Let $p_{k+1}=a_kp_k+b_kp_{k-1}, \ k \in \mathbb{N}$ be a recurrence relation with coefficients $a_k, b_k \in \mathbb{R}$ and start values $p_0, p_1 \in \mathbb{R}$.
So the values of starting and the coefficients are bugged:
$\tilde{p}_0=p_0(1+\delta_0), \ \tilde{p}_1=p_1(1+\delta_1), \ \tilde{a}_k=a_k(1+\alpha_k), \ \tilde{b}_k=b_k(1+\beta_k)$
So bugged values $\tilde{p}_k$ emerge.
How to find a recurrence relation for the absolute and relative error:
$\Delta_{Abs}(p_{k+1}):=\tilde{p}_{k+1}-p_{k+1}, \ \Delta_{Rel}(p_{k+1}):=\frac{\Delta_{abs}(p_{k+1})}{p_{k+1}}$?
I used this for the absolute error:
$\Delta(p_{k+1})=a_k\Delta p_k+b_k\Delta p_{k-1}$
So with the start values $\Delta p_0=\delta_0 p_0$ and $\Delta p_1= \delta_1p_1$ I get the recursion:
$\alpha_ka_k\tilde{p}_{k-1}+\beta_kb_k\tilde{p}_{k-1}=\alpha_ka_kp_k+\beta_kb_kp_{k-1}$
Same for the relative error.
Is this recurrence relation correct or has it to be done differently?
You forgot to consider the absolute error of the coefficients: $$\Delta(p_{k+1})=a_k\Delta p_k +b_k\Delta p_{k-1}+p_k\Delta a_k+p_{k-1}\Delta b_k.$$
The problem with the relative error is the possible cancellation hidden in the addition (while for the single summands, the relative error of $a_kp_k$ is just the sum of the relative errors of the factors). If we knew for example that all numbers involved are positive, we'd be in a much better shape, namely, $\max\{\Delta_{rel}(a_k)+\Delta_{rel}(p_k),\Delta_{rel}(b_k)+\Delta_{rel}(p_{k-1})\}$, and this is readily bounded by $(k+1)\delta$ (induction on $k$) if all involved numbers have relative error $\le\delta$.