The expectation of absolute value of the sum of n i.i.d. random variables
$begingroup$
Let $varphi_i$ be a Gaussian random variable such that
$$varphi_i sim N(0,sigma^2), quad i = 1,2,ldots,n.$$
What's the expectation:
$$Eleft(left | sum_{i=1}^n e^{j varphi_i} right |right) $$
where $left | cdot right |$ is the absolute value operation and $j = sqrt{-1}$.
probability statistics absolute-value expected-value
asked Jan 1 at 15:32
Gertsen YuanGertsen Yuan
84
$endgroup$
add a comment |
$begingroup$
Let $varphi_i$ be a Gaussian random variable such that
$$varphi_i sim N(0,sigma^2), quad i = 1,2,ldots,n.$$
What's the expectation:
$$Eleft(left | sum_{i=1}^n e^{j varphi_i} right |right) $$
where $left | cdot right |$ is the absolute value operation and $j = sqrt{-1}$.
probability statistics absolute-value expected-value
asked Jan 1 at 15:32
Gertsen YuanGertsen Yuan
84
$endgroup$
$begingroup$
I don't know if there is a nice closed form for the absolute value but for the square of the absolute value the expectation is $frac{n (n-1)}{exp left(sigma ^2right)}+n$.
$endgroup$
– JimB
Jan 1 at 19:55
add a comment |
$begingroup$
Let $varphi_i$ be a Gaussian random variable such that
$$varphi_i sim N(0,sigma^2), quad i = 1,2,ldots,n.$$
What's the expectation:
$$Eleft(left | sum_{i=1}^n e^{j varphi_i} right |right) $$
where $left | cdot right |$ is the absolute value operation and $j = sqrt{-1}$.
probability statistics absolute-value expected-value
asked Jan 1 at 15:32
Gertsen YuanGertsen Yuan
84
$endgroup$
Let $varphi_i$ be a Gaussian random variable such that
$$varphi_i sim N(0,sigma^2), quad i = 1,2,ldots,n.$$
What's the expectation:
$$Eleft(left | sum_{i=1}^n e^{j varphi_i} right |right) $$
where $left | cdot right |$ is the absolute value operation and $j = sqrt{-1}$.
probability statistics absolute-value expected-value
probability statistics absolute-value expected-value
asked Jan 1 at 15:32
Gertsen YuanGertsen Yuan
84
asked Jan 1 at 15:32
Gertsen YuanGertsen Yuan
84
asked Jan 1 at 15:32
Gertsen YuanGertsen Yuan
84
asked Jan 1 at 15:32
Gertsen YuanGertsen Yuan
84
asked Jan 1 at 15:32
Gertsen YuanGertsen Yuan
84
84
$begingroup$
I don't know if there is a nice closed form for the absolute value but for the square of the absolute value the expectation is $frac{n (n-1)}{exp left(sigma ^2right)}+n$.
$endgroup$
– JimB
Jan 1 at 19:55
add a comment |
$begingroup$
I don't know if there is a nice closed form for the absolute value but for the square of the absolute value the expectation is $frac{n (n-1)}{exp left(sigma ^2right)}+n$.
$endgroup$
– JimB
Jan 1 at 19:55
$begingroup$
I don't know if there is a nice closed form for the absolute value but for the square of the absolute value the expectation is $frac{n (n-1)}{exp left(sigma ^2right)}+n$.
$endgroup$
– JimB
Jan 1 at 19:55
$begingroup$
I don't know if there is a nice closed form for the absolute value but for the square of the absolute value the expectation is $frac{n (n-1)}{exp left(sigma ^2right)}+n$.
$endgroup$
– JimB
Jan 1 at 19:55
add a comment |
1 Answer
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$begingroup$
The expectation you are interested in is equivalent to
$$Eleft[sqrt{n+2sum_{j=1}^{n-1}sum_{k=j+1}^n (cos varphi_j cos varphi_k+sinvarphi_j sinvarphi_k)}right]$$
I don't believe there is a nice closed-form solution for that expectation. So I think you'll need to perform simulations or numerically integrate to get solutions.
If you were interested in
$$Eleft[n+2sum_{j=1}^{n-1}sum_{k=j+1}^n (cos varphi_j cos varphi_k+sinvarphi_j sinvarphi_k)right]$$
then there is a closed-form solution:
$$n+n(n-1)e^{-sigma^2}$$
answered Jan 2 at 0:46
JimBJimB
55537
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The expectation you are interested in is equivalent to
$$Eleft[sqrt{n+2sum_{j=1}^{n-1}sum_{k=j+1}^n (cos varphi_j cos varphi_k+sinvarphi_j sinvarphi_k)}right]$$
I don't believe there is a nice closed-form solution for that expectation. So I think you'll need to perform simulations or numerically integrate to get solutions.
If you were interested in
$$Eleft[n+2sum_{j=1}^{n-1}sum_{k=j+1}^n (cos varphi_j cos varphi_k+sinvarphi_j sinvarphi_k)right]$$
then there is a closed-form solution:
$$n+n(n-1)e^{-sigma^2}$$
answered Jan 2 at 0:46
JimBJimB
55537
$endgroup$
add a comment |
$begingroup$
The expectation you are interested in is equivalent to
$$Eleft[sqrt{n+2sum_{j=1}^{n-1}sum_{k=j+1}^n (cos varphi_j cos varphi_k+sinvarphi_j sinvarphi_k)}right]$$
I don't believe there is a nice closed-form solution for that expectation. So I think you'll need to perform simulations or numerically integrate to get solutions.
If you were interested in
$$Eleft[n+2sum_{j=1}^{n-1}sum_{k=j+1}^n (cos varphi_j cos varphi_k+sinvarphi_j sinvarphi_k)right]$$
then there is a closed-form solution:
$$n+n(n-1)e^{-sigma^2}$$
answered Jan 2 at 0:46
JimBJimB
55537
$endgroup$
add a comment |
$begingroup$
The expectation you are interested in is equivalent to
$$Eleft[sqrt{n+2sum_{j=1}^{n-1}sum_{k=j+1}^n (cos varphi_j cos varphi_k+sinvarphi_j sinvarphi_k)}right]$$
I don't believe there is a nice closed-form solution for that expectation. So I think you'll need to perform simulations or numerically integrate to get solutions.
If you were interested in
$$Eleft[n+2sum_{j=1}^{n-1}sum_{k=j+1}^n (cos varphi_j cos varphi_k+sinvarphi_j sinvarphi_k)right]$$
then there is a closed-form solution:
$$n+n(n-1)e^{-sigma^2}$$
answered Jan 2 at 0:46
JimBJimB
55537
$endgroup$
The expectation you are interested in is equivalent to
$$Eleft[sqrt{n+2sum_{j=1}^{n-1}sum_{k=j+1}^n (cos varphi_j cos varphi_k+sinvarphi_j sinvarphi_k)}right]$$
I don't believe there is a nice closed-form solution for that expectation. So I think you'll need to perform simulations or numerically integrate to get solutions.
If you were interested in
$$Eleft[n+2sum_{j=1}^{n-1}sum_{k=j+1}^n (cos varphi_j cos varphi_k+sinvarphi_j sinvarphi_k)right]$$
then there is a closed-form solution:
$$n+n(n-1)e^{-sigma^2}$$
answered Jan 2 at 0:46
JimBJimB
55537
answered Jan 2 at 0:46
JimBJimB
55537
answered Jan 2 at 0:46
JimBJimB
55537
answered Jan 2 at 0:46
JimBJimB
55537
55537
add a comment |
add a comment |
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$begingroup$
I don't know if there is a nice closed form for the absolute value but for the square of the absolute value the expectation is $frac{n (n-1)}{exp left(sigma ^2right)}+n$.
$endgroup$
– JimB
Jan 1 at 19:55