When does a random walk's spread reach a target?

Question

A random walk is Xt = X{t-1} + \varepsilont with X0 = 0 and shock variance \sigma^2 = 9. At what time t does the standard deviation of Xt first equal 15?

Accepted Answer

The variance of a random walk at time t is \operatorname{Var}(Xt) = t\,\sigma^2 = 9t, so its standard deviation is \sqrt{9t} = 3\sqrt{t}. Set that equal to 15: 3\sqrt{t} = 15 \quad\Longrightarrow\quad \sqrt{t} = 5 \quad\Longrightarrow\quad t = 25. So it takes 25 steps for the spread to reach 15. Note the standard deviation grows only as \sqrt{t}, not linearly, so pushing the spread higher takes progressively more time. Takeaway. A random walk's standard deviation grows like \sigma\sqrt{t}; inverting that square-root law tells you exactly when the spread reaches any target, here at t = 25.

When does a random walk's spread reach a target?

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