Start by defining the Symmetric Random Walk:

$$ M_t = \sum_{i=1}^{t}X_i, ~~ \text{with}~X_0=0 $$

where $X_i$ is equal to 1 or -1 with $p=(1-p)=0.5$.

Consider $t > s$, the variance of its increments is:

\begin{equation} \begin{split} VAR(M_t - M_s) & = VAR \Big( \sum_{i=s+1}^tX_i\Big)~\text{by indepence of} ~ X_i \\ & = \sum_{i=s+1}^t VAR(X_i) = t-s \end{split} \end{equation}

Now define the Scaled Symmetric Random Walk:

$$ W^{(n)}(t) = \begin{cases} \dfrac{1}{\sqrt{n}}M_{nt}, & \mbox{if }~t=\dfrac{m}{n}~\in~ \mathbb{N} \\ \mbox{linear interpolation}, & \mbox{if } t~\in~ \Big( \dfrac{m}{n}, \dfrac{m+1}{n} \Big) \end{cases} $$

Now suppose $~t > s~$ and $~t = l/n~$ and $~s = m/n~$ are integers, now how is possible to obtain the following variance?

$$ VAR(W^{n}(t) - W^{n}(s)) $$

I tried this way:

\begin{equation} \begin{split} VAR(W^{n}(t) - W^{n}(s)) & = VAR\Big( \dfrac{1}{\sqrt{n}} (M_l - M_m) \Big) \\ & = \dfrac{1}{n}VAR(M_l - M_m) \end{split} \end{equation}

Then by applying the passages of the normal random walk reported above, I obtain:

$$ VAR(W^{n}(t) - W^{n}(s)) = \dfrac{1}{n} (t-s) $$

But I know that the correct formula should be $t-s$. Where is my error?


2 Answers 2


You forget an $n$ in the index. After "I tried this way", second equivalence, you should apply:

$$ M_l - M_m = M_{tn} - M_{sn} $$

  • $\begingroup$ No because it is $M_{tn}$ that becomes $M_{l}$ since $t = l/n$. $\endgroup$
    – Archimede
    Commented Aug 30, 2017 at 14:14
  • $\begingroup$ Okay then the problem is later, see above. $\endgroup$
    – Gijs
    Commented Aug 30, 2017 at 14:25

In the last step of the derivation, you applied the preliminary result incorrectly. The correct substitution is $l$ and $m$ instead of $t$ and $s$. \begin{align} \frac{1}{n}\operatorname{Var}(M_l - M_m) = \frac{1}{n}(l - m). \tag{1} \end{align} Now substitute $l = tn,~ m = sn$ back to $(1)$ yields the desired formula.

  • $\begingroup$ I edited your post by using the \operatorname as it was only used once and subsequently removed the space at the beginning (& +1). $\endgroup$ Commented Feb 24, 2023 at 14:52
  • $\begingroup$ Thank you. I have been noticing the extra space or line at the beginning when preamble is used for a while. Don't like it either. $\endgroup$
    – Zhanxiong
    Commented Feb 24, 2023 at 19:24

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