# what is the expected value of the dot product of two vectors

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!

• 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. – BruceET Sep 16 at 23:51
• Thanks a lot for your answer. I've added more information. – gengar123 Sep 17 at 0:15
• All random variable independent and with 0 mean. So isn't $E(x_iy_i) = E(x_i)E(y_i) = 0(0) = 0?$ – BruceET Sep 17 at 0:19
• Yes, since $E(\vec{X}) = (E(x_1),E(x_2), E(x_3)) = \vec{0} = (0,0,0)$ – gengar123 Sep 17 at 0:33
• 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. – whuber Sep 17 at 14:53