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Let's define a helper: $y(n) = w(n) + w(n-1)$. Then, your $x$ becomes

\begin{align} x(n) &= \frac15 x(n-1) + y(n)\\ &= \frac15 \left(\frac 15x(n-2)+y(n-1)\right) + y(n)\\ &=\frac 1{25}x(n-2)+\frac15 y(n-1)+ y(n)\\ &=\frac1{5^3}x(n-3)+\frac 1{25}y(n-2)+\frac15 y(n-1)+ y(n)\\ &=\sum_{p=0}^\infty 5^{-p}y(n-p)\\ &=\sum_{p=0}^\infty\left( 5^{-p}w(n-p)+5^{-p}w(n-p-1)\right)\\ &=\sum_{p=0}^\infty 5^{-p}w(n-p)+\sum_{q=1}^\infty5^{-q+1}w(n-q)\\ &=5^{-0}w(n)+\sum_{p=1}^\infty 5^{-p}w(n-p)+\sum_{q=1}^\infty5^{-q+1}w(n-q)\\ &=w(n)+\sum_{p=1}^\infty \left(5^{-p}w(n-p)+5^{-p+1}w(n-p)\right)\\ &=w(n)+\sum_{p=1}^\infty w(n-p)\left(5^{-p}+5^{-p+1}\right)\\ &=w(n)+6\sum_{p=1}^\infty w(n-p)5^{-p}\\[2em] % % E\left[x^2(n)\right] &= E\left[\left(w(n) +\underbrace{\sum_{p=1}^\infty w(n-p)6\cdot5^{-p}}_{s(n)}\right)^2\right]\\ &= E[w^2(n)] + E[2w(n)s(n)] + E[s^2(n)]\\ &= \frac32 + 2E[w(n)s(n)] + E[s^2(n)]\\ &= \frac32 + 2E\left[\sum_{p=1}^\infty w(n)w(n-p)6\cdot5^{-p}\right] + E[s^2(n)]\\[2.5em] % % E[s^2(n)] &= E[s(n)\cdot s(n)]\\ &\text{Cauchy product formula:}\\ &= E\left[\sum_{k=0}^\infty\sum_{l=1}^{k-1} w(n-l)\left(5^{-l}+5^{-l+1}\right) w(n-k+l)\left(5^{-k+l}+5^{-k+l+1}\right)\right]\\ &= E\left[\sum_{k=0}^\infty \sum_{l=1}^{k-1}\left(5^{-l}+5^{-l+1}\right)\left(5^{-k+l}+5^{-k+l+1}\right) w(n-l) w(n-k+l)\right]\\ &= E\left[\sum_{k=0}^\infty \sum_{l=1}^{k-1}36\cdot 5^{-k} w(n-l) w(n-k+l)\right]\\ &= E\left[36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1} w(n-l) w(n-k+l)\right]\\[2em] % % E\left[x^2(n)\right]&= \frac32 + 2E\left[\sum_{p=1}^\infty w(n)w(n-p)6\cdot5^{-p}\right] +E\left[36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1} w(n-l) w(n-k+l)\right]\\ &= \frac32 + 12E\left[\sum_{p=1}^\infty w(n)w(n-p)\cdot5^{-p}\right] + 36 E\left[\sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1} w(n-l) w(n-k+l)\right] \end{align}

At this point, things get very hairy: We can proceed if we allow ourselves to "pull" the $E$ operator into these infinite sums. But that only works if the result is finite, otherwise we'd be making a false statement.

Therefore, we must assume that $\left\lvert\sum_{p=1}^\infty E\left[w(n)w(n-p)\cdot5^{-p}\right]\right\rvert<\infty$. A gut feeling is that as long as $w(n)w(n-p) < 5^p$, we're fine, but bear in mind that this might not be the case for just any noise! An example of cases where that would not be the case is an exponentially decaying noisy oscillator of unknown phase with an offset. (due to the offset: non-zero mean, the exponential envelope gives us a bounded variance if we observe across all times, but looking back into the past for $p\to\infty$, we get infinitely many values $>5^p$.)

So!!!! Check! YOUR! Model for $w(n)$!111eleven!

If you trust you're not in one of the cases where this goes wrong (please, be sure):

\begin{align} E\left[x^2(n)\right]&= \frac32 \\ &\phantom= + 12\sum_{p=1}^\infty 5^{-p}E\left[w(n)w(n-p)\right]\\ &\phantom= + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1} E\left[w(n-l) w(n-k+l)\right] \end{align}

Standard trick: $v(n) = w(n) - m_w$, that makes $v(n)$ have a zero mean. Also, $\sum_{k=0}^{\infty} 5^{-k}=\frac54$:

\begin{align} E\left[x^2(n)\right]&= \frac32 \\ &\phantom= + 12\sum_{p=1}^\infty 5^{-p}\left(E\left[m_w^2 + m_w(v(n)+v(n-p))+v(n)v(n-p)\right]\right)\\ &\phantom= + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1}\left( E\left[m_w^2 + m_w(v(n-l)+v(n-k+l)) + v(n-l)v(n-k+l)\right]\right)\\ &= \frac32 \\ &\phantom= + 12\sum_{p=1}^\infty\ 5^{-p}\left(E\left[m_w^2\right] +E\left[ m_w(v(n)+v(n-p))\right]+E\left[v(n)v(n-p)\right]\right)\\ &\phantom= + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1}\left( E\left[m_w^2\right] + E\left[m_w(v(n-l)+v(n-k+l))\right] + E\left[v(n-l)v(n-k+l)\right]\right)\\ % &= \frac32 +12\cdot\frac54 m_w^2 +36\cdot \frac54 m_w^2 \\ &\phantom= + 12m_w\sum_{p=1}^\infty\ 5^{-p}(\underbrace{E\left[v(n)\right]}_{=0}+\underbrace{E\left[v(n-p)\right]}_{=0})\\ &\phantom= + 12\sum_{p=1}^\infty\ 5^{-p}E\left[v(n)v(n-p)\right]\\ &\phantom= + 36 m_w\sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1}E[(v(n-l)]+E[v(n-k+l)] \\ &\phantom= + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1} E[v(n-l)v(n-k+l)]\\ &= \frac32 +60 m_w^2 \\ &\phantom= + 12\sum_{p=1}^\infty\ 5^{-p}E\left[v(n)v(n-p)\right]\\ &\phantom= + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1} E[v(n-l)v(n-k+l)]\\ &= 16.5 \\ &\phantom= + 12\sum_{p=1}^\infty\ 5^{-p}E\left[v(n)v(n-p)\right]\\ &\phantom= + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1}. E[v(n-l)v(n-k+l) \end{align}

That's quite a nice result, innit?

We can now look at a few special cases:

1. Different points in time of the "mean-removed" $w(n)-m_w = v(n)$ are uncorrelated: $E[x^2(n)] = 16.5$.
2. $E[v(n)v(n-\tau)] = \alpha^k$:
3. $k<5$: $E[x^2(n)]$ converges, value needs to be derived
4. $k\ge5$: $E[x^2(n)]$ diverges

Let's define a helper: $y(n) = w(n) + w(n-1)$. Then, your $x$ becomes

\begin{align} x(n) &= \frac15 x(n-1) + y(n)\\ &= \frac15 \left(\frac 15x(n-2)+y(n-1)\right) + y(n)\\ &=\frac 1{25}x(n-2)+\frac15 y(n-1)+ y(n)\\ &=\frac1{5^3}x(n-3)+\frac 1{25}y(n-2)+\frac15 y(n-1)+ y(n)\\ &=\sum_{p=0}^\infty 5^{-p}y(n-p)\\ &=\sum_{p=0}^\infty\left( 5^{-p}w(n-p)+5^{-p}w(n-p-1)\right)\\ &=\sum_{p=0}^\infty 5^{-p}w(n-p)+\sum_{q=1}^\infty5^{-q+1}w(n-q)\\ &=5^{-0}w(n)+\sum_{p=1}^\infty 5^{-p}w(n-p)+\sum_{q=1}^\infty5^{-q+1}w(n-q)\\ &=w(n)+\sum_{p=1}^\infty \left(5^{-p}w(n-p)+5^{-p+1}w(n-p)\right)\\ &=w(n)+\sum_{p=1}^\infty w(n-p)\left(5^{-p}+5^{-p+1}\right)\\ &=w(n)+6\sum_{p=1}^\infty w(n-p)5^{-p}\\[2em] % % E\left[x^2(n)\right] &= E\Big[\Big(w(n) +\underbrace{\sum_{p=1}^\infty w(n-p)6\cdot5^{-p}}_{s(n)}\Big)^2\Big]\\ &= E[w^2(n)] + E[2w(n)s(n)] + E[s^2(n)]\\ &= \frac32 + 2E[w(n)s(n)] + E[s^2(n)]\\ &= \frac32 + 2E\left[\sum_{p=1}^\infty w(n)w(n-p)6\cdot5^{-p}\right] + E[s^2(n)]\\[2.5em] % % E[s^2(n)] &= E[s(n)\cdot s(n)]\\ &\text{Cauchy product formula:}\\ &= E\left[\sum_{k=0}^\infty\sum_{l=1}^{k-1} w(n-l)\left(5^{-l}+5^{-l+1}\right) w(n-k+l)\left(5^{-k+l}+5^{-k+l+1}\right)\right]\\ &= E\left[\sum_{k=0}^\infty \sum_{l=1}^{k-1}\left(5^{-l}+5^{-l+1}\right)\left(5^{-k+l}+5^{-k+l+1}\right) w(n-l) w(n-k+l)\right]\\ &= E\left[\sum_{k=0}^\infty \sum_{l=1}^{k-1}36\cdot 5^{-k} w(n-l) w(n-k+l)\right]\\ &= E\left[36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1} w(n-l) w(n-k+l)\right]\\[2em] % % E\left[x^2(n)\right]&= \frac32 + 2E\left[\sum_{p=1}^\infty w(n)w(n-p)6\cdot5^{-p}\right] +E\left[36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1} w(n-l) w(n-k+l)\right]\\ &= \frac32 + 12E\left[\sum_{p=1}^\infty w(n)w(n-p)\cdot5^{-p}\right] + 36 E\left[\sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1} w(n-l) w(n-k+l)\right] \end{align}

At this point, things get very hairy: We can proceed if we allow ourselves to "pull" the $E$ operator into these infinite sums. But that only works if the result is finite, otherwise we'd be making a false statement.

Therefore, we must assume that $\left\lvert\sum_{p=1}^\infty E\left[w(n)w(n-p)\cdot5^{-p}\right]\right\rvert<\infty$. A gut feeling is that as long as $w(n)w(n-p) < 5^p$, we're fine, but bear in mind that this might not be the case for just any noise! An example of cases where that would not be the case is an exponentially decaying noisy oscillator of unknown phase with an offset. (due to the offset: non-zero mean, the exponential envelope gives us a bounded variance if we observe across all times, but looking back into the past for $p\to\infty$, we get infinitely many values $>5^p$.)

So!!!! Check! YOUR! Model for $w(n)$!111eleven!

If you trust you're not in one of the cases where this goes wrong (please, be sure):

\begin{align} E\left[x^2(n)\right]&= \frac32 \\ &\phantom= + 12\sum_{p=1}^\infty 5^{-p}E\left[w(n)w(n-p)\right]\\ &\phantom= + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1} E\left[w(n-l) w(n-k+l)\right] \end{align}

Standard trick: $v(n) = w(n) - m_w$, that makes $v(n)$ have a zero mean. Also, $\sum_{k=0}^{\infty} 5^{-k}=\frac54$:

\begin{align} &= \frac32 \\ &\phantom= + 12\sum_{p=1}^\infty 5^{-p}\left(E\left[m_w^2 + m_w(v(n)+v(n-p))+v(n)v(n-p)\right]\right)\\ &\phantom= + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1}\left( E\left[m_w^2 + m_w(v(n-l)+v(n-k+l)) + v(n-l)v(n-k+l)\right]\right)\\ &= \frac32 \\ &\phantom= + 12\sum_{p=1}^\infty\ 5^{-p}\left(E\left[m_w^2\right] +E\left[ m_w(v(n)+v(n-p))\right]+E\left[v(n)v(n-p)\right]\right)\\ &\phantom= + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1}\left( E\left[m_w^2\right] + E\left[m_w(v(n-l)+v(n-k+l))\right] + E\left[v(n-l)v(n-k+l)\right]\right)\\ % &= \frac32 +12\cdot\frac54 m_w^2 +36\cdot \frac54 m_w^2 \\ &\phantom= + 12m_w\sum_{p=1}^\infty\ 5^{-p}(\underbrace{E\left[v(n)\right]}_{=0}+\underbrace{E\left[v(n-p)\right]}_{=0})\\ &\phantom= + 12\sum_{p=1}^\infty\ 5^{-p}E\left[v(n)v(n-p)\right]\\ &\phantom= + 36 m_w\sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1}E[(v(n-l)]+E[v(n-k+l)] \\ &\phantom= + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1} E[v(n-l)v(n-k+l)]\\ &= \frac32 +60 m_w^2 \\ &\phantom= + 12\sum_{p=1}^\infty\ 5^{-p}E\left[v(n)v(n-p)\right]\\ &\phantom= + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1} E[v(n-l)v(n-k+l)]\\ &= 16.5 \\ &\phantom= + 12\sum_{p=1}^\infty\ 5^{-p}E\left[v(n)v(n-p)\right]\\ &\phantom= + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1}. E[v(n-l)v(n-k+l) \end{align}

That's quite a nice result, innit?

We can now look at a few special cases:

1. Different points in time of the "mean-removed" $w(n)-m_w = v(n)$ are uncorrelated: $E[x^2(n)] = 16.5$.
2. $E[v(n)v(n-\tau)] = \alpha^k$:
3. $k<5$: $E[x^2(n)]$ converges, value needs to be derived
4. $k\ge5$: $E[x^2(n)]$ diverges

\begin{align} x(n) &= \frac15 x(n-1) + y(n)\ &= \frac15 \left(\frac 15x(n-2)+y(n-1)\right) + y(n)\ &=\frac 1{25}x(n-2)+\frac15 y(n-1)+ y(n)\ &=\frac1{5^3}x(n-3)+\frac 1{25}y(n-2)+\frac15 y(n-1)+ y(n)\ &=\sum_{p=0}^\infty 5^{-p}y(n-p)\ &=\sum_{p=0}^\infty\left( 5^{-p}w(n-p)+5^{-p}w(n-p-1)\right)\ &=\sum_{p=0}^\infty 5^{-p}w(n-p)+\sum_{q=1}^\infty5^{-q+1}w(n-q)\ &=5^{-0}w(n)+\sum_{p=1}^\infty 5^{-p}w(n-p)+\sum_{q=1}^\infty5^{-q+1}w(n-q)\ &=w(n)+\sum_{p=1}^\infty \left(5^{-p}w(n-p)+5^{-p+1}w(n-p)\right)\ &=w(n)+\sum_{p=1}^\infty w(n-p)\left(5^{-p}+5^{-p+1}\right)\ &=w(n)+6\sum_{p=1}^\infty w(n-p)5^{-p}\[2em] % % E\left[x^2(n)\right] &= E\left[\left(w(n) +\underbrace{\sum_{p=1}^\infty w(n-p)6\cdot5^{-p}}{s(n)}\right)^2\right]\ &= E[w^2(n)] + E[2w(n)s(n)] + E[s^2(n)]\ &= \frac32 + 2E[w(n)s(n)] + E[s^2(n)]\ &= \frac32 + 2E\left[\sum{p=1}^\infty w(n)w(n-p)6\cdot5^{-p}\right] + E[s^2(n)]\[2.5em] % % E[s^2(n)] &= E[s(n)\cdot s(n)]\ &\text{Cauchy product formula:}\ &= E\left[\sum_{k=0}^\infty\sum_{l=1}^{k-1} w(n-l)\left(5^{-l}+5^{-l+1}\right) w(n-k+l)\left(5^{-k+l}+5^{-k+l+1}\right)\right]\ &= E\left[\sum_{k=0}^\infty \sum_{l=1}^{k-1}\left(5^{-l}+5^{-l+1}\right)\left(5^{-k+l}+5^{-k+l+1}\right) w(n-l) w(n-k+l)\right]\ &= E\left[\sum_{k=0}^\infty \sum_{l=1}^{k-1}36\cdot 5^{-k} w(n-l) w(n-k+l)\right]\ &= E\left[36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1} w(n-l) w(n-k+l)\right]\[2em] % % E\left[x^2(n)\right]&= \frac32

• 2E\left[\sum_{p=1}^\infty w(n)w(n-p)6\cdot5^{-p}\right] +E\left[36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1} w(n-l) w(n-k+l)\right]\ &= \frac32
• 12E\left[\sum_{p=1}^\infty w(n)w(n-p)\cdot5^{-p}\right]
• 36 E\left[\sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1} w(n-l) w(n-k+l)\right]

\end{align}

Standard trick: $v(n) = w(n) - m_w$, that makes $v(n)$ have a zero mean:

\begin{align} E\left[x^2(n)\right]&= \frac32 \\ &\phantom= + 12\sum_{p=1}^\infty 5^{-p}\left(E\left[m_w^2 + m_w(v(n)+v(n-p))+v(n)v(n-p)\right]\right)\\ &\phantom= + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1}\left( E\left[m_w^2 + m_w(v(n-l)+v(n-k+l)) + v(n-l)v(n-k+l)\right]\right)\\ &= \frac32 \\ &\phantom= + 12\sum_{p=1}^\infty\ 5^{-p}\left(E\left[m_w^2\right] +E\left[ m_w(v(n)+v(n-p))\right]+E\left[v(n)v(n-p)\right]\right)\\ &\phantom= + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1}\left( E\left[m_w^2\right] + E\left[m_w(v(n-l)+v(n-k+l))\right] + E\left[v(n-l)v(n-k+l)\right]\right) \end{align}

\begin{align} x(n) &= \frac15 x(n-1) + y(n)\\ &= \frac15 \left(\frac 15x(n-2)+y(n-1)\right) + y(n)\\ &=\frac 1{25}x(n-2)+\frac15 y(n-1)+ y(n)\\ &=\frac1{5^3}x(n-3)+\frac 1{25}y(n-2)+\frac15 y(n-1)+ y(n)\\ &=\sum_{p=0}^\infty 5^{-p}y(n-p)\\ &=\sum_{p=0}^\infty\left( 5^{-p}w(n-p)+5^{-p}w(n-p-1)\right)\\ &=\sum_{p=0}^\infty 5^{-p}w(n-p)+\sum_{q=1}^\infty5^{-q+1}w(n-q)\\ &=5^{-0}w(n)+\sum_{p=1}^\infty 5^{-p}w(n-p)+\sum_{q=1}^\infty5^{-q+1}w(n-q)\\ &=w(n)+\sum_{p=1}^\infty \left(5^{-p}w(n-p)+5^{-p+1}w(n-p)\right)\\ &=w(n)+\sum_{p=1}^\infty w(n-p)\left(5^{-p}+5^{-p+1}\right)\\ &=w(n)+6\sum_{p=1}^\infty w(n-p)5^{-p}\\[2em] % % E\left[x^2(n)\right] &= E\left[\left(w(n) +\underbrace{\sum_{p=1}^\infty w(n-p)6\cdot5^{-p}}_{s(n)}\right)^2\right]\\ &= E[w^2(n)] + E[2w(n)s(n)] + E[s^2(n)]\\ &= \frac32 + 2E[w(n)s(n)] + E[s^2(n)]\\ &= \frac32 + 2E\left[\sum_{p=1}^\infty w(n)w(n-p)6\cdot5^{-p}\right] + E[s^2(n)]\\[2.5em] % % E[s^2(n)] &= E[s(n)\cdot s(n)]\\ &\text{Cauchy product formula:}\\ &= E\left[\sum_{k=0}^\infty\sum_{l=1}^{k-1} w(n-l)\left(5^{-l}+5^{-l+1}\right) w(n-k+l)\left(5^{-k+l}+5^{-k+l+1}\right)\right]\\ &= E\left[\sum_{k=0}^\infty \sum_{l=1}^{k-1}\left(5^{-l}+5^{-l+1}\right)\left(5^{-k+l}+5^{-k+l+1}\right) w(n-l) w(n-k+l)\right]\\ &= E\left[\sum_{k=0}^\infty \sum_{l=1}^{k-1}36\cdot 5^{-k} w(n-l) w(n-k+l)\right]\\ &= E\left[36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1} w(n-l) w(n-k+l)\right]\\[2em] % % E\left[x^2(n)\right]&= \frac32 + 2E\left[\sum_{p=1}^\infty w(n)w(n-p)6\cdot5^{-p}\right] +E\left[36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1} w(n-l) w(n-k+l)\right]\\ &= \frac32 + 12E\left[\sum_{p=1}^\infty w(n)w(n-p)\cdot5^{-p}\right] + 36 E\left[\sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1} w(n-l) w(n-k+l)\right] \end{align}

Standard trick: $v(n) = w(n) - m_w$, that makes $v(n)$ have a zero mean. Also, $\sum_{k=0}^{\infty} 5^{-k}=\frac54$:

\begin{align} E\left[x^2(n)\right]&= \frac32 \\ &\phantom= + 12\sum_{p=1}^\infty 5^{-p}\left(E\left[m_w^2 + m_w(v(n)+v(n-p))+v(n)v(n-p)\right]\right)\\ &\phantom= + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1}\left( E\left[m_w^2 + m_w(v(n-l)+v(n-k+l)) + v(n-l)v(n-k+l)\right]\right)\\ &= \frac32 \\ &\phantom= + 12\sum_{p=1}^\infty\ 5^{-p}\left(E\left[m_w^2\right] +E\left[ m_w(v(n)+v(n-p))\right]+E\left[v(n)v(n-p)\right]\right)\\ &\phantom= + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1}\left( E\left[m_w^2\right] + E\left[m_w(v(n-l)+v(n-k+l))\right] + E\left[v(n-l)v(n-k+l)\right]\right)\\ % &= \frac32 +12\cdot\frac54 m_w^2 +36\cdot \frac54 m_w^2 \\ &\phantom= + 12m_w\sum_{p=1}^\infty\ 5^{-p}(\underbrace{E\left[v(n)\right]}_{=0}+\underbrace{E\left[v(n-p)\right]}_{=0})\\ &\phantom= + 12\sum_{p=1}^\infty\ 5^{-p}E\left[v(n)v(n-p)\right]\\ &\phantom= + 36 m_w\sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1}E[(v(n-l)]+E[v(n-k+l)] \\ &\phantom= + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1} E[v(n-l)v(n-k+l)]\\ &= \frac32 +60 m_w^2 \\ &\phantom= + 12\sum_{p=1}^\infty\ 5^{-p}E\left[v(n)v(n-p)\right]\\ &\phantom= + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1} E[v(n-l)v(n-k+l)]\\ &= 16.5 \\ &\phantom= + 12\sum_{p=1}^\infty\ 5^{-p}E\left[v(n)v(n-p)\right]\\ &\phantom= + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1}. E[v(n-l)v(n-k+l) \end{align}

That's quite a nice result, innit?

We can now look at a few special cases:

1. Different points in time of the "mean-removed" $w(n)-m_w = v(n)$ are uncorrelated: $E[x^2(n)] = 16.5$.
2. $E[v(n)v(n-\tau)] = \alpha^k$:
3. $k<5$: $E[x^2(n)]$ converges, value needs to be derived
4. $k\ge5$: $E[x^2(n)]$ diverges

\begin{align} E\left[x^2(n)\right]&= \frac32 + 12\sum_{p=1}^\infty 5^{-p}E\left[w(n)w(n-p)\right] + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1} E\left[w(n-l) w(n-k+l)\right] \end{align}

\begin{align} E\left[x^2(n)\right]&= \frac32 \\ &\phantom= + 12\sum_{p=1}^\infty 5^{-p}E\left[w(n)w(n-p)\right]\\ &\phantom= + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1} E\left[w(n-l) w(n-k+l)\right] \end{align}

Standard trick: $v(n) = w(n) - m_w$, that makes $v(n)$ have a zero mean:

\begin{align} E\left[x^2(n)\right]&= \frac32 \\ &\phantom= + 12\sum_{p=1}^\infty 5^{-p}\left(E\left[m_w^2 + m_w(v(n)+v(n-p))+v(n)v(n-p)\right]\right)\\ &\phantom= + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1}\left( E\left[m_w^2 + m_w(v(n-l)+v(n-k+l)) + v(n-l)v(n-k+l)\right]\right)\\ &= \frac32 \\ &\phantom= + 12\sum_{p=1}^\infty\ 5^{-p}\left(E\left[m_w^2\right] +E\left[ m_w(v(n)+v(n-p))\right]+E\left[v(n)v(n-p)\right]\right)\\ &\phantom= + 36 \sum_{k=0}^\infty 5^{-k} \sum_{l=1}^{k-1}\left( E\left[m_w^2\right] + E\left[m_w(v(n-l)+v(n-k+l))\right] + E\left[v(n-l)v(n-k+l)\right]\right) \end{align}