Autocorrelation function of a random walk process What is the intuition behind the result that the autocorrelation function of a random walk process $y_{t}=y_{t-1}+e_{t}$ tends to 1 as $t\rightarrow 0$? Thank you.
By recursive substitution you obtain $y_t = \sum_{j=0}^t e_{t-j}.$ Thus $\forall p \in \mathbb{N} \hspace{.2cm} y_{t-p} = \sum_{j=0}^{t-p} e_{t-p-j}.$ 

Under the usual white noise assumption for the errors $i \neq j \Rightarrow E(e_je_i) =0\Rightarrow Cov(y_t, y_{t-p}) = E(\sum_{j=0}^{t-p} e_{t-p-j}^2) = (t-p+1)\sigma_e^2.$

$Var(y_t) = E(\sum_{j=0}^{t} e_{t}^2) = (t+1)\sigma_e^2 \hspace{.2cm} \wedge Var(y_{t-p}) = E(\sum_{j=0}^{t-p} e_{t-p-j}^2)(t-p+1)\sigma_e^2.$

$\Rightarrow \rho_t(p) = \frac{(t-p+1)\sigma_e^2}{\sqrt{(t+1)\sigma_e^2}\sqrt{(t-p+1)\sigma_e^2}} = \frac{\sqrt{(t-p+1)}}{\sqrt{(t+1)}}.$

$\Rightarrow lim_{p\rightarrow 0} \rho_t(p) = 1 \hspace{.2cm} \wedge \Rightarrow lim_{p\rightarrow t} \rho_t(p) =  \frac{1}{\sqrt{(t+1)}}$.

$\Rightarrow$ for fixed $p$ $lim_{t\rightarrow \infty} \rho_t(p) = lim_{t\rightarrow \infty} \frac{\sqrt{(t-p+1)}}{\sqrt{(t+1)}}= 1$