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I have a little question, but I don't know that well how to answer it. I have a random walker with position vector $\vec{r} = \sum_{i=1}^N \vec{r}_i$, where i is the random walker's step. Every vector component is a random number, for example $\vec{r}_2 = (x_2,y_2,z_2)$, where $x_2,y_2,z_2$ are random numbers. The steps are discrete and every step has a variable length a.

It is isotropic (it is a given condition in the problem), so I think is easy to know that $<\vec{r}> = \vec{0}$ . Now, I need to know the variances $(\Delta x)^2, (\Delta y)^2, (\Delta z)^2$.

After some operations I got that

$$ \left<|\vec{r}|^2\right> = \sum_{i=j}^N \left<|\vec{r}_i|^2\right> + \sum_{i\neq j}^N \left<\vec{r_i}\cdot \vec{r_j}\right> $$

Since there is not correlation between the steps (it is another given condition in the problem), I think $\sum_{i\neq j}^N \left<\vec{r_i}\cdot \vec{r_j}\right>$ should be 0, but I'm not that sure how to prove it.

I was thinking that, since $E[XY] = E[X]E[Y]$ when X and Y are uncorrelated, I could use this property to prove it, but I don't know how does it work when I have a dot product and vectors. Since $\left<\vec{r}\right> = \vec{0}$, I think I could say $\left<\vec{r_i}\right> = \vec{0}$ and $\left<\vec{r_j}\right> = \vec{0}$, but how can I prove that $\sum_{i\neq j}^N \left<\vec{r_i}\cdot \vec{r_j}\right> = 0$?

Thank you!

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  • $\begingroup$ If I understand the conditions correctly, you're on the right track with $E(xy) = E(x)E(y),$ for $x, y$ independent. Let $X = (x_1, x_2, x_3), Y = (y_1, y_2, y_3),$ so that $X\cdot Y = \sum_{i=1}^3 x_iy_i.$ Then $E(X\cdot Y) = \sum_{i=1}^3 E(x_iy_i) = \sum_{i=1}^3 E(x_i)E(y_i) = 0$ because $E(\vec X) = \vec 0 = (0,0,0).$ // If you meant something different, please edit to explain. $\endgroup$ – BruceET Sep 16 at 23:51
  • $\begingroup$ Thanks a lot for your answer. I've added more information. $\endgroup$ – gengar123 Sep 17 at 0:15
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    $\begingroup$ All random variable independent and with 0 mean. So isn't $E(x_iy_i) = E(x_i)E(y_i) = 0(0) = 0?$ $\endgroup$ – BruceET Sep 17 at 0:19
  • $\begingroup$ Yes, since $E(\vec{X}) = (E(x_1),E(x_2), E(x_3)) = \vec{0} = (0,0,0)$ $\endgroup$ – gengar123 Sep 17 at 0:33
  • $\begingroup$ Hint: isotropy implies the variances are equal, so sum them: you will get the squared distance. Its expectation is a very simple function of the moments of the random increments. $\endgroup$ – whuber Sep 17 at 14:53

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