AR(1) process: Finding the distribution of the prediction error I have the AR(1) process given by
$$X_t = \rho X_{t-1} + \varepsilon_t, \ \ \ t=1,\ldots,T+h,$$
where the random variables $\varepsilon_t$ are independent, 
$\varepsilon_t \sim N(0,\sigma^2)$, and $X_0$ is fixed and known. We
assume that we have data 
$X_1, \ldots, X_T$. The best forecast of $X_{T+h}$ (given the data)
is $\hat X_{T+h} = \hat \rho^h_T X_T$, where $\hat \rho_T$ is the
maximum likelihood estimator of $\rho$ based on
the sample $X_1,\ldots,X_T$.
(a) From the first assignment we know that when $|\rho|<1$, then
$$\sqrt{T}(\hat \rho_T^h - \rho^h)\stackrel{D}{\longrightarrow} 
N \left( 0,(1-\rho^2)(h \rho^{h-1})^2 \right)$$
as $T \rightarrow \infty$. Now assume instead that $\rho = 1$ and use a Taylor
expansion to show that
$$
T(\hat \rho^h_T - 1) \stackrel{D}{\longrightarrow} h U
$$
as $T \rightarrow \infty$, where
$
U = \frac{\int_0^1 W_u dW_u}{\int_0^1 W_u^2 du}.
$
Here $W$ denotes a standard Brownian motion on $[0,1]$.
I have solved exercse a). The next exercise is the one I am having troubles with. 
(b) Use question (a) and the results from the first assignment to
argue that for large values of $T$, the conditional distribution of 
$\hat X_{T+h} - X_{T+h}$ given $X_T$ is approximately equal to the 
distribution of
$$
\sqrt{h \sigma^2} V + \frac{h X_T}{T} U,
$$
where $U$ is as in question (a), and $V$ is independent of $U$ and $V
\sim N(0,1)$.
The information that is reffered to from the previous exercise is: 
From the previous assignment we have that 
$$\left(\hat X_{T+h} - X_{T+h}\right) \vert X_T = \left((\hat \rho^h_T - \rho^h)X_T - \sum_{i=0}^{h-1}
\rho^i \varepsilon_{T+h-i}\right) \vert X_T$$

I was over complicating it, here is my answer if anyone should be interested. 
From the previous assignment we have that 
$$\left(\hat X_{T+h} - X_{T+h}\right) \vert X_T = \left((\hat \rho^h_T - \rho^h)X_T - \sum_{i=0}^{h-1}
\rho^i \varepsilon_{T+h-i}\right) \vert X_T$$
with $\rho = 1$ this becomes 
$$\left(\hat X_{T+h} - X_{T+h}\right) \vert X_T = \left((\hat \rho^h_T - 1)X_T - \sum_{i=0}^{h-1} \varepsilon_{T+h-i}\right) \vert X_T\\
= (\hat \rho^h_T - 1)X_T - \sum_{i=0}^{h-1} \varepsilon_{T+h-i} \vert X_T.$$
We have that $T(\hat \rho^h_T - 1) \stackrel{D}{\longrightarrow}  hU$, hence $(\hat \rho^h_T - 1)X_T \stackrel{D}{\longrightarrow}  \frac{hX_T}{T}U$. We also have that  $\sum_{i=0}^{h-1} \varepsilon_{T+h-i} \vert X_T \stackrel{D}{\longrightarrow} N(0, h\sigma^2).$  Using these results and remembering that the two terms are independent of each other we obtain: 
$$\left(\hat X_{T+h} - X_{T+h}\right) \vert X_T \stackrel{D}{\longrightarrow} \frac{hX_T}{T}U + N(0, h\sigma^2) = \frac{hX_T}{T}U + \sqrt{h\sigma^2}V.$$
\
