Correlation of independent random processesA special case of 2 jointly Weak-Sense Stationary (WSS) stochastic processesexplanation of correlation of stationary stochastic processesUnderstanding the definition of mean/autocorrelationAutocorrelation function $R_yy(t_1,t_2)$?Is the output of function of two ergodic processes ergodic?Autocorrelation of Addition of Two Independent SignalsAre two jointly stationary white noise processes independent?Physical interpretation of 4th-order correlationsWhat is definition of independent random variableIs the expectation of a random process $X(t)$ with zero DC component necessarily zero?
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Correlation of independent random processes
A special case of 2 jointly Weak-Sense Stationary (WSS) stochastic processesexplanation of correlation of stationary stochastic processesUnderstanding the definition of mean/autocorrelationAutocorrelation function $R_yy(t_1,t_2)$?Is the output of function of two ergodic processes ergodic?Autocorrelation of Addition of Two Independent SignalsAre two jointly stationary white noise processes independent?Physical interpretation of 4th-order correlationsWhat is definition of independent random variableIs the expectation of a random process $X(t)$ with zero DC component necessarily zero?
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Suppose $X(t)$ and $Y(t)$ be two independent random processes. Is $E(X(t_1)Y(t_2))$ necessarily zero?
autocorrelation cross-correlation correlation random-process random
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Suppose $X(t)$ and $Y(t)$ be two independent random processes. Is $E(X(t_1)Y(t_2))$ necessarily zero?
autocorrelation cross-correlation correlation random-process random
edited 1 hour ago
Dilip Sarwate
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asked 11 hours ago
helloworld1e.helloworld1e.
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1
$begingroup$
No. $E<X cdot Y>$ is simply $E<X> cdot E<Y>$
$endgroup$
– Hilmar
11 hours ago
add a comment |
$begingroup$
Suppose $X(t)$ and $Y(t)$ be two independent random processes. Is $E(X(t_1)Y(t_2))$ necessarily zero?
autocorrelation cross-correlation correlation random-process random
edited 1 hour ago
Dilip Sarwate
14.1k1 gold badge26 silver badges63 bronze badges
asked 11 hours ago
helloworld1e.helloworld1e.
102 bronze badges
New contributor
helloworld1e. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Suppose $X(t)$ and $Y(t)$ be two independent random processes. Is $E(X(t_1)Y(t_2))$ necessarily zero?
autocorrelation cross-correlation correlation random-process random
autocorrelation cross-correlation correlation random-process random
edited 1 hour ago
Dilip Sarwate
14.1k1 gold badge26 silver badges63 bronze badges
asked 11 hours ago
helloworld1e.helloworld1e.
102 bronze badges
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edited 1 hour ago
Dilip Sarwate
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asked 11 hours ago
helloworld1e.helloworld1e.
102 bronze badges
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edited 1 hour ago
Dilip Sarwate
14.1k1 gold badge26 silver badges63 bronze badges
edited 1 hour ago
Dilip Sarwate
14.1k1 gold badge26 silver badges63 bronze badges
edited 1 hour ago
Dilip Sarwate
14.1k1 gold badge26 silver badges63 bronze badges
14.1k1 gold badge26 silver badges63 bronze badges
asked 11 hours ago
helloworld1e.helloworld1e.
102 bronze badges
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asked 11 hours ago
helloworld1e.helloworld1e.
102 bronze badges
asked 11 hours ago
helloworld1e.helloworld1e.
102 bronze badges
102 bronze badges
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helloworld1e. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
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1
$begingroup$
No. $E<X cdot Y>$ is simply $E<X> cdot E<Y>$
$endgroup$
– Hilmar
11 hours ago
add a comment |
1
$begingroup$
No. $E<X cdot Y>$ is simply $E<X> cdot E<Y>$
$endgroup$
– Hilmar
11 hours ago
1
1
$begingroup$
No. $E<X cdot Y>$ is simply $E<X> cdot E<Y>$
$endgroup$
– Hilmar
11 hours ago
$begingroup$
No. $E<X cdot Y>$ is simply $E<X> cdot E<Y>$
$endgroup$
– Hilmar
11 hours ago
add a comment |
1 Answer
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No. Quoting Wikipedia's article Independence (probability theory):
If $X$ and $Y$ are independent random variables,
then the expectation operator $operatornameE$ has the property
$$operatornameE[X Y] = operatornameE[X]operatornameE[Y].$$
Consider your $X(t_1)$ and $Y(t_2)$ as $X$ and $Y$ in this answer. If both $operatornameE[X] ne 0$ and $operatornameE[X] ne 0,$ then the product $operatornameE[X]operatornameE[Y]$ will be non-zero.
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$begingroup$
No. Quoting Wikipedia's article Independence (probability theory):
If $X$ and $Y$ are independent random variables,
then the expectation operator $operatornameE$ has the property
$$operatornameE[X Y] = operatornameE[X]operatornameE[Y].$$
Consider your $X(t_1)$ and $Y(t_2)$ as $X$ and $Y$ in this answer. If both $operatornameE[X] ne 0$ and $operatornameE[X] ne 0,$ then the product $operatornameE[X]operatornameE[Y]$ will be non-zero.
$endgroup$
add a comment |
$begingroup$
No. Quoting Wikipedia's article Independence (probability theory):
If $X$ and $Y$ are independent random variables,
then the expectation operator $operatornameE$ has the property
$$operatornameE[X Y] = operatornameE[X]operatornameE[Y].$$
Consider your $X(t_1)$ and $Y(t_2)$ as $X$ and $Y$ in this answer. If both $operatornameE[X] ne 0$ and $operatornameE[X] ne 0,$ then the product $operatornameE[X]operatornameE[Y]$ will be non-zero.
$endgroup$
add a comment |
$begingroup$
No. Quoting Wikipedia's article Independence (probability theory):
If $X$ and $Y$ are independent random variables,
then the expectation operator $operatornameE$ has the property
$$operatornameE[X Y] = operatornameE[X]operatornameE[Y].$$
Consider your $X(t_1)$ and $Y(t_2)$ as $X$ and $Y$ in this answer. If both $operatornameE[X] ne 0$ and $operatornameE[X] ne 0,$ then the product $operatornameE[X]operatornameE[Y]$ will be non-zero.
$endgroup$
No. Quoting Wikipedia's article Independence (probability theory):
If $X$ and $Y$ are independent random variables,
then the expectation operator $operatornameE$ has the property
$$operatornameE[X Y] = operatornameE[X]operatornameE[Y].$$
Consider your $X(t_1)$ and $Y(t_2)$ as $X$ and $Y$ in this answer. If both $operatornameE[X] ne 0$ and $operatornameE[X] ne 0,$ then the product $operatornameE[X]operatornameE[Y]$ will be non-zero.
edited 11 hours ago
community wiki
2 revs, 2 users 95%
Olli Niemitalo
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$begingroup$
No. $E<X cdot Y>$ is simply $E<X> cdot E<Y>$
$endgroup$
– Hilmar
11 hours ago