mean and covariance of a random process
I'm looking at an example from a book I'm reading,
How does it formulate the mean?
I am thinking $$E[X[n]] = sumlimits^infty_{-infty} X[n]P[X[n]] =\
sumlimits^infty_{n=-infty} X[n = even] frac{1}{2} + X[n = even] frac{1}{2} + X[n = odd] frac{9}{10} + X[n = odd] frac{1}{10} \
= sumlimits^infty_{-infty} frac{1}{2}-frac{1}{2}+frac{1}{3}frac{9}{10}-3frac{1}{10} = 0$$
Does this formulation look right? I'm confused when X[n] can be probabilistically assigned values.
random-variables random-walk
asked Nov 19 '18 at 6:01
drerD
1519
|
show 2 more comments
I'm looking at an example from a book I'm reading,
How does it formulate the mean?
I am thinking $$E[X[n]] = sumlimits^infty_{-infty} X[n]P[X[n]] =\
sumlimits^infty_{n=-infty} X[n = even] frac{1}{2} + X[n = even] frac{1}{2} + X[n = odd] frac{9}{10} + X[n = odd] frac{1}{10} \
= sumlimits^infty_{-infty} frac{1}{2}-frac{1}{2}+frac{1}{3}frac{9}{10}-3frac{1}{10} = 0$$
Does this formulation look right? I'm confused when X[n] can be probabilistically assigned values.
random-variables random-walk
asked Nov 19 '18 at 6:01
drerD
1519
What does $X[n]$ mean? Did you mean $X_n$?
– Will M.
Nov 19 '18 at 6:08
@WillM. Yes, I think that makes it easier to write.
– drerD
Nov 19 '18 at 6:10
1
The formulae you wrote does not make sense to me. $mathbf{E}(X_n)$ is clearly zero for $n$ even and it is $dfrac{1}{3} times dfrac{9}{10} - 3 times dfrac{1}{10} = 0$ for $n$ odd.
– Will M.
Nov 19 '18 at 6:13
It seems like you are thinking of $X$ as the random variable. That is not correct, $X_n$ is a random variable, $X$ is a "random sequence."
– Will M.
Nov 19 '18 at 6:14
1
Ok, stop using $X[k]$ because that IS confusing. For a random sequence $X$ we can define a function $m_X$ such that $m_X(n) = mathbf{E}(X_n).$
– Will M.
Nov 19 '18 at 6:24
|
show 2 more comments
I'm looking at an example from a book I'm reading,
How does it formulate the mean?
I am thinking $$E[X[n]] = sumlimits^infty_{-infty} X[n]P[X[n]] =\
sumlimits^infty_{n=-infty} X[n = even] frac{1}{2} + X[n = even] frac{1}{2} + X[n = odd] frac{9}{10} + X[n = odd] frac{1}{10} \
= sumlimits^infty_{-infty} frac{1}{2}-frac{1}{2}+frac{1}{3}frac{9}{10}-3frac{1}{10} = 0$$
Does this formulation look right? I'm confused when X[n] can be probabilistically assigned values.
random-variables random-walk
asked Nov 19 '18 at 6:01
drerD
1519
I'm looking at an example from a book I'm reading,
How does it formulate the mean?
I am thinking $$E[X[n]] = sumlimits^infty_{-infty} X[n]P[X[n]] =\
sumlimits^infty_{n=-infty} X[n = even] frac{1}{2} + X[n = even] frac{1}{2} + X[n = odd] frac{9}{10} + X[n = odd] frac{1}{10} \
= sumlimits^infty_{-infty} frac{1}{2}-frac{1}{2}+frac{1}{3}frac{9}{10}-3frac{1}{10} = 0$$
Does this formulation look right? I'm confused when X[n] can be probabilistically assigned values.
random-variables random-walk
random-variables random-walk
asked Nov 19 '18 at 6:01
drerD
1519
asked Nov 19 '18 at 6:01
drerD
1519
asked Nov 19 '18 at 6:01
drerD
1519
asked Nov 19 '18 at 6:01
drerD
1519
asked Nov 19 '18 at 6:01
drerD
1519
1519
What does $X[n]$ mean? Did you mean $X_n$?
– Will M.
Nov 19 '18 at 6:08
@WillM. Yes, I think that makes it easier to write.
– drerD
Nov 19 '18 at 6:10
1
The formulae you wrote does not make sense to me. $mathbf{E}(X_n)$ is clearly zero for $n$ even and it is $dfrac{1}{3} times dfrac{9}{10} - 3 times dfrac{1}{10} = 0$ for $n$ odd.
– Will M.
Nov 19 '18 at 6:13
It seems like you are thinking of $X$ as the random variable. That is not correct, $X_n$ is a random variable, $X$ is a "random sequence."
– Will M.
Nov 19 '18 at 6:14
1
Ok, stop using $X[k]$ because that IS confusing. For a random sequence $X$ we can define a function $m_X$ such that $m_X(n) = mathbf{E}(X_n).$
– Will M.
Nov 19 '18 at 6:24
|
show 2 more comments
What does $X[n]$ mean? Did you mean $X_n$?
– Will M.
Nov 19 '18 at 6:08
@WillM. Yes, I think that makes it easier to write.
– drerD
Nov 19 '18 at 6:10
1
The formulae you wrote does not make sense to me. $mathbf{E}(X_n)$ is clearly zero for $n$ even and it is $dfrac{1}{3} times dfrac{9}{10} - 3 times dfrac{1}{10} = 0$ for $n$ odd.
– Will M.
Nov 19 '18 at 6:13
It seems like you are thinking of $X$ as the random variable. That is not correct, $X_n$ is a random variable, $X$ is a "random sequence."
– Will M.
Nov 19 '18 at 6:14
1
Ok, stop using $X[k]$ because that IS confusing. For a random sequence $X$ we can define a function $m_X$ such that $m_X(n) = mathbf{E}(X_n).$
– Will M.
Nov 19 '18 at 6:24
What does $X[n]$ mean? Did you mean $X_n$?
– Will M.
Nov 19 '18 at 6:08
What does $X[n]$ mean? Did you mean $X_n$?
– Will M.
Nov 19 '18 at 6:08
@WillM. Yes, I think that makes it easier to write.
– drerD
Nov 19 '18 at 6:10
@WillM. Yes, I think that makes it easier to write.
– drerD
Nov 19 '18 at 6:10
1
1
The formulae you wrote does not make sense to me. $mathbf{E}(X_n)$ is clearly zero for $n$ even and it is $dfrac{1}{3} times dfrac{9}{10} - 3 times dfrac{1}{10} = 0$ for $n$ odd.
– Will M.
Nov 19 '18 at 6:13
The formulae you wrote does not make sense to me. $mathbf{E}(X_n)$ is clearly zero for $n$ even and it is $dfrac{1}{3} times dfrac{9}{10} - 3 times dfrac{1}{10} = 0$ for $n$ odd.
– Will M.
Nov 19 '18 at 6:13
It seems like you are thinking of $X$ as the random variable. That is not correct, $X_n$ is a random variable, $X$ is a "random sequence."
– Will M.
Nov 19 '18 at 6:14
It seems like you are thinking of $X$ as the random variable. That is not correct, $X_n$ is a random variable, $X$ is a "random sequence."
– Will M.
Nov 19 '18 at 6:14
1
1
Ok, stop using $X[k]$ because that IS confusing. For a random sequence $X$ we can define a function $m_X$ such that $m_X(n) = mathbf{E}(X_n).$
– Will M.
Nov 19 '18 at 6:24
Ok, stop using $X[k]$ because that IS confusing. For a random sequence $X$ we can define a function $m_X$ such that $m_X(n) = mathbf{E}(X_n).$
– Will M.
Nov 19 '18 at 6:24
|
show 2 more comments
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What does $X[n]$ mean? Did you mean $X_n$?
– Will M.
Nov 19 '18 at 6:08
@WillM. Yes, I think that makes it easier to write.
– drerD
Nov 19 '18 at 6:10
1
The formulae you wrote does not make sense to me. $mathbf{E}(X_n)$ is clearly zero for $n$ even and it is $dfrac{1}{3} times dfrac{9}{10} - 3 times dfrac{1}{10} = 0$ for $n$ odd.
– Will M.
Nov 19 '18 at 6:13
It seems like you are thinking of $X$ as the random variable. That is not correct, $X_n$ is a random variable, $X$ is a "random sequence."
– Will M.
Nov 19 '18 at 6:14
1
Ok, stop using $X[k]$ because that IS confusing. For a random sequence $X$ we can define a function $m_X$ such that $m_X(n) = mathbf{E}(X_n).$
– Will M.
Nov 19 '18 at 6:24