For my statistics class I need to prove that:
$\frac{1}{T^{5/2}} \sum\limits_{t=1}^{T}tY_{t-1} \rightarrow \sigma \int\limits_0^1 rW(r) dr$
where $Y_t$ is a RW process and $W(t)$ is a Brownian Motion process.
I know that $\frac{1}{T^2} \sum\limits_{t=1}^{T}Y_{t-1} = \int\limits_0^1 X_T(r) dr$, i.e. $\frac{1}{T^{3/2}}\sum\limits_{t=1}^{T}Y_{t-1} = \int\limits_0^1\sqrt{T}X_T(r) dr$.
Applying the FCLT yields that $\frac{1}{T^{3/2}}\sum\limits_{t=1}^{T}Y_{t-1} = \int\limits_0^1\sqrt{T}X_T(r) dr \rightarrow \sigma \int\limits_0^1 W(r)dr$, but how do I use this to prove the "extended" version?