prove that this expression converges in probability to zero Apparently $T^{-3/2} \sum\limits_{t=1}^T{y_{t-1}}u_t$ converges by law to $0.5\times T^{-1/2}\sigma^2 (X-1)$, where $X$ is a $\chi^2(1)$ random variable. $u_t$ is white noise and $y_t$ is an AR(1) process, that is $y_t=y_{t-1}+u_t$ with $u_t\sim iid N(0,\sigma^2)$.
I want to prove that the above expression also converges in probability to zero.
Now it appears the variance of this expression is of order $1/T$. I deduct this by noting that the variance of a a $\chi^2$ variable is finite and we have something finite divided by $\sqrt{T}$. This implies that the variance of the whole expression converges to zero. If I am not mistaken, this then also implies that the above expression converges in probability to its mean (?)  
Now am a little unsure now how to obtain the mean of this assymptotic expression. 
To prove  $0.5\times T^{-1/2}\sigma^2 (X-1)$ converges in probability to zero, can I simply take the expectation this expression by assumging that the expecation of a $\chi^2(n)$ variable is $n$? That would be then $0.5\times T^{-1/2}\sigma^2 (1-1)=0$. Therefore we have $0.5\times T^{-1/2}\sigma^2 (X-1)\rightarrow_p0$? Is this rigorous enough?
 A: We have the standard random-walk model
$$y_t = \rho y_{t-1} + u_t,\;\; y_0=u_0=0$$ and a sample of $T$ observations on $y$.
Under the null hypothesis that $\rho =1$, we have
$$y_t^2 = (y_{t-1} + u_t)^2 = y_{t-1}^2 + 2y_{t-1}u_t + u_t^2$$
$$\Rightarrow y_{t-1}u_t  =\frac 12(y_t^2-y_{t-1}^2-u_t^2)$$
Summing over $T$ we obtain
$$\sum_{t=1}^T{y_{t-1}}u_t = \frac 12 (y_T^2 - y_0^2) - \frac 12\sum_{t=1}^Tu_t^2=\frac 12 y_T^2 - \frac 12\sum_{t=1}^Tu_t^2$$
Divide both sides by $\sigma^2T$ and manipulate to obtain
$$\frac 1{\sigma^2T}\sum_{t=1}^T{y_{t-1}}u_t = \frac 1{2} \left(\frac{y_T}{\sigma\sqrt T}\right)^2 - \frac 1{2\sigma^2T}\sum_{t=1}^Tu_t^2$$
Since under $\rho=1$ we have $y_T = \sum_{t=1}^Tu_t$, $y_T$ is a normal random variable with mean zero and variance $\sigma^2T$. So, as an exact result, $\left(\frac{y_T}{\sigma\sqrt T}\right)^2$ is a chi-square with one degree of freedom, and this does not depend on the sample size. Call this random variable $Z$. Divide both sides further by $\sqrt T$ and re-arrange to obtain
$$\frac 1{T^{3/2}}\sum_{t=1}^T{y_{t-1}}u_t = \frac {\sigma^2}{2\sqrt{T}}Z - \frac 1{2T^{3/2}}\sum_{t=1}^Tu_t^2$$
Convergence to zero in probability means that the following must hold
$$\lim_{T\rightarrow \infty}\text {Pr}\left(\Big|\frac {\sigma^2}{2\sqrt{T}}Z - \frac 1{2T^{3/2}}\sum_{t=1}^Tu_t^2\Big|>\epsilon\right) =0, \;\; \forall \epsilon >0$$
$$\Rightarrow \lim_{T\rightarrow \infty}\text {Pr}\left(\Big|\frac {\sigma^2}{2}Z - \frac 1{2T}\sum_{t=1}^Tu_t^2\Big|>\sqrt T\epsilon\right) =0$$
$\frac {\sigma^2}{2}Z$ remains finite as $T\rightarrow \infty$ since $Z$ remains a chi-square with one d.f. (hence finite variance). Also, $\frac 1{2T}\sum_{t=1}^Tu_t^2$ goes to $(1/2)\sigma^2$, so over all the absolute value goes to a finite quantity, while the term $\sqrt T\epsilon$ explodes. Therefore the limit of the probability of this event is indeed zero, and convergence in probability is established.
Note that this is the quantity for which you want to establish a zero plim, not for its limiting distribution.
