I was trying to understand the intuition behind this. I understand how the following reccurence was done:
$$X_{t} = \alpha X_{t-1} + C^{2}$$ $$ \alpha X_{t-1} = \alpha X_{t-2} + C^{2} $$ $$ \alpha^{2} X_{t-2} = \alpha X_{t-3} + C^{2}$$ $$... $$ $$\alpha^{t-1} X_{1} = \alpha X_{0} + C^{2}$$
What I don't understand is how do we go from that to this: $$X_{t} = \alpha C^{2} + \alpha^{2} C^{2} + \alpha^{3} C^{2} + ... + \alpha^{t-1} C^{2} + \alpha^{t}X_{0}$$
What's the logic here? Thanks.
On
Note: as written, your expression is slightly wrong, It should be $X_{t} = C^2+\alpha C^{2} + \alpha^{2} C^{2} + \alpha^{3} C^{2} + ... + \alpha^{t-1} C^{2} + \alpha^{t}X_{0}$
Our recurrence is the first order linear recurrence:
$X_k=\alpha X_{k-1} +C^2 \quad\forall k\geq 1 $
In practice, for fixed t we can find the given expression by substituting the recurrence "t times". Formally, we can prove your claim by induction:
The base case is clear: $X_1=C^2+\alpha X_0$
Suppose $X_{t} = C^2 +\alpha C^{2} + \alpha^{2} C^{2} + \alpha^{3} C^{2} + ... + \alpha^{t-1} C^{2} + \alpha^{t}X_{0}$ for some fixed t. Then, by our recurrence,
$X_{t+1}= \alpha X_t +C^2 = C^2 +\alpha(C^2+\alpha C^{2} + \alpha^{2} C^{2} + ... + \alpha^{t-1} C^{2} + \alpha^{t}X_{0}) = C^2 + \alpha C^2 + \cdots \alpha^t C^2 +\alpha^{t+1} X_0$
which is what we wanted to show.
Thus, our claim holds true by mathematical induction.
It must be: $$\begin{align}X_t=&\alpha X_{t-1}+C^2=\\ &=\alpha(\alpha X_{t-2}+C^2)+C^2=\\ &=\alpha^2(\alpha X_{t-3}+C^2)+\alpha C^2+C^2=\cdots=\\ &=\alpha^tX_0+(\alpha^{t-1}+\alpha^{t-2}+\cdots+1)C^2.\end{align}$$