Solved – Definition of random walk as a summation of independent random processes

I have a complete beginner question on random walk.
As per this paper

``Random walk – the stochastic process formed by successive summation of independent, identically distributed random variable.... ``

I really cannot get past the first line. I thought that the central limit theorem and law of large numbers state that the mean of a large number of independent random processes will approximate a normal process. But the paper seems to indicate that the summation of these random processes is a random walk? So the mean is a normal process but the sum is a random walk? IS this correct?

Contents

You need to know what a stochastic process is. In this context, it's just a collection of random variables \$(X_0, X_1, X_2, ldots)\$.

Seeing a simple worked example may help. Let's set it up. Suppose you have a collection of independent variables \$mathbf Y = (Y_0, Y_1, ldots)\$, all with the same distribution. For instance, each \$Y_i\$ could represent the flip of a fair coin using (say) \$1\$ for heads and \$0\$ for tails. That's a stochastic process (which we could call a "Bernoulli process").

You can construct new processes out of old. One way is to convert \$mathbf Y\$ into its cumulative sum

\$\$mathbf X = (Y_0, Y_0+Y_1, Y_0+Y_1+Y_2, ldots)\$\$

This is a random walk.

As an example, let's consider a finite random walk of length \$3\$ based on fair coins. That Bernoulli process \$mathbf Y\$ has eight possible outcomes, aka "walks" or "paths," each with equal probabilities of \$1/8\$:

\$\$(0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0), (1,1,1).\$\$

The associated paths of \$mathbf X\$, computed by taking cumulative sums, are therefore

\$\$(0,0,0), (0,0,1), (0,1,1), (0,1,2), (1,1,1), (1,1,2), (1,2,2), (1,2,3).\$\$

If you like, you can now identify the component variables \$X_i\$. For instance, \$X_0\$ takes on the value \$0\$ four times, for a total probability of \$4times 1/8=1/2\$, and the value \$1\$ four times, for a total probability of \$1/2\$. \$X_1\$ takes on the values \$0, 1,\$ and \$2\$ with probabilities \$1/4, 1/2, 1/4\$, respectively. And \$X_2\$ takes on the values \$0,1,2,3\$ with probabilities \$1/8, 3/8, 3/8, 1/8\$, respectively. Notice that these three variables do not have identical distributions. The distributions have different means and variances, too: their means are \$1/2, 1, 3/2\$ (in order) and their variances are \$1/4, 1/2, 3/4\$ (in order).

The component variables in a random walk also are dependent. For instance, given that \$X_1=0\$ (which occurs only in the paths \$(0,0,0)\$ and \$(0,0,1)\$), the chance that \$X_2=0\$ is \$1/2\$. But given that \$X_1=1\$, the chance that \$X_2=0\$ is now zero: it's just not possible. Because these conditional probabilities vary with the value of \$X_1\$, \$X_1\$ and \$X_2\$ are not independent. In fact, no pair of these component variables is independent.

The Central Limit Theorem makes a statement about the distribution of \$X_n\$ when \$n\$ gets very large. Besides assuming the \$Y_i\$ (out of which the \$X_n\$ are constructed) are independent and identically distributed, it has to assume that this common distribution has a finite variance. The concept of a stochastic process is separate from any idea of limits (which wouldn't even make sense for a finite one, as in the example). The CLT holds only for very special processes.

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