In a reference, it is stated that
$W^{(n)}(t)=\frac {1}{\sqrt{n}}M_{nt}$
with :
$W^{(n)}(t)$ as scaled random walk and
$M_{nt}=\sum_{j=1}^{nt}X_j$.
Where does $\sqrt{n}$ come from? Would you please explain with its relationship with $\sqrt{\frac{T}{n}}$?
While I'm not entirely sure about your notation here, if the $X_j$ are iid then your $W^n$ have constant variance. This is in turn because of the basic properties
$$\text{Var} \left ( \sum_{j=1}^n X_j \right ) = n \text{Var}(X_1) \\ \text{Var}(cX)=|c|^2 \text{Var}(X)$$
when the $X_j$ are iid and $c$ is a constant. Combining these gives
$$\text{Var} \left ( \frac{1}{\sqrt{n}} \sum_{j=1}^n X_j \right ) = \frac{1}{n} n \text{Var}(X_1)=\text{Var}(X_1).$$
This is probably why you would want the walk to be scaled in that way. This same scaling is used in, for example, the central limit theorem.
