Expectation of random 2d walk
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Start at the origin, take $n$ independent steps of length 1 in the direction of $theta_i$, which is uniformly distributed on $[0,2pi]$.
If $X,Y$ is the position after $n$ steps and $D = X^2 + Y^2$, what is $E[D]$?
probability probability-distributions
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Start at the origin, take $n$ independent steps of length 1 in the direction of $theta_i$, which is uniformly distributed on $[0,2pi]$.
If $X,Y$ is the position after $n$ steps and $D = X^2 + Y^2$, what is $E[D]$?
probability probability-distributions
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Please do not delete posts after you got an answer.
– quid♦
Nov 18 at 23:40
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up vote
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down vote
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Start at the origin, take $n$ independent steps of length 1 in the direction of $theta_i$, which is uniformly distributed on $[0,2pi]$.
If $X,Y$ is the position after $n$ steps and $D = X^2 + Y^2$, what is $E[D]$?
probability probability-distributions
Start at the origin, take $n$ independent steps of length 1 in the direction of $theta_i$, which is uniformly distributed on $[0,2pi]$.
If $X,Y$ is the position after $n$ steps and $D = X^2 + Y^2$, what is $E[D]$?
probability probability-distributions
probability probability-distributions
edited Nov 20 at 3:35
asked Nov 18 at 4:11
user616954
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Please do not delete posts after you got an answer.
– quid♦
Nov 18 at 23:40
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Please do not delete posts after you got an answer.
– quid♦
Nov 18 at 23:40
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Please do not delete posts after you got an answer.
– quid♦
Nov 18 at 23:40
Please do not delete posts after you got an answer.
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1 Answer
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Observe that a move of length $1$ in the direction $theta$ is a translation by the vector $e^{itheta}$, where $i^2 = -1$.
Thus, if $D: = (X,Y)$ is the position after $n$ moves on the (complex) plane, then
$$
D = e^{itheta_1} + ... + e^{i theta_n} in mathbb{C},
$$
hence
$$
|D|^2 = D overline{D} = sumlimits_{k,m=1}^n e^{i(theta_k - theta_m)} = n + sumlimits_{1leq k neq mleq n}^n e^{i(theta_k - theta_m)}.
$$
It follows that
$$
mathbb{E} |D|^2 = n + sumlimits_{1leq k neq mleq n}^n mathbb{E} e^{i(theta_k - theta_m)} = n + sumlimits_{1leq k neq mleq n}^n mathbb{E}e^{itheta_k} mathbb{E}e^{-i theta_m} = n,
$$
where we used the independence of ${theta_k}$, and the fact that $theta_k sim U[0,2pi]$ for concluding that $mathbb{E}(theta_k) = frac{1}{2pi} intlimits_0^{2pi} e^{ix} dx = 0$.
answered Nov 18 at 5:58
Hayk
2,012211
Can you clarify the indexes on your second line of equations?
– user616954
Nov 18 at 17:08
@jdoe, first we multiply all pairs $e^{itheta_k}$ and $e^{-itheta_m}$ where $1leq k,m leq n$. When in these pairs $k=m$, the complex exponential becomes $1$ and we have $n$ such instances of such pairs. Hence, the second sum on the second displayed formula runs over all $k,m$ from the range $1...n$ which does not coincide.
– Hayk
Nov 18 at 17:18
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Observe that a move of length $1$ in the direction $theta$ is a translation by the vector $e^{itheta}$, where $i^2 = -1$.
Thus, if $D: = (X,Y)$ is the position after $n$ moves on the (complex) plane, then
$$
D = e^{itheta_1} + ... + e^{i theta_n} in mathbb{C},
$$
hence
$$
|D|^2 = D overline{D} = sumlimits_{k,m=1}^n e^{i(theta_k - theta_m)} = n + sumlimits_{1leq k neq mleq n}^n e^{i(theta_k - theta_m)}.
$$
It follows that
$$
mathbb{E} |D|^2 = n + sumlimits_{1leq k neq mleq n}^n mathbb{E} e^{i(theta_k - theta_m)} = n + sumlimits_{1leq k neq mleq n}^n mathbb{E}e^{itheta_k} mathbb{E}e^{-i theta_m} = n,
$$
where we used the independence of ${theta_k}$, and the fact that $theta_k sim U[0,2pi]$ for concluding that $mathbb{E}(theta_k) = frac{1}{2pi} intlimits_0^{2pi} e^{ix} dx = 0$.
answered Nov 18 at 5:58
Hayk
2,012211
Can you clarify the indexes on your second line of equations?
– user616954
Nov 18 at 17:08
@jdoe, first we multiply all pairs $e^{itheta_k}$ and $e^{-itheta_m}$ where $1leq k,m leq n$. When in these pairs $k=m$, the complex exponential becomes $1$ and we have $n$ such instances of such pairs. Hence, the second sum on the second displayed formula runs over all $k,m$ from the range $1...n$ which does not coincide.
– Hayk
Nov 18 at 17:18
add a comment |
up vote
0
down vote
Observe that a move of length $1$ in the direction $theta$ is a translation by the vector $e^{itheta}$, where $i^2 = -1$.
Thus, if $D: = (X,Y)$ is the position after $n$ moves on the (complex) plane, then
$$
D = e^{itheta_1} + ... + e^{i theta_n} in mathbb{C},
$$
hence
$$
|D|^2 = D overline{D} = sumlimits_{k,m=1}^n e^{i(theta_k - theta_m)} = n + sumlimits_{1leq k neq mleq n}^n e^{i(theta_k - theta_m)}.
$$
It follows that
$$
mathbb{E} |D|^2 = n + sumlimits_{1leq k neq mleq n}^n mathbb{E} e^{i(theta_k - theta_m)} = n + sumlimits_{1leq k neq mleq n}^n mathbb{E}e^{itheta_k} mathbb{E}e^{-i theta_m} = n,
$$
where we used the independence of ${theta_k}$, and the fact that $theta_k sim U[0,2pi]$ for concluding that $mathbb{E}(theta_k) = frac{1}{2pi} intlimits_0^{2pi} e^{ix} dx = 0$.
answered Nov 18 at 5:58
Hayk
2,012211
Can you clarify the indexes on your second line of equations?
– user616954
Nov 18 at 17:08
@jdoe, first we multiply all pairs $e^{itheta_k}$ and $e^{-itheta_m}$ where $1leq k,m leq n$. When in these pairs $k=m$, the complex exponential becomes $1$ and we have $n$ such instances of such pairs. Hence, the second sum on the second displayed formula runs over all $k,m$ from the range $1...n$ which does not coincide.
– Hayk
Nov 18 at 17:18
add a comment |
up vote
0
down vote
up vote
0
down vote
Observe that a move of length $1$ in the direction $theta$ is a translation by the vector $e^{itheta}$, where $i^2 = -1$.
Thus, if $D: = (X,Y)$ is the position after $n$ moves on the (complex) plane, then
$$
D = e^{itheta_1} + ... + e^{i theta_n} in mathbb{C},
$$
hence
$$
|D|^2 = D overline{D} = sumlimits_{k,m=1}^n e^{i(theta_k - theta_m)} = n + sumlimits_{1leq k neq mleq n}^n e^{i(theta_k - theta_m)}.
$$
It follows that
$$
mathbb{E} |D|^2 = n + sumlimits_{1leq k neq mleq n}^n mathbb{E} e^{i(theta_k - theta_m)} = n + sumlimits_{1leq k neq mleq n}^n mathbb{E}e^{itheta_k} mathbb{E}e^{-i theta_m} = n,
$$
where we used the independence of ${theta_k}$, and the fact that $theta_k sim U[0,2pi]$ for concluding that $mathbb{E}(theta_k) = frac{1}{2pi} intlimits_0^{2pi} e^{ix} dx = 0$.
answered Nov 18 at 5:58
Hayk
2,012211
Observe that a move of length $1$ in the direction $theta$ is a translation by the vector $e^{itheta}$, where $i^2 = -1$.
Thus, if $D: = (X,Y)$ is the position after $n$ moves on the (complex) plane, then
$$
D = e^{itheta_1} + ... + e^{i theta_n} in mathbb{C},
$$
hence
$$
|D|^2 = D overline{D} = sumlimits_{k,m=1}^n e^{i(theta_k - theta_m)} = n + sumlimits_{1leq k neq mleq n}^n e^{i(theta_k - theta_m)}.
$$
It follows that
$$
mathbb{E} |D|^2 = n + sumlimits_{1leq k neq mleq n}^n mathbb{E} e^{i(theta_k - theta_m)} = n + sumlimits_{1leq k neq mleq n}^n mathbb{E}e^{itheta_k} mathbb{E}e^{-i theta_m} = n,
$$
where we used the independence of ${theta_k}$, and the fact that $theta_k sim U[0,2pi]$ for concluding that $mathbb{E}(theta_k) = frac{1}{2pi} intlimits_0^{2pi} e^{ix} dx = 0$.
answered Nov 18 at 5:58
Hayk
2,012211
edited Nov 18 at 7:02
answered Nov 18 at 5:58
Hayk
2,012211
answered Nov 18 at 5:58
Hayk
2,012211
answered Nov 18 at 5:58
Hayk
2,012211
2,012211
Can you clarify the indexes on your second line of equations?
– user616954
Nov 18 at 17:08
@jdoe, first we multiply all pairs $e^{itheta_k}$ and $e^{-itheta_m}$ where $1leq k,m leq n$. When in these pairs $k=m$, the complex exponential becomes $1$ and we have $n$ such instances of such pairs. Hence, the second sum on the second displayed formula runs over all $k,m$ from the range $1...n$ which does not coincide.
– Hayk
Nov 18 at 17:18
add a comment |
Can you clarify the indexes on your second line of equations?
– user616954
Nov 18 at 17:08
@jdoe, first we multiply all pairs $e^{itheta_k}$ and $e^{-itheta_m}$ where $1leq k,m leq n$. When in these pairs $k=m$, the complex exponential becomes $1$ and we have $n$ such instances of such pairs. Hence, the second sum on the second displayed formula runs over all $k,m$ from the range $1...n$ which does not coincide.
– Hayk
Nov 18 at 17:18
Can you clarify the indexes on your second line of equations?
– user616954
Nov 18 at 17:08
Can you clarify the indexes on your second line of equations?
– user616954
Nov 18 at 17:08
@jdoe, first we multiply all pairs $e^{itheta_k}$ and $e^{-itheta_m}$ where $1leq k,m leq n$. When in these pairs $k=m$, the complex exponential becomes $1$ and we have $n$ such instances of such pairs. Hence, the second sum on the second displayed formula runs over all $k,m$ from the range $1...n$ which does not coincide.
– Hayk
Nov 18 at 17:18
@jdoe, first we multiply all pairs $e^{itheta_k}$ and $e^{-itheta_m}$ where $1leq k,m leq n$. When in these pairs $k=m$, the complex exponential becomes $1$ and we have $n$ such instances of such pairs. Hence, the second sum on the second displayed formula runs over all $k,m$ from the range $1...n$ which does not coincide.
– Hayk
Nov 18 at 17:18
add a comment |
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– quid♦
Nov 18 at 23:40