Expectation value of the average displacement squared in a random walk
$begingroup$
Consider a simple 1D random walk, with equal probability of going to the right (toward positive x) by one unit of distance and to the left (toward negative x) with one unit of distance. Let x=0 be the initial position of the particle and D be the position of the particle at the end of the walk. If the random walk consist of N steps, then $<D^2>=N$.
So my question is why exactly is$<D^2>$ equal to N. I have seen mathematical proofs as to why this is the case and so I am not really looking for one. I am more looking for an intuitive answer, as my intuition tells me that $<D^2>$ should be 0, since the particle shouldn't really be moving away from the origin as the probability of going to the right or left is the same, so it should really oscillate near the origin. Does anyone have an intuitive reason as to why this is not the case?
Thank you.
probability brownian-motion
asked Dec 26 '18 at 17:26
ShamazShamaz
212
$endgroup$
add a comment |
$begingroup$
Consider a simple 1D random walk, with equal probability of going to the right (toward positive x) by one unit of distance and to the left (toward negative x) with one unit of distance. Let x=0 be the initial position of the particle and D be the position of the particle at the end of the walk. If the random walk consist of N steps, then $<D^2>=N$.
So my question is why exactly is$<D^2>$ equal to N. I have seen mathematical proofs as to why this is the case and so I am not really looking for one. I am more looking for an intuitive answer, as my intuition tells me that $<D^2>$ should be 0, since the particle shouldn't really be moving away from the origin as the probability of going to the right or left is the same, so it should really oscillate near the origin. Does anyone have an intuitive reason as to why this is not the case?
Thank you.
probability brownian-motion
asked Dec 26 '18 at 17:26
ShamazShamaz
212
$endgroup$
1
$begingroup$
You're right that the expected position is zero. However, $D^2$ is the same as $|D|^2$, so you can think of it as a measure of how far from the starting point you are (below or above). Since $D^2$ is always positive, it doesn't really make sense for the expectation to be zero unless $D$ is always 0 (which it is not). As to why it is exactly $N$, I'm also interested in an intuitive explication.
$endgroup$
– tch
Dec 26 '18 at 17:38
add a comment |
$begingroup$
Consider a simple 1D random walk, with equal probability of going to the right (toward positive x) by one unit of distance and to the left (toward negative x) with one unit of distance. Let x=0 be the initial position of the particle and D be the position of the particle at the end of the walk. If the random walk consist of N steps, then $<D^2>=N$.
So my question is why exactly is$<D^2>$ equal to N. I have seen mathematical proofs as to why this is the case and so I am not really looking for one. I am more looking for an intuitive answer, as my intuition tells me that $<D^2>$ should be 0, since the particle shouldn't really be moving away from the origin as the probability of going to the right or left is the same, so it should really oscillate near the origin. Does anyone have an intuitive reason as to why this is not the case?
Thank you.
probability brownian-motion
asked Dec 26 '18 at 17:26
ShamazShamaz
212
$endgroup$
Consider a simple 1D random walk, with equal probability of going to the right (toward positive x) by one unit of distance and to the left (toward negative x) with one unit of distance. Let x=0 be the initial position of the particle and D be the position of the particle at the end of the walk. If the random walk consist of N steps, then $<D^2>=N$.
So my question is why exactly is$<D^2>$ equal to N. I have seen mathematical proofs as to why this is the case and so I am not really looking for one. I am more looking for an intuitive answer, as my intuition tells me that $<D^2>$ should be 0, since the particle shouldn't really be moving away from the origin as the probability of going to the right or left is the same, so it should really oscillate near the origin. Does anyone have an intuitive reason as to why this is not the case?
Thank you.
probability brownian-motion
probability brownian-motion
asked Dec 26 '18 at 17:26
ShamazShamaz
212
asked Dec 26 '18 at 17:26
ShamazShamaz
212
edited Dec 26 '18 at 17:40
Shamaz
asked Dec 26 '18 at 17:26
ShamazShamaz
212
asked Dec 26 '18 at 17:26
ShamazShamaz
212
asked Dec 26 '18 at 17:26
ShamazShamaz
212
212
1
$begingroup$
You're right that the expected position is zero. However, $D^2$ is the same as $|D|^2$, so you can think of it as a measure of how far from the starting point you are (below or above). Since $D^2$ is always positive, it doesn't really make sense for the expectation to be zero unless $D$ is always 0 (which it is not). As to why it is exactly $N$, I'm also interested in an intuitive explication.
$endgroup$
– tch
Dec 26 '18 at 17:38
add a comment |
1
$begingroup$
You're right that the expected position is zero. However, $D^2$ is the same as $|D|^2$, so you can think of it as a measure of how far from the starting point you are (below or above). Since $D^2$ is always positive, it doesn't really make sense for the expectation to be zero unless $D$ is always 0 (which it is not). As to why it is exactly $N$, I'm also interested in an intuitive explication.
$endgroup$
– tch
Dec 26 '18 at 17:38
1
1
$begingroup$
You're right that the expected position is zero. However, $D^2$ is the same as $|D|^2$, so you can think of it as a measure of how far from the starting point you are (below or above). Since $D^2$ is always positive, it doesn't really make sense for the expectation to be zero unless $D$ is always 0 (which it is not). As to why it is exactly $N$, I'm also interested in an intuitive explication.
$endgroup$
– tch
Dec 26 '18 at 17:38
$begingroup$
You're right that the expected position is zero. However, $D^2$ is the same as $|D|^2$, so you can think of it as a measure of how far from the starting point you are (below or above). Since $D^2$ is always positive, it doesn't really make sense for the expectation to be zero unless $D$ is always 0 (which it is not). As to why it is exactly $N$, I'm also interested in an intuitive explication.
$endgroup$
– tch
Dec 26 '18 at 17:38
add a comment |
1 Answer
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$begingroup$
Let $s_i$ denote the steps $pm 1$, with $D=sum s_i$, clearly $E{D}=0$
$E{D^2}=text{Var}(D)=text{Var}(s_1+ldots +s_n) = n text{Var}(s_1) = n$
since all $s_i$ are independent.
answered Dec 26 '18 at 18:18
karakfakarakfa
2,005811
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$begingroup$
Let $s_i$ denote the steps $pm 1$, with $D=sum s_i$, clearly $E{D}=0$
$E{D^2}=text{Var}(D)=text{Var}(s_1+ldots +s_n) = n text{Var}(s_1) = n$
since all $s_i$ are independent.
answered Dec 26 '18 at 18:18
karakfakarakfa
2,005811
$endgroup$
add a comment |
$begingroup$
Let $s_i$ denote the steps $pm 1$, with $D=sum s_i$, clearly $E{D}=0$
$E{D^2}=text{Var}(D)=text{Var}(s_1+ldots +s_n) = n text{Var}(s_1) = n$
since all $s_i$ are independent.
answered Dec 26 '18 at 18:18
karakfakarakfa
2,005811
$endgroup$
add a comment |
$begingroup$
Let $s_i$ denote the steps $pm 1$, with $D=sum s_i$, clearly $E{D}=0$
$E{D^2}=text{Var}(D)=text{Var}(s_1+ldots +s_n) = n text{Var}(s_1) = n$
since all $s_i$ are independent.
answered Dec 26 '18 at 18:18
karakfakarakfa
2,005811
$endgroup$
Let $s_i$ denote the steps $pm 1$, with $D=sum s_i$, clearly $E{D}=0$
$E{D^2}=text{Var}(D)=text{Var}(s_1+ldots +s_n) = n text{Var}(s_1) = n$
since all $s_i$ are independent.
answered Dec 26 '18 at 18:18
karakfakarakfa
2,005811
answered Dec 26 '18 at 18:18
karakfakarakfa
2,005811
answered Dec 26 '18 at 18:18
karakfakarakfa
2,005811
answered Dec 26 '18 at 18:18
karakfakarakfa
2,005811
2,005811
add a comment |
add a comment |
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$begingroup$
You're right that the expected position is zero. However, $D^2$ is the same as $|D|^2$, so you can think of it as a measure of how far from the starting point you are (below or above). Since $D^2$ is always positive, it doesn't really make sense for the expectation to be zero unless $D$ is always 0 (which it is not). As to why it is exactly $N$, I'm also interested in an intuitive explication.
$endgroup$
– tch
Dec 26 '18 at 17:38