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My professor explained the math behind a simple random walk process but I was unable to follow one of the steps.

The first equation was: Xt=(Xt-1)+et

He then said "we need to back substitute in for Xt. This resulted in:

Xt=Xt+(et-1)+et

Can someone explain the step by step process of this "back substitution"?

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  • $\begingroup$ Can you clarify the subscripts and superscripts in your formula? $\endgroup$ Apr 7, 2015 at 4:15
  • $\begingroup$ Sorry about that. I just figured out how to do that in html. I edited the question with actual subscripts. $\endgroup$
    – Amaziah
    Apr 7, 2015 at 4:28

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Backward substitution is a method for solving certain recurrence relations. A recurrence relation is an equation defined by itself (i.e. recursively). For an example, take the Fibonacci numbers. The $n$-th Fibonacci number is defined as the sum of it's two predecessors. That is, $F_n = F_{n-1} + F_{n-2}. (\star)$

Now perhaps we want to find an expression for $F_{n-1}$. Well, we can just substitute $n\mapsto (n-1)$ into the LHS of $(\star)$. Then $F_{\color{red}{n-1}} = F_{\color{red}{n-1} -1} + F_{\color{red}{n-1} - 2}.$ That is, $F_{n-1}= F_{n-2} + F_{n-3}.$

Similarly, notice that our $X_t$ is defined in terms of itself. That is, $X_t = \color{blue}{X_{t-1}} + e_t. (\circ)$

Then we may find an expression for $X_{t-1}$ by substituting $t\mapsto (t-1)$ into $X_t$ on the LHS of $(\circ)$.

That is, $X_\color{red}{{t-1}} = X_{\color{red}{{t-1}}-1} + e_{\color{red}{{t-1}}}$.

Putting this back into $(\circ)$, we have $X_t = \color{blue}{X_{t-2} + e_{t-1}} + e_t$.

Doing this backward substitution allows us to see a pattern in the recurrence relation, perhaps finding a formula for a general $X_{t-k}$. Then, with the set of initial conditions, we will be able to find a formula for $X_t$ that doesn't rely on itself.

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    $\begingroup$ I see. Thank you very much. $\endgroup$
    – Amaziah
    Apr 7, 2015 at 4:33

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