Expectation of the stochastic process
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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
asked Dec 9 '18 at 3:11
user3509199user3509199
316
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add a comment |
$begingroup$
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
asked Dec 9 '18 at 3:11
user3509199user3509199
316
$endgroup$
$begingroup$
Is this a discrete time process? You might be able to use induction.
$endgroup$
– Sean Roberson
Dec 9 '18 at 3:33
add a comment |
$begingroup$
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
asked Dec 9 '18 at 3:11
user3509199user3509199
316
$endgroup$
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
asked Dec 9 '18 at 3:11
user3509199user3509199
316
asked Dec 9 '18 at 3:11
user3509199user3509199
316
asked Dec 9 '18 at 3:11
user3509199user3509199
316
asked Dec 9 '18 at 3:11
user3509199user3509199
316
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
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oldest
votes
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It’s simple induction. From $E[X(t)]=E[X(t-1)]+1 $ you can easily get the desired result.
answered Dec 9 '18 at 3:35
mineiromineiro
813
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$begingroup$
Thanks! Could you elaborate little bit more?
$endgroup$
– user3509199
Dec 9 '18 at 3:47
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$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
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It’s simple induction. From $E[X(t)]=E[X(t-1)]+1 $ you can easily get the desired result.
answered Dec 9 '18 at 3:35
mineiromineiro
813
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
It’s simple induction. From $E[X(t)]=E[X(t-1)]+1 $ you can easily get the desired result.
answered Dec 9 '18 at 3:35
mineiromineiro
813
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
It’s simple induction. From $E[X(t)]=E[X(t-1)]+1 $ you can easily get the desired result.
answered Dec 9 '18 at 3:35
mineiromineiro
813
$endgroup$
It’s simple induction. From $E[X(t)]=E[X(t-1)]+1 $ you can easily get the desired result.
answered Dec 9 '18 at 3:35
mineiromineiro
813
answered Dec 9 '18 at 3:35
mineiromineiro
813
answered Dec 9 '18 at 3:35
mineiromineiro
813
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
Is this a discrete time process? You might be able to use induction.
$endgroup$
– Sean Roberson
Dec 9 '18 at 3:33