Expectation of the stochastic process












0












$begingroup$


Consider the following stochastic process.





  • $X(t)=Y(t-1)$ where $Y(t-1)sim Normal(X(t-1)+1,sigma^2)$

  • $X(0)=1$


Now I am interested in the $E[X(n)]$.
Intuitively, I think $E[X(n)]$ is equal to $n$.
Is this the correct answer? If it is, how can I show this mathematically?






expected-value







share|cite|improve this question









$endgroup$












  • $begingroup$
    Is this a discrete time process? You might be able to use induction.
    $endgroup$
    – Sean Roberson
    Dec 9 '18 at 3:33
















0












$begingroup$


Consider the following stochastic process.





  • $X(t)=Y(t-1)$ where $Y(t-1)sim Normal(X(t-1)+1,sigma^2)$

  • $X(0)=1$


Now I am interested in the $E[X(n)]$.
Intuitively, I think $E[X(n)]$ is equal to $n$.
Is this the correct answer? If it is, how can I show this mathematically?






expected-value







share|cite|improve this question









$endgroup$












  • $begingroup$
    Is this a discrete time process? You might be able to use induction.
    $endgroup$
    – Sean Roberson
    Dec 9 '18 at 3:33














0












0








0





$begingroup$


Consider the following stochastic process.





  • $X(t)=Y(t-1)$ where $Y(t-1)sim Normal(X(t-1)+1,sigma^2)$

  • $X(0)=1$


Now I am interested in the $E[X(n)]$.
Intuitively, I think $E[X(n)]$ is equal to $n$.
Is this the correct answer? If it is, how can I show this mathematically?






expected-value







share|cite|improve this question









$endgroup$




Consider the following stochastic process.





  • $X(t)=Y(t-1)$ where $Y(t-1)sim Normal(X(t-1)+1,sigma^2)$

  • $X(0)=1$


Now I am interested in the $E[X(n)]$.
Intuitively, I think $E[X(n)]$ is equal to $n$.
Is this the correct answer? If it is, how can I show this mathematically?






expected-value




expected-value






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 9 '18 at 3:11









user3509199user3509199

316




316












  • $begingroup$
    Is this a discrete time process? You might be able to use induction.
    $endgroup$
    – Sean Roberson
    Dec 9 '18 at 3:33


















  • $begingroup$
    Is this a discrete time process? You might be able to use induction.
    $endgroup$
    – Sean Roberson
    Dec 9 '18 at 3:33
















$begingroup$
Is this a discrete time process? You might be able to use induction.
$endgroup$
– Sean Roberson
Dec 9 '18 at 3:33




$begingroup$
Is this a discrete time process? You might be able to use induction.
$endgroup$
– Sean Roberson
Dec 9 '18 at 3:33










1 Answer
1






active

oldest

votes


















2












$begingroup$

It’s simple induction. From $E[X(t)]=E[X(t-1)]+1 $ you can easily get the desired result.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! Could you elaborate little bit more?
    $endgroup$
    – user3509199
    Dec 9 '18 at 3:47










  • $begingroup$
    $E[X(t)] = E[X(t-2)]+2=E[X(t-3)]+3=...=E[X(0)]+n
    $endgroup$
    – mineiro
    Dec 9 '18 at 3:50











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1 Answer
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oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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2












$begingroup$

It’s simple induction. From $E[X(t)]=E[X(t-1)]+1 $ you can easily get the desired result.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! Could you elaborate little bit more?
    $endgroup$
    – user3509199
    Dec 9 '18 at 3:47










  • $begingroup$
    $E[X(t)] = E[X(t-2)]+2=E[X(t-3)]+3=...=E[X(0)]+n
    $endgroup$
    – mineiro
    Dec 9 '18 at 3:50
















2












$begingroup$

It’s simple induction. From $E[X(t)]=E[X(t-1)]+1 $ you can easily get the desired result.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! Could you elaborate little bit more?
    $endgroup$
    – user3509199
    Dec 9 '18 at 3:47










  • $begingroup$
    $E[X(t)] = E[X(t-2)]+2=E[X(t-3)]+3=...=E[X(0)]+n
    $endgroup$
    – mineiro
    Dec 9 '18 at 3:50














2












2








2





$begingroup$

It’s simple induction. From $E[X(t)]=E[X(t-1)]+1 $ you can easily get the desired result.






share|cite|improve this answer









$endgroup$



It’s simple induction. From $E[X(t)]=E[X(t-1)]+1 $ you can easily get the desired result.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 '18 at 3:35









mineiromineiro

813




813












  • $begingroup$
    Thanks! Could you elaborate little bit more?
    $endgroup$
    – user3509199
    Dec 9 '18 at 3:47










  • $begingroup$
    $E[X(t)] = E[X(t-2)]+2=E[X(t-3)]+3=...=E[X(0)]+n
    $endgroup$
    – mineiro
    Dec 9 '18 at 3:50


















  • $begingroup$
    Thanks! Could you elaborate little bit more?
    $endgroup$
    – user3509199
    Dec 9 '18 at 3:47










  • $begingroup$
    $E[X(t)] = E[X(t-2)]+2=E[X(t-3)]+3=...=E[X(0)]+n
    $endgroup$
    – mineiro
    Dec 9 '18 at 3:50
















$begingroup$
Thanks! Could you elaborate little bit more?
$endgroup$
– user3509199
Dec 9 '18 at 3:47




$begingroup$
Thanks! Could you elaborate little bit more?
$endgroup$
– user3509199
Dec 9 '18 at 3:47












$begingroup$
$E[X(t)] = E[X(t-2)]+2=E[X(t-3)]+3=...=E[X(0)]+n
$endgroup$
– mineiro
Dec 9 '18 at 3:50




$begingroup$
$E[X(t)] = E[X(t-2)]+2=E[X(t-3)]+3=...=E[X(0)]+n
$endgroup$
– mineiro
Dec 9 '18 at 3:50


















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