If $a\_1,a\_2,\dots$ are IID random bits (correction as per Anthony Quas: these "bits" are $+1$ and $-1$ with equal probability), then with probability 1, the set of natural numbers $n$ such that $a\_1+a\_2+\dots+a\_n \leq 0$ has lower density 0 and upper density 1, so it has no density in the ordinary sense. Still, I wonder if there is a principled way to generalize the manner in which we assign "densities" to subsets of the natural numbers in such a fashion that, with probability 1, the aforementioned set has generalized density 1/2 -- and, more generally, for every real $t$, the set of $n$ such that $(a\_1+a\_2+\dots+a\_n)/\sqrt{n} \leq t$ has generalized density equal to the probability that the relevant Gaussian random variable has value less than $t$.



Generalized density functions on the natural numbers
So I think a logarithmic average will do the trick for you. If you define $Y\_n$ to be the sign of $a\_1+\ldots+a\_n$, then calculations with Brownian motion in place of random walk suggest the covariance of $Y\_n$ and $Y\_m$ with $m<n$ is approximately $(1/2\pi)\arctan\sqrt{m/(m-n)}$. Now define $S\_N=(1/\log N)(Y\_1/1+\ldots +Y\_N/N)$. This has expectation 0 and variance $\approx 1/\log N$, which gives a systematic way of saying that the random walk is "positive half the time". 


