Help calculating second order distribution for a continuous stochastic process
I'm really stuck on this problem, and appreciate any help.
Problem: Let $gamma$ be a random number, chosen with uniform probability in the interval $[0,2pi]$. We define the stochastic process $X(t)=cos(t+gamma)$. Calculate the first and second order distribution.
My attempt at an answer:
I want to calculate the first and second order distributions, which are given, respectively, by:
$P(X(t_{1})<x_{1})$
and
$P(X(t_{1})<x_{1},X(t_{2})<x_{2})$
1st order distribution calculation
I can calculate the first order distribution easily (at least I think this is correct):
$P(X(t_{1})<x_{1})=P(cos(t_{1}+gamma)<x_{1})=P(t_{1}+gamma<cos^{-1}(x_{1}))=P(gamma<cos^{-1}(x_{1})-t_{1})$
Since $gamma$ is uniformly distributed I know that :
$P(gamma<y)=frac{y}{2pi}$
Thus:
$P(X(t_{1})<x_{1})=P(gamma<cos^{-1}(x_{1})-t_{1})=frac{cos^{-1}(x_{1})-t_{1}}{2pi}$
(that is if $-1>x_{1}>1$ )
2nd order distribution calculation
I am having problems calculating the second order distribution, this is what I have done up till now:
$P(X(t_{1})<x_{1},X(t_{2}<x_2{}))=P(X(t_{2})<x_{2}|X(t_{1})<x_{1})P(X(t_{1})<x_{1})$
The second factor is the first order distribution, which I have already calculated. But how do I get $P(X(t_{2})<x_{2}|X(t_{1})<x_{1})$? Clearly if $X(t_{1})$ is already defined (which imples $gamma$ is fixed) $X(t_{2})$ is already defined, but don't see how to reach this answer.
Thank you for reading!
statistics random
asked Nov 25 '18 at 23:16
Aye Prado
215
add a comment |
I'm really stuck on this problem, and appreciate any help.
Problem: Let $gamma$ be a random number, chosen with uniform probability in the interval $[0,2pi]$. We define the stochastic process $X(t)=cos(t+gamma)$. Calculate the first and second order distribution.
My attempt at an answer:
I want to calculate the first and second order distributions, which are given, respectively, by:
$P(X(t_{1})<x_{1})$
and
$P(X(t_{1})<x_{1},X(t_{2})<x_{2})$
1st order distribution calculation
I can calculate the first order distribution easily (at least I think this is correct):
$P(X(t_{1})<x_{1})=P(cos(t_{1}+gamma)<x_{1})=P(t_{1}+gamma<cos^{-1}(x_{1}))=P(gamma<cos^{-1}(x_{1})-t_{1})$
Since $gamma$ is uniformly distributed I know that :
$P(gamma<y)=frac{y}{2pi}$
Thus:
$P(X(t_{1})<x_{1})=P(gamma<cos^{-1}(x_{1})-t_{1})=frac{cos^{-1}(x_{1})-t_{1}}{2pi}$
(that is if $-1>x_{1}>1$ )
2nd order distribution calculation
I am having problems calculating the second order distribution, this is what I have done up till now:
$P(X(t_{1})<x_{1},X(t_{2}<x_2{}))=P(X(t_{2})<x_{2}|X(t_{1})<x_{1})P(X(t_{1})<x_{1})$
The second factor is the first order distribution, which I have already calculated. But how do I get $P(X(t_{2})<x_{2}|X(t_{1})<x_{1})$? Clearly if $X(t_{1})$ is already defined (which imples $gamma$ is fixed) $X(t_{2})$ is already defined, but don't see how to reach this answer.
Thank you for reading!
statistics random
asked Nov 25 '18 at 23:16
Aye Prado
215
add a comment |
I'm really stuck on this problem, and appreciate any help.
Problem: Let $gamma$ be a random number, chosen with uniform probability in the interval $[0,2pi]$. We define the stochastic process $X(t)=cos(t+gamma)$. Calculate the first and second order distribution.
My attempt at an answer:
I want to calculate the first and second order distributions, which are given, respectively, by:
$P(X(t_{1})<x_{1})$
and
$P(X(t_{1})<x_{1},X(t_{2})<x_{2})$
1st order distribution calculation
I can calculate the first order distribution easily (at least I think this is correct):
$P(X(t_{1})<x_{1})=P(cos(t_{1}+gamma)<x_{1})=P(t_{1}+gamma<cos^{-1}(x_{1}))=P(gamma<cos^{-1}(x_{1})-t_{1})$
Since $gamma$ is uniformly distributed I know that :
$P(gamma<y)=frac{y}{2pi}$
Thus:
$P(X(t_{1})<x_{1})=P(gamma<cos^{-1}(x_{1})-t_{1})=frac{cos^{-1}(x_{1})-t_{1}}{2pi}$
(that is if $-1>x_{1}>1$ )
2nd order distribution calculation
I am having problems calculating the second order distribution, this is what I have done up till now:
$P(X(t_{1})<x_{1},X(t_{2}<x_2{}))=P(X(t_{2})<x_{2}|X(t_{1})<x_{1})P(X(t_{1})<x_{1})$
The second factor is the first order distribution, which I have already calculated. But how do I get $P(X(t_{2})<x_{2}|X(t_{1})<x_{1})$? Clearly if $X(t_{1})$ is already defined (which imples $gamma$ is fixed) $X(t_{2})$ is already defined, but don't see how to reach this answer.
Thank you for reading!
statistics random
asked Nov 25 '18 at 23:16
Aye Prado
215
I'm really stuck on this problem, and appreciate any help.
Problem: Let $gamma$ be a random number, chosen with uniform probability in the interval $[0,2pi]$. We define the stochastic process $X(t)=cos(t+gamma)$. Calculate the first and second order distribution.
My attempt at an answer:
I want to calculate the first and second order distributions, which are given, respectively, by:
$P(X(t_{1})<x_{1})$
and
$P(X(t_{1})<x_{1},X(t_{2})<x_{2})$
1st order distribution calculation
I can calculate the first order distribution easily (at least I think this is correct):
$P(X(t_{1})<x_{1})=P(cos(t_{1}+gamma)<x_{1})=P(t_{1}+gamma<cos^{-1}(x_{1}))=P(gamma<cos^{-1}(x_{1})-t_{1})$
Since $gamma$ is uniformly distributed I know that :
$P(gamma<y)=frac{y}{2pi}$
Thus:
$P(X(t_{1})<x_{1})=P(gamma<cos^{-1}(x_{1})-t_{1})=frac{cos^{-1}(x_{1})-t_{1}}{2pi}$
(that is if $-1>x_{1}>1$ )
2nd order distribution calculation
I am having problems calculating the second order distribution, this is what I have done up till now:
$P(X(t_{1})<x_{1},X(t_{2}<x_2{}))=P(X(t_{2})<x_{2}|X(t_{1})<x_{1})P(X(t_{1})<x_{1})$
The second factor is the first order distribution, which I have already calculated. But how do I get $P(X(t_{2})<x_{2}|X(t_{1})<x_{1})$? Clearly if $X(t_{1})$ is already defined (which imples $gamma$ is fixed) $X(t_{2})$ is already defined, but don't see how to reach this answer.
Thank you for reading!
statistics random
statistics random
asked Nov 25 '18 at 23:16
Aye Prado
215
asked Nov 25 '18 at 23:16
Aye Prado
215
asked Nov 25 '18 at 23:16
Aye Prado
215
asked Nov 25 '18 at 23:16
Aye Prado
215
asked Nov 25 '18 at 23:16
Aye Prado
215
215
add a comment |
add a comment |
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