Understanding the Variance of a One-Dimensional Random Walk

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The expectation E(Sn) of a one-dimensional simple random walk is zero, while the variance is given by E(Sn²) = n. This can be proven by recognizing that S_n is the sum of independent random variables Z_i, each with mean zero and variance one. The correct formulation for S_n² involves the sum of products of these variables, but the independence allows for the calculation of E(Sn²) to still yield n. Thus, the variance of a one-dimensional random walk is confirmed to be n.
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Hi,

I know that the expectation E(Sn) for a one-dimensional simple random walk is zero. But what about the variance?

I read in http://en.wikipedia.org/wiki/Random_walk#One-dimensional_random_walk" that the variance should be E(Sn2) = n.

Why is that? Can anyone prove it?

Thank you very much!
 
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Just write down the definition of S_n and you will be able to answer your question yourself.
 
Var(Sn) = E(Sn2) = E(Z12 + Z22 + Z32 + ... + Zn2) =* E(Z12) + E(Z22) + ... + E(Zn2) = 1 + 1 + ... + 1 (n times) = n

*variables are independent and uncorrelated

Is this correct then?
 
This is almost correct. S_n is defined to be Z_1+\ldots +Z_n, where the Z_i are independent (or at least uncorrelated) with mean zero and variance one. It follows that
<br /> S_n^2 = \sum_{i,j=1}^n{Z_i Z_j}<br />
and not, as you wrote,
<br /> S_n^2 = \sum_{i=1}^n{Z_i^2}<br />

However, using independence of the Z_i you can still do a similar computation to prove \mathbb{E}\left[S_n^2\right]=n.
 
Thank you!
 
You're welcome:smile:
 
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