Expected value of sum of N sines with random phase differences
$begingroup$
This is a follow-up to this question. It discusses the amplitude of a sum of $N$ functions of the form $asin(kx+c)$
$sum_{i=1}^{N} asin(kx+c_i)$
with $a$ and $k$ constant and $c_i$ random numbers between $0$ and $2π$. Let's assume for the following that the $c_i$ are uniformly distributed within those limits. I am interested in the total result of that sum.
It is clear to me, according to for instance this link, that the sum of sine/cosine functions with the same frequency but different phases is again a sine/cosine function of the same frequency, but with different amplitude and phase. The expected value of the new amplitude amounts to $Na^2$. However, what happens to the phase of the sum? Intuitively, I would assume that for large $N$, the summation of all those sine/cosine functions with different phase tends to zero, as we will have all different shifts of them and there will always be pairs that completely cancel. This means, while we have a non-zero amplitude, the expected value of the phase should be $0$, $π$ or $2π$ when dealing with sine functions, and the total sum turns zero then. Is that assumption correct? How is it possible to calculate the expected value of the total phase, which, again from this link, can be calculated by
$tan c=frac{sum_{i=1}^{N} sin c_i}{sum_{i=1}^{N} cos c_i}$?
In addition, I am not sure if it would be the proper way to calculate the expected value of the sum of sines by considering resulting amplitude and phase separately. Wouldn't be the proper way to use something like the law of the unconscious statistician?
Any help and/or literature recommendations are greatly appreciated. I feel this is a rather common problem, but was not able to find useful references.
statistics trigonometry average expected-value
asked Nov 29 '18 at 17:28
MinowMinow
214
$endgroup$
add a comment |
$begingroup$
This is a follow-up to this question. It discusses the amplitude of a sum of $N$ functions of the form $asin(kx+c)$
$sum_{i=1}^{N} asin(kx+c_i)$
with $a$ and $k$ constant and $c_i$ random numbers between $0$ and $2π$. Let's assume for the following that the $c_i$ are uniformly distributed within those limits. I am interested in the total result of that sum.
It is clear to me, according to for instance this link, that the sum of sine/cosine functions with the same frequency but different phases is again a sine/cosine function of the same frequency, but with different amplitude and phase. The expected value of the new amplitude amounts to $Na^2$. However, what happens to the phase of the sum? Intuitively, I would assume that for large $N$, the summation of all those sine/cosine functions with different phase tends to zero, as we will have all different shifts of them and there will always be pairs that completely cancel. This means, while we have a non-zero amplitude, the expected value of the phase should be $0$, $π$ or $2π$ when dealing with sine functions, and the total sum turns zero then. Is that assumption correct? How is it possible to calculate the expected value of the total phase, which, again from this link, can be calculated by
$tan c=frac{sum_{i=1}^{N} sin c_i}{sum_{i=1}^{N} cos c_i}$?
In addition, I am not sure if it would be the proper way to calculate the expected value of the sum of sines by considering resulting amplitude and phase separately. Wouldn't be the proper way to use something like the law of the unconscious statistician?
Any help and/or literature recommendations are greatly appreciated. I feel this is a rather common problem, but was not able to find useful references.
statistics trigonometry average expected-value
asked Nov 29 '18 at 17:28
MinowMinow
214
$endgroup$
add a comment |
$begingroup$
This is a follow-up to this question. It discusses the amplitude of a sum of $N$ functions of the form $asin(kx+c)$
$sum_{i=1}^{N} asin(kx+c_i)$
with $a$ and $k$ constant and $c_i$ random numbers between $0$ and $2π$. Let's assume for the following that the $c_i$ are uniformly distributed within those limits. I am interested in the total result of that sum.
It is clear to me, according to for instance this link, that the sum of sine/cosine functions with the same frequency but different phases is again a sine/cosine function of the same frequency, but with different amplitude and phase. The expected value of the new amplitude amounts to $Na^2$. However, what happens to the phase of the sum? Intuitively, I would assume that for large $N$, the summation of all those sine/cosine functions with different phase tends to zero, as we will have all different shifts of them and there will always be pairs that completely cancel. This means, while we have a non-zero amplitude, the expected value of the phase should be $0$, $π$ or $2π$ when dealing with sine functions, and the total sum turns zero then. Is that assumption correct? How is it possible to calculate the expected value of the total phase, which, again from this link, can be calculated by
$tan c=frac{sum_{i=1}^{N} sin c_i}{sum_{i=1}^{N} cos c_i}$?
In addition, I am not sure if it would be the proper way to calculate the expected value of the sum of sines by considering resulting amplitude and phase separately. Wouldn't be the proper way to use something like the law of the unconscious statistician?
Any help and/or literature recommendations are greatly appreciated. I feel this is a rather common problem, but was not able to find useful references.
statistics trigonometry average expected-value
asked Nov 29 '18 at 17:28
MinowMinow
214
$endgroup$
This is a follow-up to this question. It discusses the amplitude of a sum of $N$ functions of the form $asin(kx+c)$
$sum_{i=1}^{N} asin(kx+c_i)$
with $a$ and $k$ constant and $c_i$ random numbers between $0$ and $2π$. Let's assume for the following that the $c_i$ are uniformly distributed within those limits. I am interested in the total result of that sum.
It is clear to me, according to for instance this link, that the sum of sine/cosine functions with the same frequency but different phases is again a sine/cosine function of the same frequency, but with different amplitude and phase. The expected value of the new amplitude amounts to $Na^2$. However, what happens to the phase of the sum? Intuitively, I would assume that for large $N$, the summation of all those sine/cosine functions with different phase tends to zero, as we will have all different shifts of them and there will always be pairs that completely cancel. This means, while we have a non-zero amplitude, the expected value of the phase should be $0$, $π$ or $2π$ when dealing with sine functions, and the total sum turns zero then. Is that assumption correct? How is it possible to calculate the expected value of the total phase, which, again from this link, can be calculated by
$tan c=frac{sum_{i=1}^{N} sin c_i}{sum_{i=1}^{N} cos c_i}$?
In addition, I am not sure if it would be the proper way to calculate the expected value of the sum of sines by considering resulting amplitude and phase separately. Wouldn't be the proper way to use something like the law of the unconscious statistician?
Any help and/or literature recommendations are greatly appreciated. I feel this is a rather common problem, but was not able to find useful references.
statistics trigonometry average expected-value
statistics trigonometry average expected-value
asked Nov 29 '18 at 17:28
MinowMinow
214
asked Nov 29 '18 at 17:28
MinowMinow
214
asked Nov 29 '18 at 17:28
MinowMinow
214
asked Nov 29 '18 at 17:28
MinowMinow
214
asked Nov 29 '18 at 17:28
MinowMinow
214
214
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add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you want the expectation of the sum $sum_i asin(kx+c_i)$, where $a$ and $k$ are constant, and each $c_i$ has uniform distribution between $0$ and $2pi$, then the expectation is zero for each $x$, since the $i$th term in the sum has expectation
$$Eleft[asin(kx+c_i)right]=int_0^{2pi}asin(kx+t)frac1{2pi},dt=0.$$
So yes, you can get the expectation via Unconscious Statistician. If you attempt to convert the sum to the form $Asin(kx+c')$, you have to contend with the fact that both amplitude $A$ and phase $c'$ are random quantities.
answered Nov 29 '18 at 18:03
grand_chatgrand_chat
20.2k11226
$endgroup$
$begingroup$
that makes sense from a mathematical point of view and kind of confirms my intuition. However, I just came across this very clever question: math.stackexchange.com/questions/3010475/why-do-choirs-work; basically, it deals with the same mathematical problem, but applies that to a choir: every singer sings the same tone, but due to the different path lengths from the singer to my ear, and due to the fact all singers start from a different time base, there is a phase shift that can be treated as a random number. Following your argument: why do I hear anything at all!?
$endgroup$
– Minow
Nov 29 '18 at 18:30
$begingroup$
@Minow It's because the expectation is averaging over all realizations of the experiment. For any given realization of phase, we perceive a sine wave of frequency $k$, but averaging over all possible phases returns zero. Consider the degenerate case of only one singer in your choir, located a randomly chosen distance from your ear. What this means is that the expectation of the experiment isn't really as interesting an entity as each realization of the experiment.
$endgroup$
– grand_chat
Nov 29 '18 at 18:42
$begingroup$
Ok, that sounds reasonable. To come back to the choir example, I think there is something else that needs to be considered: one does not have an infinitely small ear, but it has some area. Also one needs to consider that the ear within that area is most likely (?) an intensity detector. This means that at some position within your "hearing area", you might capture a realization that is zero, but other points within that area actually receive nonzero contributions. On average across the area, you will probably hear the same sum of intensities, even when listening to multiple realizations...
$endgroup$
– Minow
Nov 29 '18 at 19:26
add a comment |
$begingroup$
Lets start with an identity: $$a sin(theta)+bcos(theta)=rsin(theta + arctan(b/a))$$
with $r=sqrt{a^2+b^2}$
Now we look at $$sum sin(kx+c_i)=Im sum e^{i(kx+c_i)}$$
$$=Re (cos(kx)+isin(kx))sum (cos(c_i)+isin(c_i))$$
Collecting the imaginary part, we are left with
$$left[sum sin(c_i)right] cos(kx) + left[sum cos(c_i)right]sin(kx)$$ using the first identity completes the job.
answered Nov 29 '18 at 18:09
user619894user619894
111
$endgroup$
$begingroup$
Thanks; that's interesting, as it derives how the resulting phase of the sum can be calculated. How can I calculate the phase's expected value then? That seems not so obvious to me
$endgroup$
– Minow
Nov 29 '18 at 18:24
$begingroup$
Lets simplify and look at $sum e^{i c_j}$ the phase is just the angle of this complex vector. If you consider this sum as a random walk, the distribution of the end of the walk is distributed equally in all angles, so the expectation is indeed zero.
$endgroup$
– user619894
Nov 29 '18 at 21:57
add a comment |
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2 Answers
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2 Answers
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active
oldest
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$begingroup$
If you want the expectation of the sum $sum_i asin(kx+c_i)$, where $a$ and $k$ are constant, and each $c_i$ has uniform distribution between $0$ and $2pi$, then the expectation is zero for each $x$, since the $i$th term in the sum has expectation
$$Eleft[asin(kx+c_i)right]=int_0^{2pi}asin(kx+t)frac1{2pi},dt=0.$$
So yes, you can get the expectation via Unconscious Statistician. If you attempt to convert the sum to the form $Asin(kx+c')$, you have to contend with the fact that both amplitude $A$ and phase $c'$ are random quantities.
answered Nov 29 '18 at 18:03
grand_chatgrand_chat
20.2k11226
$endgroup$
$begingroup$
that makes sense from a mathematical point of view and kind of confirms my intuition. However, I just came across this very clever question: math.stackexchange.com/questions/3010475/why-do-choirs-work; basically, it deals with the same mathematical problem, but applies that to a choir: every singer sings the same tone, but due to the different path lengths from the singer to my ear, and due to the fact all singers start from a different time base, there is a phase shift that can be treated as a random number. Following your argument: why do I hear anything at all!?
$endgroup$
– Minow
Nov 29 '18 at 18:30
$begingroup$
@Minow It's because the expectation is averaging over all realizations of the experiment. For any given realization of phase, we perceive a sine wave of frequency $k$, but averaging over all possible phases returns zero. Consider the degenerate case of only one singer in your choir, located a randomly chosen distance from your ear. What this means is that the expectation of the experiment isn't really as interesting an entity as each realization of the experiment.
$endgroup$
– grand_chat
Nov 29 '18 at 18:42
$begingroup$
Ok, that sounds reasonable. To come back to the choir example, I think there is something else that needs to be considered: one does not have an infinitely small ear, but it has some area. Also one needs to consider that the ear within that area is most likely (?) an intensity detector. This means that at some position within your "hearing area", you might capture a realization that is zero, but other points within that area actually receive nonzero contributions. On average across the area, you will probably hear the same sum of intensities, even when listening to multiple realizations...
$endgroup$
– Minow
Nov 29 '18 at 19:26
add a comment |
$begingroup$
If you want the expectation of the sum $sum_i asin(kx+c_i)$, where $a$ and $k$ are constant, and each $c_i$ has uniform distribution between $0$ and $2pi$, then the expectation is zero for each $x$, since the $i$th term in the sum has expectation
$$Eleft[asin(kx+c_i)right]=int_0^{2pi}asin(kx+t)frac1{2pi},dt=0.$$
So yes, you can get the expectation via Unconscious Statistician. If you attempt to convert the sum to the form $Asin(kx+c')$, you have to contend with the fact that both amplitude $A$ and phase $c'$ are random quantities.
answered Nov 29 '18 at 18:03
grand_chatgrand_chat
20.2k11226
$endgroup$
$begingroup$
that makes sense from a mathematical point of view and kind of confirms my intuition. However, I just came across this very clever question: math.stackexchange.com/questions/3010475/why-do-choirs-work; basically, it deals with the same mathematical problem, but applies that to a choir: every singer sings the same tone, but due to the different path lengths from the singer to my ear, and due to the fact all singers start from a different time base, there is a phase shift that can be treated as a random number. Following your argument: why do I hear anything at all!?
$endgroup$
– Minow
Nov 29 '18 at 18:30
$begingroup$
@Minow It's because the expectation is averaging over all realizations of the experiment. For any given realization of phase, we perceive a sine wave of frequency $k$, but averaging over all possible phases returns zero. Consider the degenerate case of only one singer in your choir, located a randomly chosen distance from your ear. What this means is that the expectation of the experiment isn't really as interesting an entity as each realization of the experiment.
$endgroup$
– grand_chat
Nov 29 '18 at 18:42
$begingroup$
Ok, that sounds reasonable. To come back to the choir example, I think there is something else that needs to be considered: one does not have an infinitely small ear, but it has some area. Also one needs to consider that the ear within that area is most likely (?) an intensity detector. This means that at some position within your "hearing area", you might capture a realization that is zero, but other points within that area actually receive nonzero contributions. On average across the area, you will probably hear the same sum of intensities, even when listening to multiple realizations...
$endgroup$
– Minow
Nov 29 '18 at 19:26
add a comment |
$begingroup$
If you want the expectation of the sum $sum_i asin(kx+c_i)$, where $a$ and $k$ are constant, and each $c_i$ has uniform distribution between $0$ and $2pi$, then the expectation is zero for each $x$, since the $i$th term in the sum has expectation
$$Eleft[asin(kx+c_i)right]=int_0^{2pi}asin(kx+t)frac1{2pi},dt=0.$$
So yes, you can get the expectation via Unconscious Statistician. If you attempt to convert the sum to the form $Asin(kx+c')$, you have to contend with the fact that both amplitude $A$ and phase $c'$ are random quantities.
answered Nov 29 '18 at 18:03
grand_chatgrand_chat
20.2k11226
$endgroup$
If you want the expectation of the sum $sum_i asin(kx+c_i)$, where $a$ and $k$ are constant, and each $c_i$ has uniform distribution between $0$ and $2pi$, then the expectation is zero for each $x$, since the $i$th term in the sum has expectation
$$Eleft[asin(kx+c_i)right]=int_0^{2pi}asin(kx+t)frac1{2pi},dt=0.$$
So yes, you can get the expectation via Unconscious Statistician. If you attempt to convert the sum to the form $Asin(kx+c')$, you have to contend with the fact that both amplitude $A$ and phase $c'$ are random quantities.
answered Nov 29 '18 at 18:03
grand_chatgrand_chat
20.2k11226
answered Nov 29 '18 at 18:03
grand_chatgrand_chat
20.2k11226
answered Nov 29 '18 at 18:03
grand_chatgrand_chat
20.2k11226
answered Nov 29 '18 at 18:03
grand_chatgrand_chat
20.2k11226
20.2k11226
$begingroup$
that makes sense from a mathematical point of view and kind of confirms my intuition. However, I just came across this very clever question: math.stackexchange.com/questions/3010475/why-do-choirs-work; basically, it deals with the same mathematical problem, but applies that to a choir: every singer sings the same tone, but due to the different path lengths from the singer to my ear, and due to the fact all singers start from a different time base, there is a phase shift that can be treated as a random number. Following your argument: why do I hear anything at all!?
$endgroup$
– Minow
Nov 29 '18 at 18:30
$begingroup$
@Minow It's because the expectation is averaging over all realizations of the experiment. For any given realization of phase, we perceive a sine wave of frequency $k$, but averaging over all possible phases returns zero. Consider the degenerate case of only one singer in your choir, located a randomly chosen distance from your ear. What this means is that the expectation of the experiment isn't really as interesting an entity as each realization of the experiment.
$endgroup$
– grand_chat
Nov 29 '18 at 18:42
$begingroup$
Ok, that sounds reasonable. To come back to the choir example, I think there is something else that needs to be considered: one does not have an infinitely small ear, but it has some area. Also one needs to consider that the ear within that area is most likely (?) an intensity detector. This means that at some position within your "hearing area", you might capture a realization that is zero, but other points within that area actually receive nonzero contributions. On average across the area, you will probably hear the same sum of intensities, even when listening to multiple realizations...
$endgroup$
– Minow
Nov 29 '18 at 19:26
add a comment |
$begingroup$
that makes sense from a mathematical point of view and kind of confirms my intuition. However, I just came across this very clever question: math.stackexchange.com/questions/3010475/why-do-choirs-work; basically, it deals with the same mathematical problem, but applies that to a choir: every singer sings the same tone, but due to the different path lengths from the singer to my ear, and due to the fact all singers start from a different time base, there is a phase shift that can be treated as a random number. Following your argument: why do I hear anything at all!?
$endgroup$
– Minow
Nov 29 '18 at 18:30
$begingroup$
@Minow It's because the expectation is averaging over all realizations of the experiment. For any given realization of phase, we perceive a sine wave of frequency $k$, but averaging over all possible phases returns zero. Consider the degenerate case of only one singer in your choir, located a randomly chosen distance from your ear. What this means is that the expectation of the experiment isn't really as interesting an entity as each realization of the experiment.
$endgroup$
– grand_chat
Nov 29 '18 at 18:42
$begingroup$
Ok, that sounds reasonable. To come back to the choir example, I think there is something else that needs to be considered: one does not have an infinitely small ear, but it has some area. Also one needs to consider that the ear within that area is most likely (?) an intensity detector. This means that at some position within your "hearing area", you might capture a realization that is zero, but other points within that area actually receive nonzero contributions. On average across the area, you will probably hear the same sum of intensities, even when listening to multiple realizations...
$endgroup$
– Minow
Nov 29 '18 at 19:26
$begingroup$
that makes sense from a mathematical point of view and kind of confirms my intuition. However, I just came across this very clever question: math.stackexchange.com/questions/3010475/why-do-choirs-work; basically, it deals with the same mathematical problem, but applies that to a choir: every singer sings the same tone, but due to the different path lengths from the singer to my ear, and due to the fact all singers start from a different time base, there is a phase shift that can be treated as a random number. Following your argument: why do I hear anything at all!?
$endgroup$
– Minow
Nov 29 '18 at 18:30
$begingroup$
that makes sense from a mathematical point of view and kind of confirms my intuition. However, I just came across this very clever question: math.stackexchange.com/questions/3010475/why-do-choirs-work; basically, it deals with the same mathematical problem, but applies that to a choir: every singer sings the same tone, but due to the different path lengths from the singer to my ear, and due to the fact all singers start from a different time base, there is a phase shift that can be treated as a random number. Following your argument: why do I hear anything at all!?
$endgroup$
– Minow
Nov 29 '18 at 18:30
$begingroup$
@Minow It's because the expectation is averaging over all realizations of the experiment. For any given realization of phase, we perceive a sine wave of frequency $k$, but averaging over all possible phases returns zero. Consider the degenerate case of only one singer in your choir, located a randomly chosen distance from your ear. What this means is that the expectation of the experiment isn't really as interesting an entity as each realization of the experiment.
$endgroup$
– grand_chat
Nov 29 '18 at 18:42
$begingroup$
@Minow It's because the expectation is averaging over all realizations of the experiment. For any given realization of phase, we perceive a sine wave of frequency $k$, but averaging over all possible phases returns zero. Consider the degenerate case of only one singer in your choir, located a randomly chosen distance from your ear. What this means is that the expectation of the experiment isn't really as interesting an entity as each realization of the experiment.
$endgroup$
– grand_chat
Nov 29 '18 at 18:42
$begingroup$
Ok, that sounds reasonable. To come back to the choir example, I think there is something else that needs to be considered: one does not have an infinitely small ear, but it has some area. Also one needs to consider that the ear within that area is most likely (?) an intensity detector. This means that at some position within your "hearing area", you might capture a realization that is zero, but other points within that area actually receive nonzero contributions. On average across the area, you will probably hear the same sum of intensities, even when listening to multiple realizations...
$endgroup$
– Minow
Nov 29 '18 at 19:26
$begingroup$
Ok, that sounds reasonable. To come back to the choir example, I think there is something else that needs to be considered: one does not have an infinitely small ear, but it has some area. Also one needs to consider that the ear within that area is most likely (?) an intensity detector. This means that at some position within your "hearing area", you might capture a realization that is zero, but other points within that area actually receive nonzero contributions. On average across the area, you will probably hear the same sum of intensities, even when listening to multiple realizations...
$endgroup$
– Minow
Nov 29 '18 at 19:26
add a comment |
$begingroup$
Lets start with an identity: $$a sin(theta)+bcos(theta)=rsin(theta + arctan(b/a))$$
with $r=sqrt{a^2+b^2}$
Now we look at $$sum sin(kx+c_i)=Im sum e^{i(kx+c_i)}$$
$$=Re (cos(kx)+isin(kx))sum (cos(c_i)+isin(c_i))$$
Collecting the imaginary part, we are left with
$$left[sum sin(c_i)right] cos(kx) + left[sum cos(c_i)right]sin(kx)$$ using the first identity completes the job.
answered Nov 29 '18 at 18:09
user619894user619894
111
$endgroup$
$begingroup$
Thanks; that's interesting, as it derives how the resulting phase of the sum can be calculated. How can I calculate the phase's expected value then? That seems not so obvious to me
$endgroup$
– Minow
Nov 29 '18 at 18:24
$begingroup$
Lets simplify and look at $sum e^{i c_j}$ the phase is just the angle of this complex vector. If you consider this sum as a random walk, the distribution of the end of the walk is distributed equally in all angles, so the expectation is indeed zero.
$endgroup$
– user619894
Nov 29 '18 at 21:57
add a comment |
$begingroup$
Lets start with an identity: $$a sin(theta)+bcos(theta)=rsin(theta + arctan(b/a))$$
with $r=sqrt{a^2+b^2}$
Now we look at $$sum sin(kx+c_i)=Im sum e^{i(kx+c_i)}$$
$$=Re (cos(kx)+isin(kx))sum (cos(c_i)+isin(c_i))$$
Collecting the imaginary part, we are left with
$$left[sum sin(c_i)right] cos(kx) + left[sum cos(c_i)right]sin(kx)$$ using the first identity completes the job.
answered Nov 29 '18 at 18:09
user619894user619894
111
$endgroup$
$begingroup$
Thanks; that's interesting, as it derives how the resulting phase of the sum can be calculated. How can I calculate the phase's expected value then? That seems not so obvious to me
$endgroup$
– Minow
Nov 29 '18 at 18:24
$begingroup$
Lets simplify and look at $sum e^{i c_j}$ the phase is just the angle of this complex vector. If you consider this sum as a random walk, the distribution of the end of the walk is distributed equally in all angles, so the expectation is indeed zero.
$endgroup$
– user619894
Nov 29 '18 at 21:57
add a comment |
$begingroup$
Lets start with an identity: $$a sin(theta)+bcos(theta)=rsin(theta + arctan(b/a))$$
with $r=sqrt{a^2+b^2}$
Now we look at $$sum sin(kx+c_i)=Im sum e^{i(kx+c_i)}$$
$$=Re (cos(kx)+isin(kx))sum (cos(c_i)+isin(c_i))$$
Collecting the imaginary part, we are left with
$$left[sum sin(c_i)right] cos(kx) + left[sum cos(c_i)right]sin(kx)$$ using the first identity completes the job.
answered Nov 29 '18 at 18:09
user619894user619894
111
$endgroup$
Lets start with an identity: $$a sin(theta)+bcos(theta)=rsin(theta + arctan(b/a))$$
with $r=sqrt{a^2+b^2}$
Now we look at $$sum sin(kx+c_i)=Im sum e^{i(kx+c_i)}$$
$$=Re (cos(kx)+isin(kx))sum (cos(c_i)+isin(c_i))$$
Collecting the imaginary part, we are left with
$$left[sum sin(c_i)right] cos(kx) + left[sum cos(c_i)right]sin(kx)$$ using the first identity completes the job.
answered Nov 29 '18 at 18:09
user619894user619894
111
answered Nov 29 '18 at 18:09
user619894user619894
111
answered Nov 29 '18 at 18:09
user619894user619894
111
answered Nov 29 '18 at 18:09
user619894user619894
111
111
$begingroup$
Thanks; that's interesting, as it derives how the resulting phase of the sum can be calculated. How can I calculate the phase's expected value then? That seems not so obvious to me
$endgroup$
– Minow
Nov 29 '18 at 18:24
$begingroup$
Lets simplify and look at $sum e^{i c_j}$ the phase is just the angle of this complex vector. If you consider this sum as a random walk, the distribution of the end of the walk is distributed equally in all angles, so the expectation is indeed zero.
$endgroup$
– user619894
Nov 29 '18 at 21:57
add a comment |
$begingroup$
Thanks; that's interesting, as it derives how the resulting phase of the sum can be calculated. How can I calculate the phase's expected value then? That seems not so obvious to me
$endgroup$
– Minow
Nov 29 '18 at 18:24
$begingroup$
Lets simplify and look at $sum e^{i c_j}$ the phase is just the angle of this complex vector. If you consider this sum as a random walk, the distribution of the end of the walk is distributed equally in all angles, so the expectation is indeed zero.
$endgroup$
– user619894
Nov 29 '18 at 21:57
$begingroup$
Thanks; that's interesting, as it derives how the resulting phase of the sum can be calculated. How can I calculate the phase's expected value then? That seems not so obvious to me
$endgroup$
– Minow
Nov 29 '18 at 18:24
$begingroup$
Thanks; that's interesting, as it derives how the resulting phase of the sum can be calculated. How can I calculate the phase's expected value then? That seems not so obvious to me
$endgroup$
– Minow
Nov 29 '18 at 18:24
$begingroup$
Lets simplify and look at $sum e^{i c_j}$ the phase is just the angle of this complex vector. If you consider this sum as a random walk, the distribution of the end of the walk is distributed equally in all angles, so the expectation is indeed zero.
$endgroup$
– user619894
Nov 29 '18 at 21:57
$begingroup$
Lets simplify and look at $sum e^{i c_j}$ the phase is just the angle of this complex vector. If you consider this sum as a random walk, the distribution of the end of the walk is distributed equally in all angles, so the expectation is indeed zero.
$endgroup$
– user619894
Nov 29 '18 at 21:57
add a comment |
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