



      One of my friends says it will be $X_n=(eta_1+eta_2+dots+eta_n)x$ mod $1$, where $eta_i$ are i.i.d taking values $0,1$but I don't understand his explanations.



      This is some random experiment I understand but how the process will go on, I can not visualize. Each step of the random experiment one of the above function will play their role and with probability fifty-fifty, right? So If I consider a diagram according to the experiment or iteration, I will get some curve also? Thanks for describing me mathematically.







      probability random-variables random-walk random-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 27 at 21:41









      David G. Stork

      12.2k41836




      12.2k41836










      asked Mar 27 at 21:25









      Ding DongDing Dong

      17.4k1162186




      17.4k1162186




















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Your friend is wrong, but has the right idea. Let $eta_j = 1$ if $g$ is picked at the $j$th iteration, and $0$ otherwise. (I.e. $eta_j$ is the indicator variable.) Then



          $$X_j = 2^eta_j X_j-1 pmod 1$$



          and so



          $$X_n = 2^eta_n 2^eta_n-1 cdots 2^eta_2 2^eta_1 X_0 = 2^(eta_n + cdots + eta_2 + eta_1) X_0 pmod 1$$



          As for visualization... If $f$ is picked then $X$ didn't change, but if $g$ is picked then $X$ doubles, and then "wraps around" $pmod 1$ if the doubled value $> 1$. If you plot successive $X_n$ there are some flat stretches ($f$ being chosen) then some jumps which may look hard to comprehend.



          For some starting values the series is easily described. E.g. starting at $X_0=1/2$ or indeed any $2^-k$ you keep doubling (whenever $g$ is chosen) until getting stuck at $1=0$. Alternatively starting at $X_0 = 1/3$ you just toggle between $1/3$ and $2/3$ (whenever $g$ is chosen). Somewhat similar things happen for a rational $X_0$ whose denominator is not a power of $2$ (and in particular the series eventually repeats). But for an irrational $X_0$ the series will go all over the place and will never repeat, and (I am not an expert on this last point) I believe in some technical sense may become "uniform" over the $(0,1)$ interval...?






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thanksssssssssssssssssss very muchhhhhhhhhhhhhh I understood.
            $endgroup$
            – Ding Dong
            Mar 28 at 8:38











          Your Answer








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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          2












          $begingroup$

          Your friend is wrong, but has the right idea. Let $eta_j = 1$ if $g$ is picked at the $j$th iteration, and $0$ otherwise. (I.e. $eta_j$ is the indicator variable.) Then



          $$X_j = 2^eta_j X_j-1 pmod 1$$



          and so



          $$X_n = 2^eta_n 2^eta_n-1 cdots 2^eta_2 2^eta_1 X_0 = 2^(eta_n + cdots + eta_2 + eta_1) X_0 pmod 1$$



          As for visualization... If $f$ is picked then $X$ didn't change, but if $g$ is picked then $X$ doubles, and then "wraps around" $pmod 1$ if the doubled value $> 1$. If you plot successive $X_n$ there are some flat stretches ($f$ being chosen) then some jumps which may look hard to comprehend.



          For some starting values the series is easily described. E.g. starting at $X_0=1/2$ or indeed any $2^-k$ you keep doubling (whenever $g$ is chosen) until getting stuck at $1=0$. Alternatively starting at $X_0 = 1/3$ you just toggle between $1/3$ and $2/3$ (whenever $g$ is chosen). Somewhat similar things happen for a rational $X_0$ whose denominator is not a power of $2$ (and in particular the series eventually repeats). But for an irrational $X_0$ the series will go all over the place and will never repeat, and (I am not an expert on this last point) I believe in some technical sense may become "uniform" over the $(0,1)$ interval...?






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thanksssssssssssssssssss very muchhhhhhhhhhhhhh I understood.
            $endgroup$
            – Ding Dong
            Mar 28 at 8:38















          2












          $begingroup$

          Your friend is wrong, but has the right idea. Let $eta_j = 1$ if $g$ is picked at the $j$th iteration, and $0$ otherwise. (I.e. $eta_j$ is the indicator variable.) Then



          $$X_j = 2^eta_j X_j-1 pmod 1$$



          and so



          $$X_n = 2^eta_n 2^eta_n-1 cdots 2^eta_2 2^eta_1 X_0 = 2^(eta_n + cdots + eta_2 + eta_1) X_0 pmod 1$$



          As for visualization... If $f$ is picked then $X$ didn't change, but if $g$ is picked then $X$ doubles, and then "wraps around" $pmod 1$ if the doubled value $> 1$. If you plot successive $X_n$ there are some flat stretches ($f$ being chosen) then some jumps which may look hard to comprehend.



          For some starting values the series is easily described. E.g. starting at $X_0=1/2$ or indeed any $2^-k$ you keep doubling (whenever $g$ is chosen) until getting stuck at $1=0$. Alternatively starting at $X_0 = 1/3$ you just toggle between $1/3$ and $2/3$ (whenever $g$ is chosen). Somewhat similar things happen for a rational $X_0$ whose denominator is not a power of $2$ (and in particular the series eventually repeats). But for an irrational $X_0$ the series will go all over the place and will never repeat, and (I am not an expert on this last point) I believe in some technical sense may become "uniform" over the $(0,1)$ interval...?






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thanksssssssssssssssssss very muchhhhhhhhhhhhhh I understood.
            $endgroup$
            – Ding Dong
            Mar 28 at 8:38













          2












          2








          2





          $begingroup$

          Your friend is wrong, but has the right idea. Let $eta_j = 1$ if $g$ is picked at the $j$th iteration, and $0$ otherwise. (I.e. $eta_j$ is the indicator variable.) Then



          $$X_j = 2^eta_j X_j-1 pmod 1$$



          and so



          $$X_n = 2^eta_n 2^eta_n-1 cdots 2^eta_2 2^eta_1 X_0 = 2^(eta_n + cdots + eta_2 + eta_1) X_0 pmod 1$$



          As for visualization... If $f$ is picked then $X$ didn't change, but if $g$ is picked then $X$ doubles, and then "wraps around" $pmod 1$ if the doubled value $> 1$. If you plot successive $X_n$ there are some flat stretches ($f$ being chosen) then some jumps which may look hard to comprehend.



          For some starting values the series is easily described. E.g. starting at $X_0=1/2$ or indeed any $2^-k$ you keep doubling (whenever $g$ is chosen) until getting stuck at $1=0$. Alternatively starting at $X_0 = 1/3$ you just toggle between $1/3$ and $2/3$ (whenever $g$ is chosen). Somewhat similar things happen for a rational $X_0$ whose denominator is not a power of $2$ (and in particular the series eventually repeats). But for an irrational $X_0$ the series will go all over the place and will never repeat, and (I am not an expert on this last point) I believe in some technical sense may become "uniform" over the $(0,1)$ interval...?






          share|cite|improve this answer









          $endgroup$



          Your friend is wrong, but has the right idea. Let $eta_j = 1$ if $g$ is picked at the $j$th iteration, and $0$ otherwise. (I.e. $eta_j$ is the indicator variable.) Then



          $$X_j = 2^eta_j X_j-1 pmod 1$$



          and so



          $$X_n = 2^eta_n 2^eta_n-1 cdots 2^eta_2 2^eta_1 X_0 = 2^(eta_n + cdots + eta_2 + eta_1) X_0 pmod 1$$



          As for visualization... If $f$ is picked then $X$ didn't change, but if $g$ is picked then $X$ doubles, and then "wraps around" $pmod 1$ if the doubled value $> 1$. If you plot successive $X_n$ there are some flat stretches ($f$ being chosen) then some jumps which may look hard to comprehend.



          For some starting values the series is easily described. E.g. starting at $X_0=1/2$ or indeed any $2^-k$ you keep doubling (whenever $g$ is chosen) until getting stuck at $1=0$. Alternatively starting at $X_0 = 1/3$ you just toggle between $1/3$ and $2/3$ (whenever $g$ is chosen). Somewhat similar things happen for a rational $X_0$ whose denominator is not a power of $2$ (and in particular the series eventually repeats). But for an irrational $X_0$ the series will go all over the place and will never repeat, and (I am not an expert on this last point) I believe in some technical sense may become "uniform" over the $(0,1)$ interval...?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 27 at 23:07









          antkamantkam

          3,279412




          3,279412











          • $begingroup$
            Thanksssssssssssssssssss very muchhhhhhhhhhhhhh I understood.
            $endgroup$
            – Ding Dong
            Mar 28 at 8:38
















          • $begingroup$
            Thanksssssssssssssssssss very muchhhhhhhhhhhhhh I understood.
            $endgroup$
            – Ding Dong
            Mar 28 at 8:38















          $begingroup$
          Thanksssssssssssssssssss very muchhhhhhhhhhhhhh I understood.
          $endgroup$
          – Ding Dong
          Mar 28 at 8:38




          $begingroup$
          Thanksssssssssssssssssss very muchhhhhhhhhhhhhh I understood.
          $endgroup$
          – Ding Dong
          Mar 28 at 8:38

















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