conditional vs unconditional forecast variance in AR(1) I have trouble showing that conditional forecast error of AR(1) has smaller variance than the unconditional one. 
I can show that cond. forecast error is: 
$$ 
Y_{T+1}=aY_{T}+\epsilon_{T+1} $$
$$
\hat{Y}_{T+1}=E[aY_T+\epsilon_{T+1}|Y_T,Y_{T-1},...]=aY_T $$
$$
E[Y_{T+1}-\hat{Y}_{T+1}]=E[\epsilon_{T+1}]=0 $$
$$
Var[Y_{T+1}-\hat{Y}_{T+1}]=E[(Y_{T+1}-\hat{Y}_{T+1})^2]=E[\epsilon^2_{T+1}]=\sigma^2
$$
now trying to get unconditional forecast error (w. help from plissken's answer)
$$
\hat{Y}_{T+1}=E[aY_T+\epsilon_{T+1}]=aE[Y_T]+E[\epsilon_{T+1}]=aE[Y_T]=0 $$
$$E[Y_{T+1}]=E[aY_T+\epsilon_{T+1}]=aE[Y_T]+E[\epsilon_{T+1}]=aE[Y_T]=0$$
$$Var[Y_{T+1}]=E[Y^2_{T+1}]=aE[Y^2_T]+2aE[Y_T\epsilon_{T+1}]+E[\epsilon^2_{T+1}]=\sigma^2/(1-\alpha^2)$$
$$E[Y_{T+1}-\hat{Y}_{T+1}]=aE[Y_{T+1}]+E[\epsilon_{T+1}]-aE[Y_T]=0$$
$$Var[Y_{T+1}-\hat{Y}_{T+1}]=E[(Y_{T+1}-\hat{Y}_{T+1})^2]=E[(Y_{T+1}-E[\hat{Y}_{T+1}])^2]=E[Y^2_{T+1}]=\sigma^2/(1-\alpha^2)$$
not sure if this correct. would be nice if anyone could confirm/help.
greets
 A: First you'll need to find the unconditional mean and variance of the AR(1) process $Y_{t}=aY_{t-1}+\varepsilon_{t},\;\varepsilon_{t}\sim\left(0,\,\sigma^{2}\right)$: 
$$E\left[Y_{t}\right]=E\left[aY_{t-1}+\varepsilon_{t}\right]=aE\left[Y_{t-1}\right]+E\left[\varepsilon_{t}\right]=0$$
$$V\left[Y_{t}\right] =V\left[aY_{t-1}+\varepsilon_{t}\right]
 =a^{2}V\left[Y_{t-1}\right]+V\left[\varepsilon_{t}\right]
 =\frac{\sigma^{2}}{1-a^{2}}$$
Here I have assumed that the process is stationary, i.e. $\left|a\right|<1
 $ so I can write $E\left[Y_{t}\right]=E\left[Y_{t-1}\right]$.
Then you'll have to find the unconditional variance of the forecast error: 
$V\left[Y_{t+1}-\hat{Y}_{t+1}\right]=E\left[\left(aY_{t+1}+\varepsilon_{t+1}-E\left[Y_{t}\right]\right)\right]^{2}=X +\sigma^{2}$.
You'll need to find an expression for $X
 $. Then you'll see that $X +\sigma^{2}>\sigma^{2}$
  as $X>0$. 
EDIT
Note that $X+\sigma^{2}=V\left[aY_{t+1}+\varepsilon_{t+1}\right]$ since $E\left[Y_{t}\right]=0$. Then it is quick to find $X$ as $X=V\left[aY_{t+1}\right]=a^{2}V\left[Y_{t+1}\right]$. This is not the same as the last expression you have found above. 
