Question for covariance stationary process
$begingroup$
Given a random variable Y with characteristic function
C(w) = E[exp(iwy)] . Let the random process X(t) be defined as X(t)=cos(wt+y). Show that the process X(t) is covariance stationary if C(1)-C(2)=0.
I have tried expanding the characteristic function in the trigonometric form and using the result to prove stationarity but I'm still unable to solve it. Any hint will be helpful.
stochastic-processes covariance stationary-processes
asked Jan 12 at 18:57
Kc17Kc17
346
$endgroup$
add a comment |
$begingroup$
Given a random variable Y with characteristic function
C(w) = E[exp(iwy)] . Let the random process X(t) be defined as X(t)=cos(wt+y). Show that the process X(t) is covariance stationary if C(1)-C(2)=0.
I have tried expanding the characteristic function in the trigonometric form and using the result to prove stationarity but I'm still unable to solve it. Any hint will be helpful.
stochastic-processes covariance stationary-processes
asked Jan 12 at 18:57
Kc17Kc17
346
$endgroup$
add a comment |
$begingroup$
Given a random variable Y with characteristic function
C(w) = E[exp(iwy)] . Let the random process X(t) be defined as X(t)=cos(wt+y). Show that the process X(t) is covariance stationary if C(1)-C(2)=0.
I have tried expanding the characteristic function in the trigonometric form and using the result to prove stationarity but I'm still unable to solve it. Any hint will be helpful.
stochastic-processes covariance stationary-processes
asked Jan 12 at 18:57
Kc17Kc17
346
$endgroup$
Given a random variable Y with characteristic function
C(w) = E[exp(iwy)] . Let the random process X(t) be defined as X(t)=cos(wt+y). Show that the process X(t) is covariance stationary if C(1)-C(2)=0.
I have tried expanding the characteristic function in the trigonometric form and using the result to prove stationarity but I'm still unable to solve it. Any hint will be helpful.
stochastic-processes covariance stationary-processes
stochastic-processes covariance stationary-processes
asked Jan 12 at 18:57
Kc17Kc17
346
asked Jan 12 at 18:57
Kc17Kc17
346
asked Jan 12 at 18:57
Kc17Kc17
346
asked Jan 12 at 18:57
Kc17Kc17
346
asked Jan 12 at 18:57
Kc17Kc17
346
346
add a comment |
add a comment |
1 Answer
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$begingroup$
Hint
$$E[X(t)]{=E[cos(wt+y)]\=Eleft[{e^{iwt+iy}+e^{-iwt-iy}over 2}right]\=text{const}}$$and$$E[X(t_1)X(t_2)]{=E[cos(wt_2+y)cos(wt_1+y)]\={1over 2}E[cos(w(t_2-t_1))]\+{1over2}E[cos (wt_1+wt_2+2y)]}$$since $E[X(t_1)X(t_2)]$ must be a function of $t_1-t_2$ so $E[cos (wt_1+wt_2+2y)]=0$. The rest is easy.
answered Jan 12 at 19:07
Mostafa AyazMostafa Ayaz
15.3k3939
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1 Answer
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active
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1 Answer
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oldest
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$begingroup$
Hint
$$E[X(t)]{=E[cos(wt+y)]\=Eleft[{e^{iwt+iy}+e^{-iwt-iy}over 2}right]\=text{const}}$$and$$E[X(t_1)X(t_2)]{=E[cos(wt_2+y)cos(wt_1+y)]\={1over 2}E[cos(w(t_2-t_1))]\+{1over2}E[cos (wt_1+wt_2+2y)]}$$since $E[X(t_1)X(t_2)]$ must be a function of $t_1-t_2$ so $E[cos (wt_1+wt_2+2y)]=0$. The rest is easy.
answered Jan 12 at 19:07
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
add a comment |
$begingroup$
Hint
$$E[X(t)]{=E[cos(wt+y)]\=Eleft[{e^{iwt+iy}+e^{-iwt-iy}over 2}right]\=text{const}}$$and$$E[X(t_1)X(t_2)]{=E[cos(wt_2+y)cos(wt_1+y)]\={1over 2}E[cos(w(t_2-t_1))]\+{1over2}E[cos (wt_1+wt_2+2y)]}$$since $E[X(t_1)X(t_2)]$ must be a function of $t_1-t_2$ so $E[cos (wt_1+wt_2+2y)]=0$. The rest is easy.
answered Jan 12 at 19:07
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
add a comment |
$begingroup$
Hint
$$E[X(t)]{=E[cos(wt+y)]\=Eleft[{e^{iwt+iy}+e^{-iwt-iy}over 2}right]\=text{const}}$$and$$E[X(t_1)X(t_2)]{=E[cos(wt_2+y)cos(wt_1+y)]\={1over 2}E[cos(w(t_2-t_1))]\+{1over2}E[cos (wt_1+wt_2+2y)]}$$since $E[X(t_1)X(t_2)]$ must be a function of $t_1-t_2$ so $E[cos (wt_1+wt_2+2y)]=0$. The rest is easy.
answered Jan 12 at 19:07
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
Hint
$$E[X(t)]{=E[cos(wt+y)]\=Eleft[{e^{iwt+iy}+e^{-iwt-iy}over 2}right]\=text{const}}$$and$$E[X(t_1)X(t_2)]{=E[cos(wt_2+y)cos(wt_1+y)]\={1over 2}E[cos(w(t_2-t_1))]\+{1over2}E[cos (wt_1+wt_2+2y)]}$$since $E[X(t_1)X(t_2)]$ must be a function of $t_1-t_2$ so $E[cos (wt_1+wt_2+2y)]=0$. The rest is easy.
answered Jan 12 at 19:07
Mostafa AyazMostafa Ayaz
15.3k3939
answered Jan 12 at 19:07
Mostafa AyazMostafa Ayaz
15.3k3939
answered Jan 12 at 19:07
Mostafa AyazMostafa Ayaz
15.3k3939
answered Jan 12 at 19:07
Mostafa AyazMostafa Ayaz
15.3k3939
15.3k3939
add a comment |
add a comment |
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