问题如下：
Why does a unit root with a time trend,
not depend explicitly on t?
解释：
The time trend becomes apparent as the series is propagated backwards:
这题的回答啥意思？没看懂
NO.PZ2020011101000021问题如下Why es a unit root with a time tren Yt=δ1+Yt−1+ϵtY_t = \lta_1 + Y_{t - 1} + \epsilon_tYt=δ1+Yt−1+ϵtnot penexplicitly on t?The time trenbecomes apparent the series is propagatebackwar:Yt=δ1+Yt−1+ϵt=2δ1+Yt−2+ϵt+ϵt−1=...=tδ1+Y0+∑i=1tϵiY_t = \lta_1 + Y_{t - 1} + \epsilon_t = 2\lta_1 + Y_{t - 2} + \epsilon_t + \epsilon_{t - 1} = ... = t\lta_1 + Y_0 + \sum_{i=1}^t \epsilon_iYt=δ1+Yt−1+ϵt=2δ1+Yt−2+ϵt+ϵt−1=...=tδ1+Y0+∑i=1tϵi这个答案是想说除了t的影响，还受每次波动加总的影响吗
NO.PZ2020011101000021 问题如下 Why es a unit root with a time tren Yt=δ1+Yt−1+ϵtY_t = \lta_1 + Y_{t - 1} + \epsilon_tYt=δ1+Yt−1+ϵtnot penexplicitly on t? The time trenbecomes apparent the series is propagatebackwar:Yt=δ1+Yt−1+ϵt=2δ1+Yt−2+ϵt+ϵt−1=...=tδ1+Y0+∑i=1tϵiY_t = \lta_1 + Y_{t - 1} + \epsilon_t = 2\lta_1 + Y_{t - 2} + \epsilon_t + \epsilon_{t - 1} = ... = t\lta_1 + Y_0 + \sum_{i=1}^t \epsilon_iYt=δ1+Yt−1+ϵt=2δ1+Yt−2+ϵt+ϵt−1=...=tδ1+Y0+∑i=1tϵi not penexplicitly on t?这句话是什么意思？
NO.PZ2020011101000021 问题如下 Why es a unit root with a time tren Yt=δ1+Yt−1+ϵtY_t = \lta_1 + Y_{t - 1} + \epsilon_tYt=δ1+Yt−1+ϵtnot penexplicitly on t? The time trenbecomes apparent the series is propagatebackwar:Yt=δ1+Yt−1+ϵt=2δ1+Yt−2+ϵt+ϵt−1=...=tδ1+Y0+∑i=1tϵiY_t = \lta_1 + Y_{t - 1} + \epsilon_t = 2\lta_1 + Y_{t - 2} + \epsilon_t + \epsilon_{t - 1} = ... = t\lta_1 + Y_0 + \sum_{i=1}^t \epsilon_iYt=δ1+Yt−1+ϵt=2δ1+Yt−2+ϵt+ϵt−1=...=tδ1+Y0+∑i=1tϵi Yt中的变量其实只有t*δ1 和 每一次的随机项了t*δ1 项不是penon t 吗，为什么说not penexplicitly on t
请问什么是unit root
