Variance of positively recurrent Markov chain hitting time 
Consider the following Markov chain:
  
  How to compute the variance of $T_{1,0}$ the time to get from state $1$ to state $0$?

Now I can get the stationary distribution $\pi_i=\frac{1}{3}(\frac{2}{3})^i$. Thus we know the the time to get from state k to state k is $m_{kk}=\frac{1}{\pi_k}$. 
Also from $m_{0,0}=0.6*m_{0,0}+0.4*(1+m_{1,0})$ we can get $E[T_{1,0}]$.
I think it can be solve using $Var[X]=E[X^2]-(E[X])^2$ but how to get $E[T_{1,0}^2]$?
 A: 
(2017-10-24) Amusing downvote, purely for mathematical reasons, I am sure...

To reach the higher moments of $T_{1,0}$, one can use the Markov property of the chain after one step, which translates into the identity in distribution $$T_{1,0}=1+B\cdot(T_{2,1}+T'_{1,0})$$ where $B$ is Bernoulli with $P(B=1)=b$ and $P(B=0)=1-b$ for $b=.4$, and $T'_{1,0}$ is distributed like $T_{1,0}$ and independent of $T_{2,1}$.
Since $T_{2,1}$ is also distributed like $T_{1,0}$, one is after the solution of the identity

$$T\stackrel d=1+B\cdot(T+T')\tag{$\ast$}$$

This allows to recover the first moment $E(T_{1,0})$ since $(\ast)$ implies $$E(T)=1+2b\cdot E(T)$$ hence, using $b<\frac12$, $$E(T)=\frac1{1-2b}$$ but also the second moment since $(\ast)$ also implies $$E(T^2)=1+4b\cdot E(T)+b\cdot(2E(T^2)+2E(T))$$ hence, using again $b<\frac12$, $$E(T^2)=\frac{1+6b\cdot E(T)}{1-2b}=\frac{1+4b}{(1-2b)^2}$$ and finally, 

$$\mathrm{var}(T)=\frac{4b}{(1-2b)^2}$$ 

Note that, more generally, turning to generating functions rather than moments, $(\ast)$ would allow to reach distributions since, for every $s$ in $[0,1]$, $$E(s^T)=s\cdot(1-b+b\cdot E(s^T)^2)$$ hence

$$E(s^T)=\frac{1-\sqrt{1-4b(1-b)s^2}}{2bs}$$ 

from which the full distribution of $T_{1,0}$ follows.
