Effective conductance is symmetric Given a network $G$, and states $x$, and $y$, I want to show that 

$\mathscr{C}(x \leftrightarrow y) = \mathscr{C}(y\leftrightarrow x)$

I know that $ \mathscr{C}(x \leftrightarrow y) = \pi(x)P(x\rightarrow y) $, and so intuitively I can see that the answer can follow from something that is analogous to $\pi(u)p(u,v) = \pi(v)p(v,u)$, but I was curious if there is an easy, formal, way to see this.
Edit: The transition matrix, $P$, is $\pi$-reversible
 A: We can find the effective conductance in terms of $H(x,y) + H(y,x)$, where $H(x,y)$ is the hitting time: the expected number of steps to reach $y$ starting from $x$. This quantity is symmetric, and therefore effective conductance is symmetric.

We can think of $H(x,y) + H(y,x)$ as the expected number of steps, starting from $x$, until we reach $y$ and we return to $x$. Informally, this should be the average length of an excursion out of $x$ (a random walk that starts at $x$ and ends when it returns to $x$) multiplied by the average number of excursions we need until one of them passes through $y$.
To actually manipulate expected values in this vaguely sketchy fashion, we need to use Wald's identity. 
Take an infinite random walk starting at $x$ and let $\mathbf X_1, \mathbf X_2, \mathbf X_3$ be the numbers of steps between visits to $x$ (the lengths of the excursions out of $x$.) Let $\mathbf T$ be the index of the first excursion that visits $y$. Then
$$
    H(x,y) + H(y,x) = \mathbb E[\mathbf X_1 + \mathbf X_2 + \dots + \mathbf X_{\mathbf T}].
$$
By Wald's identity, $H(x,y) + H(y,x) = \mathbb E[\mathbf X_1] \cdot \mathbb E[\mathbf T]$. It's a well-known fact that $\mathbb E[\mathbf X_1] = \frac{1}{\pi(x)}$. (If you don't know this fact, it also has a proof using Wald's identity.) Meanwhile, $\mathbb E[\mathbf T] = \frac{1}{P(x \to y)}$, since $P(x \to y)$ is the probability that an excursion out of $x$ visits $y$, so $\mathbf T$ is just a geometric random variable for that probability.
Therefore 
$$H(x,y) + H(y,x) = \frac{1}{\pi(x)} \cdot \frac{1}{P(x\to y)} = \frac{1}{\mathscr{C}(x \leftrightarrow y)}$$
and since $\frac{1}{H(x,y) + H(y,x)}$ is symmetric in $x$ and $y$, so is effective conductance.
