Talk:Stochastic computing

I removed "Still other functions (such as the averaging operator $f(p,q)\rightarrow (p+q)/2$) require stream decimation. Since decimation discards information, it leads to the problem of attenuation. "

Because you can, of course, append two streams to average them, doubling memory, but with no loss of information. The tradeoff for mean computation is analogous to floating point tradeoffs, and there are many mechanisms for doing so. Indeed, any operation that requires decimation can be done with expansion instead, and, in the interests of preserving precision, can be selected, and then compressed, as needed. Much like one can choose to operate in doulbe precision for certain operations before returning to float.