Expectation of Independent Bernoulli Trials
$begingroup$
Say there is a sequence of IID bernoulli trials $(x_1,x_2,...,x_N)$. Where $X in mathbb{R}^{Ntimes N}$ is a diagonal matrix with these RV's on the main diagonal and $mathbb{P}left[x_k =1right]=p, 0<kle N in mathbb{N}$. I want to know what is:
$$
mathbb{E}left[ X W X right],
$$
where $W inmathbb{R}^{Ntimes N}succ 0 $ is some positive definite weightings matrix. Due to this being bernoulli trials I believe the main diagonal after the expectation is excecuted to be $W_{ii} cdot p$ and $W_{ij}cdot p^2 $ for $ineq j$. However, I am unsure.
probability expected-value
asked Dec 5 '18 at 11:15
p32fr4p32fr4
3717
$endgroup$
add a comment |
$begingroup$
Say there is a sequence of IID bernoulli trials $(x_1,x_2,...,x_N)$. Where $X in mathbb{R}^{Ntimes N}$ is a diagonal matrix with these RV's on the main diagonal and $mathbb{P}left[x_k =1right]=p, 0<kle N in mathbb{N}$. I want to know what is:
$$
mathbb{E}left[ X W X right],
$$
where $W inmathbb{R}^{Ntimes N}succ 0 $ is some positive definite weightings matrix. Due to this being bernoulli trials I believe the main diagonal after the expectation is excecuted to be $W_{ii} cdot p$ and $W_{ij}cdot p^2 $ for $ineq j$. However, I am unsure.
probability expected-value
asked Dec 5 '18 at 11:15
p32fr4p32fr4
3717
$endgroup$
add a comment |
$begingroup$
Say there is a sequence of IID bernoulli trials $(x_1,x_2,...,x_N)$. Where $X in mathbb{R}^{Ntimes N}$ is a diagonal matrix with these RV's on the main diagonal and $mathbb{P}left[x_k =1right]=p, 0<kle N in mathbb{N}$. I want to know what is:
$$
mathbb{E}left[ X W X right],
$$
where $W inmathbb{R}^{Ntimes N}succ 0 $ is some positive definite weightings matrix. Due to this being bernoulli trials I believe the main diagonal after the expectation is excecuted to be $W_{ii} cdot p$ and $W_{ij}cdot p^2 $ for $ineq j$. However, I am unsure.
probability expected-value
asked Dec 5 '18 at 11:15
p32fr4p32fr4
3717
$endgroup$
Say there is a sequence of IID bernoulli trials $(x_1,x_2,...,x_N)$. Where $X in mathbb{R}^{Ntimes N}$ is a diagonal matrix with these RV's on the main diagonal and $mathbb{P}left[x_k =1right]=p, 0<kle N in mathbb{N}$. I want to know what is:
$$
mathbb{E}left[ X W X right],
$$
where $W inmathbb{R}^{Ntimes N}succ 0 $ is some positive definite weightings matrix. Due to this being bernoulli trials I believe the main diagonal after the expectation is excecuted to be $W_{ii} cdot p$ and $W_{ij}cdot p^2 $ for $ineq j$. However, I am unsure.
probability expected-value
probability expected-value
asked Dec 5 '18 at 11:15
p32fr4p32fr4
3717
asked Dec 5 '18 at 11:15
p32fr4p32fr4
3717
asked Dec 5 '18 at 11:15
p32fr4p32fr4
3717
asked Dec 5 '18 at 11:15
p32fr4p32fr4
3717
asked Dec 5 '18 at 11:15
p32fr4p32fr4
3717
3717
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can explicitly give the elements of the product $X W X$ as, for element at coordinate $(i,j)$
begin{align*}
(X W X)_{ij} &= sum_{k} X_{ik} (WX)_{kj}\
&=sum_{k} X_{ik} sum_{ell} W_{kell} X_{ell j}\
&=sum_k sum_ell W_{kell} X_{ik}X_{ell j}
end{align*}
The expectation is linear hence we only look at $mathbb E[X_{ab} X_{cd}]$ which is $p$ if $(a,b)=(c,d)$ and $p^2$ otherwise.
Hence
begin{align*}
mathbb E[(XWX)_{ij}] &= sum_k sum_ell W_{kell} [pmathbb 1(i=ell land k=j) + p^2 (1-mathbb 1(i=ell land k=j))]\
&=sum_k sum_ell W_{kell} [mathbb 1(i=ell land k=j)(p-p^2) + p^2)]\
&=p(1-p)W_{ij} + p^2 sum_k sum_ell W_{kell}
end{align*}
Or in matrix notation $mathbb E[XWX]=p(1-p)W + p^2 mathbf 1 W mathbf 1$ where in this case $mathbf 1$ is the all one matrix.
answered Dec 5 '18 at 11:44
P. QuintonP. Quinton
1,8011213
$endgroup$
$begingroup$
$X$ was a diagonal matrix so I don't believe there is a dimensional issue. Either way i also have a column vector multiplying each side ( I didn't post this in the problem above). So it should boil down to a variation of you answer thank you.
$endgroup$
– p32fr4
Dec 5 '18 at 11:55
1
$begingroup$
True, I fixed my answer so that it answers the problem, the trace trick cannot be used anymore and so I had to do it explicitly.
$endgroup$
– P. Quinton
Dec 5 '18 at 12:12
$begingroup$
That's great thanks.
$endgroup$
– p32fr4
Dec 5 '18 at 12:19
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You can explicitly give the elements of the product $X W X$ as, for element at coordinate $(i,j)$
begin{align*}
(X W X)_{ij} &= sum_{k} X_{ik} (WX)_{kj}\
&=sum_{k} X_{ik} sum_{ell} W_{kell} X_{ell j}\
&=sum_k sum_ell W_{kell} X_{ik}X_{ell j}
end{align*}
The expectation is linear hence we only look at $mathbb E[X_{ab} X_{cd}]$ which is $p$ if $(a,b)=(c,d)$ and $p^2$ otherwise.
Hence
begin{align*}
mathbb E[(XWX)_{ij}] &= sum_k sum_ell W_{kell} [pmathbb 1(i=ell land k=j) + p^2 (1-mathbb 1(i=ell land k=j))]\
&=sum_k sum_ell W_{kell} [mathbb 1(i=ell land k=j)(p-p^2) + p^2)]\
&=p(1-p)W_{ij} + p^2 sum_k sum_ell W_{kell}
end{align*}
Or in matrix notation $mathbb E[XWX]=p(1-p)W + p^2 mathbf 1 W mathbf 1$ where in this case $mathbf 1$ is the all one matrix.
answered Dec 5 '18 at 11:44
P. QuintonP. Quinton
1,8011213
$endgroup$
$begingroup$
$X$ was a diagonal matrix so I don't believe there is a dimensional issue. Either way i also have a column vector multiplying each side ( I didn't post this in the problem above). So it should boil down to a variation of you answer thank you.
$endgroup$
– p32fr4
Dec 5 '18 at 11:55
1
$begingroup$
True, I fixed my answer so that it answers the problem, the trace trick cannot be used anymore and so I had to do it explicitly.
$endgroup$
– P. Quinton
Dec 5 '18 at 12:12
$begingroup$
That's great thanks.
$endgroup$
– p32fr4
Dec 5 '18 at 12:19
add a comment |
$begingroup$
You can explicitly give the elements of the product $X W X$ as, for element at coordinate $(i,j)$
begin{align*}
(X W X)_{ij} &= sum_{k} X_{ik} (WX)_{kj}\
&=sum_{k} X_{ik} sum_{ell} W_{kell} X_{ell j}\
&=sum_k sum_ell W_{kell} X_{ik}X_{ell j}
end{align*}
The expectation is linear hence we only look at $mathbb E[X_{ab} X_{cd}]$ which is $p$ if $(a,b)=(c,d)$ and $p^2$ otherwise.
Hence
begin{align*}
mathbb E[(XWX)_{ij}] &= sum_k sum_ell W_{kell} [pmathbb 1(i=ell land k=j) + p^2 (1-mathbb 1(i=ell land k=j))]\
&=sum_k sum_ell W_{kell} [mathbb 1(i=ell land k=j)(p-p^2) + p^2)]\
&=p(1-p)W_{ij} + p^2 sum_k sum_ell W_{kell}
end{align*}
Or in matrix notation $mathbb E[XWX]=p(1-p)W + p^2 mathbf 1 W mathbf 1$ where in this case $mathbf 1$ is the all one matrix.
answered Dec 5 '18 at 11:44
P. QuintonP. Quinton
1,8011213
$endgroup$
$begingroup$
$X$ was a diagonal matrix so I don't believe there is a dimensional issue. Either way i also have a column vector multiplying each side ( I didn't post this in the problem above). So it should boil down to a variation of you answer thank you.
$endgroup$
– p32fr4
Dec 5 '18 at 11:55
1
$begingroup$
True, I fixed my answer so that it answers the problem, the trace trick cannot be used anymore and so I had to do it explicitly.
$endgroup$
– P. Quinton
Dec 5 '18 at 12:12
$begingroup$
That's great thanks.
$endgroup$
– p32fr4
Dec 5 '18 at 12:19
add a comment |
$begingroup$
You can explicitly give the elements of the product $X W X$ as, for element at coordinate $(i,j)$
begin{align*}
(X W X)_{ij} &= sum_{k} X_{ik} (WX)_{kj}\
&=sum_{k} X_{ik} sum_{ell} W_{kell} X_{ell j}\
&=sum_k sum_ell W_{kell} X_{ik}X_{ell j}
end{align*}
The expectation is linear hence we only look at $mathbb E[X_{ab} X_{cd}]$ which is $p$ if $(a,b)=(c,d)$ and $p^2$ otherwise.
Hence
begin{align*}
mathbb E[(XWX)_{ij}] &= sum_k sum_ell W_{kell} [pmathbb 1(i=ell land k=j) + p^2 (1-mathbb 1(i=ell land k=j))]\
&=sum_k sum_ell W_{kell} [mathbb 1(i=ell land k=j)(p-p^2) + p^2)]\
&=p(1-p)W_{ij} + p^2 sum_k sum_ell W_{kell}
end{align*}
Or in matrix notation $mathbb E[XWX]=p(1-p)W + p^2 mathbf 1 W mathbf 1$ where in this case $mathbf 1$ is the all one matrix.
answered Dec 5 '18 at 11:44
P. QuintonP. Quinton
1,8011213
$endgroup$
You can explicitly give the elements of the product $X W X$ as, for element at coordinate $(i,j)$
begin{align*}
(X W X)_{ij} &= sum_{k} X_{ik} (WX)_{kj}\
&=sum_{k} X_{ik} sum_{ell} W_{kell} X_{ell j}\
&=sum_k sum_ell W_{kell} X_{ik}X_{ell j}
end{align*}
The expectation is linear hence we only look at $mathbb E[X_{ab} X_{cd}]$ which is $p$ if $(a,b)=(c,d)$ and $p^2$ otherwise.
Hence
begin{align*}
mathbb E[(XWX)_{ij}] &= sum_k sum_ell W_{kell} [pmathbb 1(i=ell land k=j) + p^2 (1-mathbb 1(i=ell land k=j))]\
&=sum_k sum_ell W_{kell} [mathbb 1(i=ell land k=j)(p-p^2) + p^2)]\
&=p(1-p)W_{ij} + p^2 sum_k sum_ell W_{kell}
end{align*}
Or in matrix notation $mathbb E[XWX]=p(1-p)W + p^2 mathbf 1 W mathbf 1$ where in this case $mathbf 1$ is the all one matrix.
answered Dec 5 '18 at 11:44
P. QuintonP. Quinton
1,8011213
edited Dec 5 '18 at 12:11
answered Dec 5 '18 at 11:44
P. QuintonP. Quinton
1,8011213
answered Dec 5 '18 at 11:44
P. QuintonP. Quinton
1,8011213
answered Dec 5 '18 at 11:44
P. QuintonP. Quinton
1,8011213
1,8011213
$begingroup$
$X$ was a diagonal matrix so I don't believe there is a dimensional issue. Either way i also have a column vector multiplying each side ( I didn't post this in the problem above). So it should boil down to a variation of you answer thank you.
$endgroup$
– p32fr4
Dec 5 '18 at 11:55
1
$begingroup$
True, I fixed my answer so that it answers the problem, the trace trick cannot be used anymore and so I had to do it explicitly.
$endgroup$
– P. Quinton
Dec 5 '18 at 12:12
$begingroup$
That's great thanks.
$endgroup$
– p32fr4
Dec 5 '18 at 12:19
add a comment |
$begingroup$
$X$ was a diagonal matrix so I don't believe there is a dimensional issue. Either way i also have a column vector multiplying each side ( I didn't post this in the problem above). So it should boil down to a variation of you answer thank you.
$endgroup$
– p32fr4
Dec 5 '18 at 11:55
1
$begingroup$
True, I fixed my answer so that it answers the problem, the trace trick cannot be used anymore and so I had to do it explicitly.
$endgroup$
– P. Quinton
Dec 5 '18 at 12:12
$begingroup$
That's great thanks.
$endgroup$
– p32fr4
Dec 5 '18 at 12:19
$begingroup$
$X$ was a diagonal matrix so I don't believe there is a dimensional issue. Either way i also have a column vector multiplying each side ( I didn't post this in the problem above). So it should boil down to a variation of you answer thank you.
$endgroup$
– p32fr4
Dec 5 '18 at 11:55
$begingroup$
$X$ was a diagonal matrix so I don't believe there is a dimensional issue. Either way i also have a column vector multiplying each side ( I didn't post this in the problem above). So it should boil down to a variation of you answer thank you.
$endgroup$
– p32fr4
Dec 5 '18 at 11:55
1
1
$begingroup$
True, I fixed my answer so that it answers the problem, the trace trick cannot be used anymore and so I had to do it explicitly.
$endgroup$
– P. Quinton
Dec 5 '18 at 12:12
$begingroup$
True, I fixed my answer so that it answers the problem, the trace trick cannot be used anymore and so I had to do it explicitly.
$endgroup$
– P. Quinton
Dec 5 '18 at 12:12
$begingroup$
That's great thanks.
$endgroup$
– p32fr4
Dec 5 '18 at 12:19
$begingroup$
That's great thanks.
$endgroup$
– p32fr4
Dec 5 '18 at 12:19
add a comment |
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