Expected value random Fibonacci sequence












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For a random Fibonacci sequence given by $x_{n}=pm x_{n-1}+x_{n-2}$ , why do we have that $E(x_n)=x_0$? I only see that $E(x_n)=frac{1}{2}(x_{n-1}+x_{n-2})+frac{1}{2}(-x_{n-1}+x_{n-2})=x_{n-2}$






probability







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  • 1




    $begingroup$
    $x_{n-2}$ is also a random variable.
    $endgroup$
    – mathworker21
    Dec 28 '18 at 2:17
















0












$begingroup$


For a random Fibonacci sequence given by $x_{n}=pm x_{n-1}+x_{n-2}$ , why do we have that $E(x_n)=x_0$? I only see that $E(x_n)=frac{1}{2}(x_{n-1}+x_{n-2})+frac{1}{2}(-x_{n-1}+x_{n-2})=x_{n-2}$






probability







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $x_{n-2}$ is also a random variable.
    $endgroup$
    – mathworker21
    Dec 28 '18 at 2:17














0












0








0





$begingroup$


For a random Fibonacci sequence given by $x_{n}=pm x_{n-1}+x_{n-2}$ , why do we have that $E(x_n)=x_0$? I only see that $E(x_n)=frac{1}{2}(x_{n-1}+x_{n-2})+frac{1}{2}(-x_{n-1}+x_{n-2})=x_{n-2}$






probability







share|cite|improve this question









$endgroup$




For a random Fibonacci sequence given by $x_{n}=pm x_{n-1}+x_{n-2}$ , why do we have that $E(x_n)=x_0$? I only see that $E(x_n)=frac{1}{2}(x_{n-1}+x_{n-2})+frac{1}{2}(-x_{n-1}+x_{n-2})=x_{n-2}$






probability




probability






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asked Dec 28 '18 at 2:10









user2795243user2795243

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54








  • 1




    $begingroup$
    $x_{n-2}$ is also a random variable.
    $endgroup$
    – mathworker21
    Dec 28 '18 at 2:17














  • 1




    $begingroup$
    $x_{n-2}$ is also a random variable.
    $endgroup$
    – mathworker21
    Dec 28 '18 at 2:17








1




1




$begingroup$
$x_{n-2}$ is also a random variable.
$endgroup$
– mathworker21
Dec 28 '18 at 2:17




$begingroup$
$x_{n-2}$ is also a random variable.
$endgroup$
– mathworker21
Dec 28 '18 at 2:17










1 Answer
1






active

oldest

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0












$begingroup$

You are forgetting that all the $x_i$ are random variables themselves (for $i>1)$. Having considered this, you should reach the following result.$$E[x_n]=frac12 (E[x_{n-1}]+ E[x_{n-2}])+frac12 (-E[x_{n-1}]+ E[x_{n-2}])=E[x_{n-2}]=cdots=begin{cases}x_1&ntext{ odd}\x_0&ntext{ even}end{cases}$$




You may be more convinced by seeing an example. Suppose $x_0=1,x_1=2$. Then $$P(x_2=3)=P(x_2=-1)=frac12$$ Then $E[x_2]=1=x_0$.$$P(x_3=5)=P(x_3=-1)=P(x_3=1)=P(x_3=3)=frac14$$Then $E[x_3]=frac{5-1+1+3}4=frac84=2=x_1$



Finally, $$P(x_4=8)=frac12P(x_4=-2)=frac12P(x_4=2)=P(x_4=4)=P(x_4=0)=P(x_4=-4)=frac18$$ Therefore $$E[x_4]=frac{8-2-2+2+2+4+0-4}8=1=x_0$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ok, thanks a lot, yes I concluded the same but I'm reading this paper where they say that $E(x_n)=x_0$ sciencedirect.com/science/article/pii/S0022314X06000229. And it's at the very beginning of the paper, they don't say anything about $n$ or $x_0$ or $x_1$ but now I know that what I thought was right, thank you!
    $endgroup$
    – user2795243
    Dec 28 '18 at 2:44










  • $begingroup$
    @user2795243 it may be possible that the author simply assumes $x_0=x_1$, like in the Fibonacci sequence $1,1,2,3,5,8,13...$. But I do agree, there is confusion there. Glad to help!
    $endgroup$
    – John Doe
    Dec 28 '18 at 2:52











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1 Answer
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0












$begingroup$

You are forgetting that all the $x_i$ are random variables themselves (for $i>1)$. Having considered this, you should reach the following result.$$E[x_n]=frac12 (E[x_{n-1}]+ E[x_{n-2}])+frac12 (-E[x_{n-1}]+ E[x_{n-2}])=E[x_{n-2}]=cdots=begin{cases}x_1&ntext{ odd}\x_0&ntext{ even}end{cases}$$




You may be more convinced by seeing an example. Suppose $x_0=1,x_1=2$. Then $$P(x_2=3)=P(x_2=-1)=frac12$$ Then $E[x_2]=1=x_0$.$$P(x_3=5)=P(x_3=-1)=P(x_3=1)=P(x_3=3)=frac14$$Then $E[x_3]=frac{5-1+1+3}4=frac84=2=x_1$



Finally, $$P(x_4=8)=frac12P(x_4=-2)=frac12P(x_4=2)=P(x_4=4)=P(x_4=0)=P(x_4=-4)=frac18$$ Therefore $$E[x_4]=frac{8-2-2+2+2+4+0-4}8=1=x_0$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ok, thanks a lot, yes I concluded the same but I'm reading this paper where they say that $E(x_n)=x_0$ sciencedirect.com/science/article/pii/S0022314X06000229. And it's at the very beginning of the paper, they don't say anything about $n$ or $x_0$ or $x_1$ but now I know that what I thought was right, thank you!
    $endgroup$
    – user2795243
    Dec 28 '18 at 2:44










  • $begingroup$
    @user2795243 it may be possible that the author simply assumes $x_0=x_1$, like in the Fibonacci sequence $1,1,2,3,5,8,13...$. But I do agree, there is confusion there. Glad to help!
    $endgroup$
    – John Doe
    Dec 28 '18 at 2:52
















0












$begingroup$

You are forgetting that all the $x_i$ are random variables themselves (for $i>1)$. Having considered this, you should reach the following result.$$E[x_n]=frac12 (E[x_{n-1}]+ E[x_{n-2}])+frac12 (-E[x_{n-1}]+ E[x_{n-2}])=E[x_{n-2}]=cdots=begin{cases}x_1&ntext{ odd}\x_0&ntext{ even}end{cases}$$




You may be more convinced by seeing an example. Suppose $x_0=1,x_1=2$. Then $$P(x_2=3)=P(x_2=-1)=frac12$$ Then $E[x_2]=1=x_0$.$$P(x_3=5)=P(x_3=-1)=P(x_3=1)=P(x_3=3)=frac14$$Then $E[x_3]=frac{5-1+1+3}4=frac84=2=x_1$



Finally, $$P(x_4=8)=frac12P(x_4=-2)=frac12P(x_4=2)=P(x_4=4)=P(x_4=0)=P(x_4=-4)=frac18$$ Therefore $$E[x_4]=frac{8-2-2+2+2+4+0-4}8=1=x_0$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ok, thanks a lot, yes I concluded the same but I'm reading this paper where they say that $E(x_n)=x_0$ sciencedirect.com/science/article/pii/S0022314X06000229. And it's at the very beginning of the paper, they don't say anything about $n$ or $x_0$ or $x_1$ but now I know that what I thought was right, thank you!
    $endgroup$
    – user2795243
    Dec 28 '18 at 2:44










  • $begingroup$
    @user2795243 it may be possible that the author simply assumes $x_0=x_1$, like in the Fibonacci sequence $1,1,2,3,5,8,13...$. But I do agree, there is confusion there. Glad to help!
    $endgroup$
    – John Doe
    Dec 28 '18 at 2:52














0












0








0





$begingroup$

You are forgetting that all the $x_i$ are random variables themselves (for $i>1)$. Having considered this, you should reach the following result.$$E[x_n]=frac12 (E[x_{n-1}]+ E[x_{n-2}])+frac12 (-E[x_{n-1}]+ E[x_{n-2}])=E[x_{n-2}]=cdots=begin{cases}x_1&ntext{ odd}\x_0&ntext{ even}end{cases}$$




You may be more convinced by seeing an example. Suppose $x_0=1,x_1=2$. Then $$P(x_2=3)=P(x_2=-1)=frac12$$ Then $E[x_2]=1=x_0$.$$P(x_3=5)=P(x_3=-1)=P(x_3=1)=P(x_3=3)=frac14$$Then $E[x_3]=frac{5-1+1+3}4=frac84=2=x_1$



Finally, $$P(x_4=8)=frac12P(x_4=-2)=frac12P(x_4=2)=P(x_4=4)=P(x_4=0)=P(x_4=-4)=frac18$$ Therefore $$E[x_4]=frac{8-2-2+2+2+4+0-4}8=1=x_0$$






share|cite|improve this answer











$endgroup$



You are forgetting that all the $x_i$ are random variables themselves (for $i>1)$. Having considered this, you should reach the following result.$$E[x_n]=frac12 (E[x_{n-1}]+ E[x_{n-2}])+frac12 (-E[x_{n-1}]+ E[x_{n-2}])=E[x_{n-2}]=cdots=begin{cases}x_1&ntext{ odd}\x_0&ntext{ even}end{cases}$$




You may be more convinced by seeing an example. Suppose $x_0=1,x_1=2$. Then $$P(x_2=3)=P(x_2=-1)=frac12$$ Then $E[x_2]=1=x_0$.$$P(x_3=5)=P(x_3=-1)=P(x_3=1)=P(x_3=3)=frac14$$Then $E[x_3]=frac{5-1+1+3}4=frac84=2=x_1$



Finally, $$P(x_4=8)=frac12P(x_4=-2)=frac12P(x_4=2)=P(x_4=4)=P(x_4=0)=P(x_4=-4)=frac18$$ Therefore $$E[x_4]=frac{8-2-2+2+2+4+0-4}8=1=x_0$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 28 '18 at 2:33

























answered Dec 28 '18 at 2:28









John DoeJohn Doe

11.2k11239




11.2k11239












  • $begingroup$
    Ok, thanks a lot, yes I concluded the same but I'm reading this paper where they say that $E(x_n)=x_0$ sciencedirect.com/science/article/pii/S0022314X06000229. And it's at the very beginning of the paper, they don't say anything about $n$ or $x_0$ or $x_1$ but now I know that what I thought was right, thank you!
    $endgroup$
    – user2795243
    Dec 28 '18 at 2:44










  • $begingroup$
    @user2795243 it may be possible that the author simply assumes $x_0=x_1$, like in the Fibonacci sequence $1,1,2,3,5,8,13...$. But I do agree, there is confusion there. Glad to help!
    $endgroup$
    – John Doe
    Dec 28 '18 at 2:52


















  • $begingroup$
    Ok, thanks a lot, yes I concluded the same but I'm reading this paper where they say that $E(x_n)=x_0$ sciencedirect.com/science/article/pii/S0022314X06000229. And it's at the very beginning of the paper, they don't say anything about $n$ or $x_0$ or $x_1$ but now I know that what I thought was right, thank you!
    $endgroup$
    – user2795243
    Dec 28 '18 at 2:44










  • $begingroup$
    @user2795243 it may be possible that the author simply assumes $x_0=x_1$, like in the Fibonacci sequence $1,1,2,3,5,8,13...$. But I do agree, there is confusion there. Glad to help!
    $endgroup$
    – John Doe
    Dec 28 '18 at 2:52
















$begingroup$
Ok, thanks a lot, yes I concluded the same but I'm reading this paper where they say that $E(x_n)=x_0$ sciencedirect.com/science/article/pii/S0022314X06000229. And it's at the very beginning of the paper, they don't say anything about $n$ or $x_0$ or $x_1$ but now I know that what I thought was right, thank you!
$endgroup$
– user2795243
Dec 28 '18 at 2:44




$begingroup$
Ok, thanks a lot, yes I concluded the same but I'm reading this paper where they say that $E(x_n)=x_0$ sciencedirect.com/science/article/pii/S0022314X06000229. And it's at the very beginning of the paper, they don't say anything about $n$ or $x_0$ or $x_1$ but now I know that what I thought was right, thank you!
$endgroup$
– user2795243
Dec 28 '18 at 2:44












$begingroup$
@user2795243 it may be possible that the author simply assumes $x_0=x_1$, like in the Fibonacci sequence $1,1,2,3,5,8,13...$. But I do agree, there is confusion there. Glad to help!
$endgroup$
– John Doe
Dec 28 '18 at 2:52




$begingroup$
@user2795243 it may be possible that the author simply assumes $x_0=x_1$, like in the Fibonacci sequence $1,1,2,3,5,8,13...$. But I do agree, there is confusion there. Glad to help!
$endgroup$
– John Doe
Dec 28 '18 at 2:52


















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