Is the average of iid random process independent stationary increments?
If $I_1, I_2, I_3, ...$ is a iid process. Then can I say that
$$M_n = frac{1}{n}sum_{i=1}^n I_i$$
Is a independent stationary increments?
I think it shouldn't, but I am not sure if my answer is right, but essentially I know that
$$M_2 - M_1 = frac{1}{2}I_1 + frac{1}{2}I_2 - I_1 = frac{1}{2}I_2-frac{1}{2}I_1$$
$$M_1-M_0 = I_1$$
And hence
$$M_2 - M_1 = frac{1}{2}I_2 - frac{1}{2}(M_1-M_0)$$
And hence $M_n$ does not have independent increments.
Does that work?
stochastic-processes random-variables
asked Dec 12 '18 at 14:59
AspiringMatAspiringMat
540518
add a comment |
If $I_1, I_2, I_3, ...$ is a iid process. Then can I say that
$$M_n = frac{1}{n}sum_{i=1}^n I_i$$
Is a independent stationary increments?
I think it shouldn't, but I am not sure if my answer is right, but essentially I know that
$$M_2 - M_1 = frac{1}{2}I_1 + frac{1}{2}I_2 - I_1 = frac{1}{2}I_2-frac{1}{2}I_1$$
$$M_1-M_0 = I_1$$
And hence
$$M_2 - M_1 = frac{1}{2}I_2 - frac{1}{2}(M_1-M_0)$$
And hence $M_n$ does not have independent increments.
Does that work?
stochastic-processes random-variables
asked Dec 12 '18 at 14:59
AspiringMatAspiringMat
540518
Indeed the increments of $(M_n)$ are neither independent nor identically distributed.
– Did
Dec 12 '18 at 15:17
add a comment |
If $I_1, I_2, I_3, ...$ is a iid process. Then can I say that
$$M_n = frac{1}{n}sum_{i=1}^n I_i$$
Is a independent stationary increments?
I think it shouldn't, but I am not sure if my answer is right, but essentially I know that
$$M_2 - M_1 = frac{1}{2}I_1 + frac{1}{2}I_2 - I_1 = frac{1}{2}I_2-frac{1}{2}I_1$$
$$M_1-M_0 = I_1$$
And hence
$$M_2 - M_1 = frac{1}{2}I_2 - frac{1}{2}(M_1-M_0)$$
And hence $M_n$ does not have independent increments.
Does that work?
stochastic-processes random-variables
asked Dec 12 '18 at 14:59
AspiringMatAspiringMat
540518
If $I_1, I_2, I_3, ...$ is a iid process. Then can I say that
$$M_n = frac{1}{n}sum_{i=1}^n I_i$$
Is a independent stationary increments?
I think it shouldn't, but I am not sure if my answer is right, but essentially I know that
$$M_2 - M_1 = frac{1}{2}I_1 + frac{1}{2}I_2 - I_1 = frac{1}{2}I_2-frac{1}{2}I_1$$
$$M_1-M_0 = I_1$$
And hence
$$M_2 - M_1 = frac{1}{2}I_2 - frac{1}{2}(M_1-M_0)$$
And hence $M_n$ does not have independent increments.
Does that work?
stochastic-processes random-variables
stochastic-processes random-variables
asked Dec 12 '18 at 14:59
AspiringMatAspiringMat
540518
asked Dec 12 '18 at 14:59
AspiringMatAspiringMat
540518
asked Dec 12 '18 at 14:59
AspiringMatAspiringMat
540518
asked Dec 12 '18 at 14:59
AspiringMatAspiringMat
540518
asked Dec 12 '18 at 14:59
AspiringMatAspiringMat
540518
540518
Indeed the increments of $(M_n)$ are neither independent nor identically distributed.
– Did
Dec 12 '18 at 15:17
add a comment |
Indeed the increments of $(M_n)$ are neither independent nor identically distributed.
– Did
Dec 12 '18 at 15:17
Indeed the increments of $(M_n)$ are neither independent nor identically distributed.
– Did
Dec 12 '18 at 15:17
Indeed the increments of $(M_n)$ are neither independent nor identically distributed.
– Did
Dec 12 '18 at 15:17
add a comment |
1 Answer
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Yes that works. To give slightly more detail, suppose that $M_2-M_1$ were independent of $M_1-M_0$. Since $I_2$ is independent of $M_1-M_0$, this would imply that $-I_2+2(M_2-M_1)=M_1-M_0$ is also independent of $M_1-M_0$. But the only way a random variable can be independent of itself is if it is constant a.s. So the claim holds except in the degenerate case where $I_1$ is a.s. constant.
answered Dec 12 '18 at 15:17
Mike HawkMike Hawk
1,500110
add a comment |
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1 Answer
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Yes that works. To give slightly more detail, suppose that $M_2-M_1$ were independent of $M_1-M_0$. Since $I_2$ is independent of $M_1-M_0$, this would imply that $-I_2+2(M_2-M_1)=M_1-M_0$ is also independent of $M_1-M_0$. But the only way a random variable can be independent of itself is if it is constant a.s. So the claim holds except in the degenerate case where $I_1$ is a.s. constant.
answered Dec 12 '18 at 15:17
Mike HawkMike Hawk
1,500110
add a comment |
Yes that works. To give slightly more detail, suppose that $M_2-M_1$ were independent of $M_1-M_0$. Since $I_2$ is independent of $M_1-M_0$, this would imply that $-I_2+2(M_2-M_1)=M_1-M_0$ is also independent of $M_1-M_0$. But the only way a random variable can be independent of itself is if it is constant a.s. So the claim holds except in the degenerate case where $I_1$ is a.s. constant.
answered Dec 12 '18 at 15:17
Mike HawkMike Hawk
1,500110
add a comment |
Yes that works. To give slightly more detail, suppose that $M_2-M_1$ were independent of $M_1-M_0$. Since $I_2$ is independent of $M_1-M_0$, this would imply that $-I_2+2(M_2-M_1)=M_1-M_0$ is also independent of $M_1-M_0$. But the only way a random variable can be independent of itself is if it is constant a.s. So the claim holds except in the degenerate case where $I_1$ is a.s. constant.
answered Dec 12 '18 at 15:17
Mike HawkMike Hawk
1,500110
Yes that works. To give slightly more detail, suppose that $M_2-M_1$ were independent of $M_1-M_0$. Since $I_2$ is independent of $M_1-M_0$, this would imply that $-I_2+2(M_2-M_1)=M_1-M_0$ is also independent of $M_1-M_0$. But the only way a random variable can be independent of itself is if it is constant a.s. So the claim holds except in the degenerate case where $I_1$ is a.s. constant.
answered Dec 12 '18 at 15:17
Mike HawkMike Hawk
1,500110
answered Dec 12 '18 at 15:17
Mike HawkMike Hawk
1,500110
answered Dec 12 '18 at 15:17
Mike HawkMike Hawk
1,500110
answered Dec 12 '18 at 15:17
Mike HawkMike Hawk
1,500110
1,500110
add a comment |
add a comment |
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Indeed the increments of $(M_n)$ are neither independent nor identically distributed.
– Did
Dec 12 '18 at 15:17