Double sum, how to proceed?

Question

I'm struggling with the following problem. I get as variance

$$\operatorname{Var}(\hat{k})=C^2\sum_{i=0}^\infty \sum_{j=0}^\infty B^{i+j} \frac{\sigma_{\varepsilon}^2 \rho^{\,j-i}}{1-\rho^2}$$

Can I simplify this further? I tried

$$\operatorname{Var}(\hat{k})=\frac{C^2\sigma_{\varepsilon}^2}{1-\rho^2} \sum_{i=0}^\infty B^i \rho^{-i}\sum_{j=0}^\infty B^{\,j}\rho^{\,j}$$

and thus

$$\operatorname{Var}(\hat{k})=\frac{C^2\sigma_{\varepsilon}^2}{\left(1-\rho^2 \right) \left(1-\frac B \rho\right) (1-B\rho)}$$

but then I get a negative variance for the parameterization of the model. Moreover, I get $\operatorname{Var}(\hat{k})=0$ for $\rho=0$ but it is given that for $\rho=0$

$$\operatorname{Var}(\hat{k})=\frac{C^2\sigma_{\varepsilon}^2}{1-B^2}$$

Is it incorrect to "pull out" $B^i\rho^{-i}$ from the second into the first sum? Or is there any other mistake I don't see?

Can anybody help? I'd be really thankful guys!!!

edit

That's how I cam up with the first line. Given $\varepsilon_{t}\sim N(0,\sigma_{\varepsilon}^{2})$ are $i.i.d.$, $\hat{a}_{t+1}=\rho\hat{a}_{t} + \varepsilon_{t+1}$, the problem states then $ \hat{a}_{t+1} = \sum_{i=0}^{\infty}\rho^{i}\varepsilon_{t-i}$ as an approximation.

I want to find

$$Var\left(C\sum_{i=0}^{\infty}B^i\hat{a}_{t-i}\right)$$ Thus, \begin{split} Var\left(C\sum_{i=0}^{\infty}B^i\hat{a}_{t-i}\right) &= Cov\left(C\sum_{i=0}^{\infty}B^i\hat{a}_{t-i},C\sum_{i=0}^{\infty}B^i\hat{a}_{t-i}\right) \\ &= C^2\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}B^{i+j}Cov\left(\hat{a}_{t-i},\hat{a}_{t-j}\right) \end{split}

Then I used \begin{equation} \begin{split} Cov\left(\hat{a}_{t-i},\hat{a}_{t-j}\right) &= Cov\left(\sum_{k=0}^{\infty}\rho^{k}\varepsilon_{t-i-k}, \sum_{l=0}^{\infty}\rho^{l}\varepsilon_{t-j-l}\right) \\ &= \sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\rho^{k+l}Cov\left(\varepsilon_{t-i-k},\varepsilon_{t-j-l}\right) \end{split} \end{equation}

Then I used that $Cov\left(\varepsilon_{t-i-k},\varepsilon_{t-j-l}\right)=0$ for $i+k \ne j+l$ so I put the double sum together in one sum with $k=j+l-i$

$$ Cov\left(\hat{a}_{t-i},\hat{a}_{t-j}\right)=\sum_{l=0}^{\infty}\rho^{2l+j-i}Cov\left(\varepsilon_{t-i-l},\varepsilon_{t-i-l}\right) = \sigma_{\varepsilon}^{2}\rho^{j-i}\sum_{l=0}^{\infty}\rho^{2l} = \frac{\sigma_{\varepsilon}^{2}\rho^{j-i}}{1-\rho^2}$$

and then I plugged this result in the double sum above to get

$$\operatorname{Var}(\hat{k})=C^2\sum_{i=0}^\infty \sum_{j=0}^\infty B^{i+j} \frac{\sigma_{\varepsilon}^2 \rho^{\,j-i}}{1-\rho^2}$$

Does anyone see the mistake??

Answer 1

1. For $\rho=0$ the starting sum will have a term $0^{j-i}$which tells that you shall sum over $j=i$ thus getting $$\operatorname{Var}(\hat{k})=\frac{C^2\sigma_\varepsilon^2}{1-B^2}$$. Then $$ \eqalign{ & Var\left( {\hat k} \right) = {{C^{\,2} \sigma _\varepsilon ^{\,2} } \over {\left( {1 - \rho ^{\,2} } \right)}}\sum\limits_{0\, \le \,i} {\sum\limits_{0\, \le \,j} {B^{\,j + i} \rho ^{\,j - i} } } = \cr & = \left\{ {\matrix{ {{{C^{\,2} \sigma _\varepsilon ^{\,2} } \over {\left( {1 - B^{\,2} } \right)}}} & {0 = \rho } \cr {{{C^{\,2} \sigma _\varepsilon ^{\,2} } \over {\left( {1 - \rho ^{\,2} } \right)\left( {1 - B/\rho } \right)\left( {1 - \rho B} \right)}}} & {0 < \left| \rho \right| < 1} \cr } } \right. \cr} $$ which for certain values of $\rho$ and $B$ (e.g. $\rho=1/2,\,B=1$) gives negative value of the variance
2. Concerning your addendum, regarding the recursion $$ \hat a_{\,t + 1} = \rho \hat a_{\,t} + \varepsilon _{\,t + 1} \quad \Rightarrow \quad \hat a_{\,t + 1\;\left( ? \right)} = \sum\limits_{i = 0}^{\infty \;\left( ? \right)} {\rho ^{\,i} \varepsilon _{\,t - i} } $$ shouldn't the LHS be $a_{\,t}$ ? and the upper bound be $t$ ? Also concerning $$ Var\left( {C\sum\limits_{i = 0}^{\infty \;?} {B^{\,i} \hat a_{\,t - i} } } \right) $$ the upper bound shall be $t$, otherwise you get negative indices for $a_{\,t-i}$

Answer 2

NEW SOLUTION PROPOSAL

Thanks for your answers guys. Now here's my new attempt with the remarks of @G Cab.

\begin{equation} \begin{split} \label{eq:khat rho} Var\left(\hat{k}_{t+1}\right) &= Var\left(C\sum_{i=0}^{t}B^i\hat{a}_{t-i}\right) = Cov\left(C\sum_{i=0}^{t}B^i\hat{a}_{t-i},C\sum_{i=0}^{t}B^i\hat{a}_{t-i}\right) \\ &= C^2\sum_{i=0}^{t}\sum_{j=0}^{t}B^{i+j}Cov\left(\hat{a}_{t-i},\hat{a}_{t-j}\right) \end{split} \end{equation}

Hence, we need $Cov\left(\hat{a}_{t-i},\hat{a}_{t-j}\right)$ as an intermediate result. Substituting for $\hat{a}_{t-i}$ and $\hat{a}_{t-j}$ in $Cov\left(\hat{a}_{t-i},\hat{a}_{t-j}\right)$ gives \begin{equation} \begin{split} \label{eq:cov1} Cov\left(\hat{a}_{t-i},\hat{a}_{t-j}\right) &= Cov\left(\sum_{k=0}^{t}\rho^{k}\varepsilon_{t-i-k}, \sum_{l=0}^{t}\rho^{l}\varepsilon_{t-j-l}\right) \\ &= \sum_{k=0}^{t}\sum_{l=0}^{t}\rho^{k+l}Cov\left(\varepsilon_{t-i-k},\varepsilon_{t-j-l}\right) \end{split} \end{equation}

Now note that $Cov\left(\varepsilon_{t-i-k},\varepsilon_{t-j-l}\right) = 0 \, \forall \, i+k \ne j+l$. Hence, put $k=j+l-i$ to reduce the above equation to

\begin{equation} \label{eq:cov2} Cov\left(\hat{a}_{t-i},\hat{a}_{t-j}\right)=\sum_{l=0}^{t}\rho^{2l+j-i}Cov\left(\varepsilon_{t-i-l},\varepsilon_{t-i-l}\right) = \sigma_{\varepsilon}^{2}\rho^{j-i}\sum_{l=0}^{t}\rho^{2l} = \frac{\sigma_{\varepsilon}^{2}\rho^{j-i}\left(1-\rho^{2(t+1)}\right)}{1-\rho^2} \end{equation}

Then substituting this back in the original equation for the variance gives

\begin{equation} \begin{split} Var\left(\hat{k}_{t+1}\right) &= C^2\sum_{i=0}^{t} \sum_{j=0}^{t} B^{i+j} \frac{\sigma_{\varepsilon}^{2}\rho^{j-i}\left(1-\rho^{2(t+1)}\right)}{1-\rho^2} \\ &= \frac{C^2\sigma_{\varepsilon}^2\left(1-\rho^{2(t+1)}\right)}{1-\rho^2} \sum_{i=0}^{t} B^i \rho^{-i}\sum_{j=0}^{t} B^{\,j}\rho^{\,j} \\ &= \frac{C^2\sigma_{\varepsilon}^2\left(1-\rho^{2(t+1)}\right)}{1-\rho^2}\sum_{i=0}^{t}\left(\frac{B}{\rho}\right)^{i}\sum_{j=0}^{t} \left(B\rho\right)^{j} \\ &= \frac{C^2\sigma_{\varepsilon}^2\left(1-\rho^{2(t+1)}\right)\left(1-\left(\frac{B}{\rho}\right)^{t+1}\right)\left(1-\left(B\rho\right)^{t+1}\right)}{\left(1-\rho^2\right)\left(1-\frac{B}{\rho}\right)\left(1-B\rho\right)} \\ \end{split} \end{equation}

What do you think about this solution? Many thans in advance!!!