Expectation of random 2d walk
up vote
1
down vote
favorite
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
add a comment |
up vote
1
down vote
favorite
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
1
Please do not delete posts after you got an answer.
– quid♦
Nov 18 at 23:40
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
1
Please do not delete posts after you got an answer.
– quid♦
Nov 18 at 23:40
add a comment |
1
Please do not delete posts after you got an answer.
– quid♦
Nov 18 at 23:40
1
1
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.
– quid♦
Nov 18 at 23:40
add a comment |
1 Answer
1
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
1,97729
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
1,97729
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
1,97729
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
1,97729
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
1,97729
edited Nov 18 at 7:02
answered Nov 18 at 5:58
Hayk
1,97729
answered Nov 18 at 5:58
Hayk
1,97729
answered Nov 18 at 5:58
Hayk
1,97729
1,97729
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 |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3003122%2fexpectation-of-random-2d-walk%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
Please do not delete posts after you got an answer.
– quid♦
Nov 18 at 23:40