Probabilities: The mean of the square

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SUMMARY

The discussion focuses on calculating the mean of the square of a particle's displacement after n random moves in a two-dimensional space. The particle can move in four directions represented by vectors r_1 = (a,0), r_2 = (-a,0), r_3 = (0,a), and r_4 = (0,-a). The initial calculation by the user, = ∑(r_i·r_i)P_i, incorrectly assumes only one move, leading to a mean of a². The correct approach, as clarified by a peer, incorporates n moves, resulting in the mean of the square of r being na².

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Niles
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Homework Statement


Hi all.

A particle can choose randomly to move in one of these directions:

[tex] r_1 = (a,0), \quad r_2 = (-a,0), \quad r_3 = (0,a) \quad \text{and}\quad r_4 = (0,-a).[/tex]

These are vectors, not coordinates! I have to find the mean of the square of r, i.e. [itex]<r>[/itex] after n moves, where the particle starts in (0,0).

What I have done is the following:

[tex] <r^2> = \sum_i {(r_i\cdot r_i)P_i},[/tex]

where Pi is 1/4, because it is random. So I believe the mean of the square of r is a2. But my teacher says it is na2. I cannot see why he wants to multiply by n, since my method is quite straightforward. Where am I wrong?

Thanks in advance.


Niles.
 
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You got to have an n somewhere, otherwise, how do you incorporate the fact that it is after n moves? Your expression currently find mean of r^2 after the first step.

For n moves, you'll have n terms and each step has a different possibility. You should think about how to arrive at the corresponding expression.

Hint:
[tex]\langle r^2 \rangle=\overbrace{\sum ... \sum}^{\rm{n times}}(r_1+r_2+...+r_n)^2 \times \rm{Probability}[/tex]

Now, how would you simplify that?
 
Thanks, I see it now.
 

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