Let $\lambda(n)$ be Liouville's function, so that for each positive integer $n = p_1^{m_1}\cdots p_r^{m_r}$, we have that $\lambda(n) = (-1)^{\sum^{r}_{k=1}{m_k}}$. In 1919, Polya conjectured that $L(x) = \sum_{n \leq x}{\lambda(n)} \leq 0$ for all $x \geq 2$; his reasoning was based on some limited numerical evidence (up to $x = 1500$, I believe), its connection to the Riemann Hypothesis (it implies RH and the simplicity of the zeroes of $\zeta(s)$), and Polya showed that for $p \equiv 3 \pmod{4}$ with class number $h(-p) = 1$, $L(p) = 0$. Unfortunately, Polya's conjecture is false; it is known that the first counterexample occurs at $x = 906150257$ (so one can't really blame Polya for trying), and that there exist infinitely many positive integers $n$ such that $L(n) \geq 0.061867 \ldots$.
Nevertheless, Polya's conjecture does seem to be usually true, in that $L(x) \leq 0$ "most" of the time. There are a couple of different arguments that give an indication of why one would expect $L(x)$ to often be negative. For example, standard methods show (under RH, of course) that $${\sum_{n \leq x}}'{\lambda(n)} = \frac{\sqrt{x}}{\zeta(1/2)} + \sum_{\rho}{\frac{\zeta(2\rho)}{\zeta'(\rho)}\frac{x^{\rho}}{\rho}} - 1 + O\left(\frac{1}{\sqrt{x}}\right),$$ and one expects the terms in the sum over the zeroes to generally be very small, whereas $1/\zeta(1/2) = -0.684765\ldots$, so it would be expected that $L(x)$ is "usually" negative. Another method is via Lambert series; I mentioned here that one can show that $$\sum_{n=1}^{\infty}{\frac{\lambda(n)}{e^{n\pi/x}+1}} = \frac{1-\sqrt{2}}{2}\sqrt{x} + \frac{1}{2} + (\psi(x)-2\psi(x/2))\sqrt{x},$$ where $\psi(x) = \sum_{n=1}^{\infty}{e^{-\pi xn^2}} = O(e^{-\pi x})$; this Lambert series is in some sense a smoothed version of $L(x)$. Again, the leading term is negative, suggesting that $L(x) \leq 0$ often.
My question is: what other methods (elementary, analytic, or probabilistic) can be used to show why we would expect $L(x)$ to usually be negative?
