I was asking myself the following question but I'm not really sure if it's in this way:
I have a first $PRG$: $$G_1: \{0,1\}^{\lambda} \rightarrow \{0,1\}^{\lambda+1}$$
Now I want to construct a second $PRG$ that is structured as follows:
$$G: \{0,1\}^{\lambda} \rightarrow \{0,1\}^{2\lambda}$$
Such that ($||$ is the concatenation symbol and for $x_i$ I'm assuming it's the $i^{th}$ bit of $x$:
$$G(x) = x_1||x_2||...||x_{\lambda-1}||G_1(x)$$
I need to prove that $G$ is a $PRG$ if $G_1$ is. But I was thinking that if I can show a third $PRG$ such that: $$G_1 (x)=x_1||x_2||...x_{\lambda/2}||G_2(x_{\lambda/2}||x_{(\lambda/2)+1}||...||x_{\lambda})$$
Right now my question is: since I'm assuming $G1$ is a $PRG$, I can't modify it right? I mean I cannot do the stuff I'm trying to do with $G_2$?
If I continue stepping on $PRGs$($G_2, G_3, ...$)in practice it's still a $PRG$ since it is the $x$ selected as input of the function is indistinguishable from a uniform but string right?