Mixing
Mean and Variance of dot product of two normalized random vector
0
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I have two N-dimensions normalized vectors, X and Y; X ~ N($mu$,$Sigma$), $mu$ is an N-1 vector and $Sigma$ is the covariance due to X is normalized. Y is a const, what is the mean and variance of dot(X, Y)? Could you help me give an expression about $mu$, $Sigma$ and Y?
linear-algebra probability statistics
asked Jan 24 at 9:10
user6897059user6897059
11
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show 2 more comments
0
$begingroup$
I have two N-dimensions normalized vectors, X and Y; X ~ N($mu$,$Sigma$), $mu$ is an N-1 vector and $Sigma$ is the covariance due to X is normalized. Y is a const, what is the mean and variance of dot(X, Y)? Could you help me give an expression about $mu$, $Sigma$ and Y?
linear-algebra probability statistics
asked Jan 24 at 9:10
user6897059user6897059
11
$endgroup$
$begingroup$
I'm sorry for mistaking you. The X is just an N-dimension vector, for example, N=3, the X is a unit sphere vector; we can drop the little i;
$endgroup$
– user6897059
Jan 24 at 9:20
$begingroup$
$mathsf{dot}(X,Y)$ is the same as $Y^TX$ and if $Y$ is constant then $mathbb EY^TX=Y^Tmathbb EX$ and $mathsf{Cov}(Y^TX)=Y^Tmathsf{Cov(X)}Y$. Does that help here?
$endgroup$
– drhab
Jan 24 at 9:24
$begingroup$
Thank you for your reply, please give me some time to comprehend this.
$endgroup$
– user6897059
Jan 24 at 9:26
$begingroup$
but X is normalized, will this not influence the result? @drhab
$endgroup$
– user6897059
Jan 24 at 9:30
$begingroup$
Unfortunately I am not familiar enough with the concept "normalized" to judge that. That is why I a gave a comment and not an answer. Personally I cannot find a reason why what I wrote in my former comment would not work.
$endgroup$
– drhab
Jan 24 at 9:35
|
show 2 more comments
0
0
0
$begingroup$
I have two N-dimensions normalized vectors, X and Y; X ~ N($mu$,$Sigma$), $mu$ is an N-1 vector and $Sigma$ is the covariance due to X is normalized. Y is a const, what is the mean and variance of dot(X, Y)? Could you help me give an expression about $mu$, $Sigma$ and Y?
linear-algebra probability statistics
asked Jan 24 at 9:10
user6897059user6897059
11
$endgroup$
I have two N-dimensions normalized vectors, X and Y; X ~ N($mu$,$Sigma$), $mu$ is an N-1 vector and $Sigma$ is the covariance due to X is normalized. Y is a const, what is the mean and variance of dot(X, Y)? Could you help me give an expression about $mu$, $Sigma$ and Y?
linear-algebra probability statistics
linear-algebra probability statistics
asked Jan 24 at 9:10
user6897059user6897059
11
asked Jan 24 at 9:10
user6897059user6897059
11
edited Jan 24 at 9:17
user6897059
asked Jan 24 at 9:10
user6897059user6897059
11
asked Jan 24 at 9:10
user6897059user6897059
11
asked Jan 24 at 9:10
user6897059user6897059
11
11
$begingroup$
I'm sorry for mistaking you. The X is just an N-dimension vector, for example, N=3, the X is a unit sphere vector; we can drop the little i;
$endgroup$
– user6897059
Jan 24 at 9:20
$begingroup$
$mathsf{dot}(X,Y)$ is the same as $Y^TX$ and if $Y$ is constant then $mathbb EY^TX=Y^Tmathbb EX$ and $mathsf{Cov}(Y^TX)=Y^Tmathsf{Cov(X)}Y$. Does that help here?
$endgroup$
– drhab
Jan 24 at 9:24
$begingroup$
Thank you for your reply, please give me some time to comprehend this.
$endgroup$
– user6897059
Jan 24 at 9:26
$begingroup$
but X is normalized, will this not influence the result? @drhab
$endgroup$
– user6897059
Jan 24 at 9:30
$begingroup$
Unfortunately I am not familiar enough with the concept "normalized" to judge that. That is why I a gave a comment and not an answer. Personally I cannot find a reason why what I wrote in my former comment would not work.
$endgroup$
– drhab
Jan 24 at 9:35
|
show 2 more comments
$begingroup$
I'm sorry for mistaking you. The X is just an N-dimension vector, for example, N=3, the X is a unit sphere vector; we can drop the little i;
$endgroup$
– user6897059
Jan 24 at 9:20
$begingroup$
$mathsf{dot}(X,Y)$ is the same as $Y^TX$ and if $Y$ is constant then $mathbb EY^TX=Y^Tmathbb EX$ and $mathsf{Cov}(Y^TX)=Y^Tmathsf{Cov(X)}Y$. Does that help here?
$endgroup$
– drhab
Jan 24 at 9:24
$begingroup$
Thank you for your reply, please give me some time to comprehend this.
$endgroup$
– user6897059
Jan 24 at 9:26
$begingroup$
but X is normalized, will this not influence the result? @drhab
$endgroup$
– user6897059
Jan 24 at 9:30
$begingroup$
Unfortunately I am not familiar enough with the concept "normalized" to judge that. That is why I a gave a comment and not an answer. Personally I cannot find a reason why what I wrote in my former comment would not work.
$endgroup$
– drhab
Jan 24 at 9:35
$begingroup$
I'm sorry for mistaking you. The X is just an N-dimension vector, for example, N=3, the X is a unit sphere vector; we can drop the little i;
$endgroup$
– user6897059
Jan 24 at 9:20
$begingroup$
I'm sorry for mistaking you. The X is just an N-dimension vector, for example, N=3, the X is a unit sphere vector; we can drop the little i;
$endgroup$
– user6897059
Jan 24 at 9:20
$begingroup$
$mathsf{dot}(X,Y)$ is the same as $Y^TX$ and if $Y$ is constant then $mathbb EY^TX=Y^Tmathbb EX$ and $mathsf{Cov}(Y^TX)=Y^Tmathsf{Cov(X)}Y$. Does that help here?
$endgroup$
– drhab
Jan 24 at 9:24
$begingroup$
$mathsf{dot}(X,Y)$ is the same as $Y^TX$ and if $Y$ is constant then $mathbb EY^TX=Y^Tmathbb EX$ and $mathsf{Cov}(Y^TX)=Y^Tmathsf{Cov(X)}Y$. Does that help here?
$endgroup$
– drhab
Jan 24 at 9:24
$begingroup$
Thank you for your reply, please give me some time to comprehend this.
$endgroup$
– user6897059
Jan 24 at 9:26
$begingroup$
Thank you for your reply, please give me some time to comprehend this.
$endgroup$
– user6897059
Jan 24 at 9:26
$begingroup$
but X is normalized, will this not influence the result? @drhab
$endgroup$
– user6897059
Jan 24 at 9:30
$begingroup$
but X is normalized, will this not influence the result? @drhab
$endgroup$
– user6897059
Jan 24 at 9:30
$begingroup$
Unfortunately I am not familiar enough with the concept "normalized" to judge that. That is why I a gave a comment and not an answer. Personally I cannot find a reason why what I wrote in my former comment would not work.
$endgroup$
– drhab
Jan 24 at 9:35
$begingroup$
Unfortunately I am not familiar enough with the concept "normalized" to judge that. That is why I a gave a comment and not an answer. Personally I cannot find a reason why what I wrote in my former comment would not work.
$endgroup$
– drhab
Jan 24 at 9:35
|
show 2 more comments
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$begingroup$
I'm sorry for mistaking you. The X is just an N-dimension vector, for example, N=3, the X is a unit sphere vector; we can drop the little i;
$endgroup$
– user6897059
Jan 24 at 9:20
$begingroup$
$mathsf{dot}(X,Y)$ is the same as $Y^TX$ and if $Y$ is constant then $mathbb EY^TX=Y^Tmathbb EX$ and $mathsf{Cov}(Y^TX)=Y^Tmathsf{Cov(X)}Y$. Does that help here?
$endgroup$
– drhab
Jan 24 at 9:24
$begingroup$
Thank you for your reply, please give me some time to comprehend this.
$endgroup$
– user6897059
Jan 24 at 9:26
$begingroup$
but X is normalized, will this not influence the result? @drhab
$endgroup$
– user6897059
Jan 24 at 9:30
$begingroup$
Unfortunately I am not familiar enough with the concept "normalized" to judge that. That is why I a gave a comment and not an answer. Personally I cannot find a reason why what I wrote in my former comment would not work.
$endgroup$
– drhab
Jan 24 at 9:35