Number of times a constrained, discrete random walk changes value
$begingroup$
I have two infinitely long series, $X$ and $Y$, and a constant, $n$.
Each element in $Y$ is either $+1$ or $-1$, with equal probability. $X$ changes as follows:
$$X_i = max(-n, min(n, X_{i-1} + Y_{i-1}))$$
In words: $X$ goes up or down by 1 each step, unless this would bring it to be outside the interval $[-n, n]$.
What I now want to do is count the expected number of times $X$ changes value. Let's call this $mathbb{E}[k]$.
Of course, if $X$ is infinitely long, this will lead to $mathbb{E}[k] = infty$.
However, this value will still be different for different values of $X_0$. (For example, if $X_0=n$ then there's a $1/2$ probability that $X_1=n$ as well, so it won't have changed.) For ease of notation, let $f$ be a function such that $f(z) = mathbb{E}[k|X_0=z]$.
Then we'll have that, in general: $f(a) neq f(b)$. Furthermore, it's clear that the difference between the two has to be finite.
How do I express the difference between these two values? That is, what is $f(a) - f(b)$? (Assume that $-n le a le n)$ and same for $b$.)
I tried solving this by considering the upper limit, $n$: if $X_0=n$ then there's a $1/2$ probability that, $X_1$ is going to change, to $n-1$, and a $1/2$ probability that it'll stay the same. Keeping in mind that $X$ is infinitely long, I get:
$$f(n) = frac{1}{2} (f(n-1) + 1) + frac{1}{2}f(n)$$
From this it follows that $f(n)=f(n-1)+1$. Looks alright. But repeating this for the next step results in:
$$f(n-1) = frac{1}{2} (f(n-2) + 1) + frac{1}{2}(f(n) + 1)$$
$$ = frac{1}{2} (f(n-2) + 1) + frac{1}{2}(f(n-1) + 2)$$
$$ = f(n-2) + 3$$
But this can't be correct: the further away from $n$, the flatter $f$ should be getting. Furthermore, the closer to $n$, the lower the result should be, not higher.
limits random-walk
asked Dec 18 '18 at 9:16
acdracdr
1356
$endgroup$
add a comment |
$begingroup$
I have two infinitely long series, $X$ and $Y$, and a constant, $n$.
Each element in $Y$ is either $+1$ or $-1$, with equal probability. $X$ changes as follows:
$$X_i = max(-n, min(n, X_{i-1} + Y_{i-1}))$$
In words: $X$ goes up or down by 1 each step, unless this would bring it to be outside the interval $[-n, n]$.
What I now want to do is count the expected number of times $X$ changes value. Let's call this $mathbb{E}[k]$.
Of course, if $X$ is infinitely long, this will lead to $mathbb{E}[k] = infty$.
However, this value will still be different for different values of $X_0$. (For example, if $X_0=n$ then there's a $1/2$ probability that $X_1=n$ as well, so it won't have changed.) For ease of notation, let $f$ be a function such that $f(z) = mathbb{E}[k|X_0=z]$.
Then we'll have that, in general: $f(a) neq f(b)$. Furthermore, it's clear that the difference between the two has to be finite.
How do I express the difference between these two values? That is, what is $f(a) - f(b)$? (Assume that $-n le a le n)$ and same for $b$.)
I tried solving this by considering the upper limit, $n$: if $X_0=n$ then there's a $1/2$ probability that, $X_1$ is going to change, to $n-1$, and a $1/2$ probability that it'll stay the same. Keeping in mind that $X$ is infinitely long, I get:
$$f(n) = frac{1}{2} (f(n-1) + 1) + frac{1}{2}f(n)$$
From this it follows that $f(n)=f(n-1)+1$. Looks alright. But repeating this for the next step results in:
$$f(n-1) = frac{1}{2} (f(n-2) + 1) + frac{1}{2}(f(n) + 1)$$
$$ = frac{1}{2} (f(n-2) + 1) + frac{1}{2}(f(n-1) + 2)$$
$$ = f(n-2) + 3$$
But this can't be correct: the further away from $n$, the flatter $f$ should be getting. Furthermore, the closer to $n$, the lower the result should be, not higher.
limits random-walk
asked Dec 18 '18 at 9:16
acdracdr
1356
$endgroup$
add a comment |
$begingroup$
I have two infinitely long series, $X$ and $Y$, and a constant, $n$.
Each element in $Y$ is either $+1$ or $-1$, with equal probability. $X$ changes as follows:
$$X_i = max(-n, min(n, X_{i-1} + Y_{i-1}))$$
In words: $X$ goes up or down by 1 each step, unless this would bring it to be outside the interval $[-n, n]$.
What I now want to do is count the expected number of times $X$ changes value. Let's call this $mathbb{E}[k]$.
Of course, if $X$ is infinitely long, this will lead to $mathbb{E}[k] = infty$.
However, this value will still be different for different values of $X_0$. (For example, if $X_0=n$ then there's a $1/2$ probability that $X_1=n$ as well, so it won't have changed.) For ease of notation, let $f$ be a function such that $f(z) = mathbb{E}[k|X_0=z]$.
Then we'll have that, in general: $f(a) neq f(b)$. Furthermore, it's clear that the difference between the two has to be finite.
How do I express the difference between these two values? That is, what is $f(a) - f(b)$? (Assume that $-n le a le n)$ and same for $b$.)
I tried solving this by considering the upper limit, $n$: if $X_0=n$ then there's a $1/2$ probability that, $X_1$ is going to change, to $n-1$, and a $1/2$ probability that it'll stay the same. Keeping in mind that $X$ is infinitely long, I get:
$$f(n) = frac{1}{2} (f(n-1) + 1) + frac{1}{2}f(n)$$
From this it follows that $f(n)=f(n-1)+1$. Looks alright. But repeating this for the next step results in:
$$f(n-1) = frac{1}{2} (f(n-2) + 1) + frac{1}{2}(f(n) + 1)$$
$$ = frac{1}{2} (f(n-2) + 1) + frac{1}{2}(f(n-1) + 2)$$
$$ = f(n-2) + 3$$
But this can't be correct: the further away from $n$, the flatter $f$ should be getting. Furthermore, the closer to $n$, the lower the result should be, not higher.
limits random-walk
asked Dec 18 '18 at 9:16
acdracdr
1356
$endgroup$
I have two infinitely long series, $X$ and $Y$, and a constant, $n$.
Each element in $Y$ is either $+1$ or $-1$, with equal probability. $X$ changes as follows:
$$X_i = max(-n, min(n, X_{i-1} + Y_{i-1}))$$
In words: $X$ goes up or down by 1 each step, unless this would bring it to be outside the interval $[-n, n]$.
What I now want to do is count the expected number of times $X$ changes value. Let's call this $mathbb{E}[k]$.
Of course, if $X$ is infinitely long, this will lead to $mathbb{E}[k] = infty$.
However, this value will still be different for different values of $X_0$. (For example, if $X_0=n$ then there's a $1/2$ probability that $X_1=n$ as well, so it won't have changed.) For ease of notation, let $f$ be a function such that $f(z) = mathbb{E}[k|X_0=z]$.
Then we'll have that, in general: $f(a) neq f(b)$. Furthermore, it's clear that the difference between the two has to be finite.
How do I express the difference between these two values? That is, what is $f(a) - f(b)$? (Assume that $-n le a le n)$ and same for $b$.)
I tried solving this by considering the upper limit, $n$: if $X_0=n$ then there's a $1/2$ probability that, $X_1$ is going to change, to $n-1$, and a $1/2$ probability that it'll stay the same. Keeping in mind that $X$ is infinitely long, I get:
$$f(n) = frac{1}{2} (f(n-1) + 1) + frac{1}{2}f(n)$$
From this it follows that $f(n)=f(n-1)+1$. Looks alright. But repeating this for the next step results in:
$$f(n-1) = frac{1}{2} (f(n-2) + 1) + frac{1}{2}(f(n) + 1)$$
$$ = frac{1}{2} (f(n-2) + 1) + frac{1}{2}(f(n-1) + 2)$$
$$ = f(n-2) + 3$$
But this can't be correct: the further away from $n$, the flatter $f$ should be getting. Furthermore, the closer to $n$, the lower the result should be, not higher.
limits random-walk
limits random-walk
asked Dec 18 '18 at 9:16
acdracdr
1356
asked Dec 18 '18 at 9:16
acdracdr
1356
edited Dec 18 '18 at 9:23
acdr
asked Dec 18 '18 at 9:16
acdracdr
1356
asked Dec 18 '18 at 9:16
acdracdr
1356
asked Dec 18 '18 at 9:16
acdracdr
1356
1356
add a comment |
add a comment |
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