Random function + nonrandom function

[blike]

Not sure if this belongs in the maths forum, so one of the mods will have to clean up after me if it doesn't. On another forum I'm involved with a discussion of random functions. One poster is arguing that if you take any random process or function and add it to a non-random process or function, that your end result is still random. Is this true? If not, can it be shown in some sort of proof that this is not the case?

[blazarwolf]

Well I only went to Algebra 1, and ive avoided math since... ill put in my uneducated guess..... (noones been able to answer my simple realitivy questions anyways)

 

If your variable is random, it could be solved in a fixed frame, but not predicted in a floating equation... In a solution: if there is a varible of radom nature... it will indeed be random... To statistics: the more non-radom varibles you add to the equation, the less radom (to statisics!, fractional discernability).. I think the proof would be one of those long hard to understand ones, I think i made up words with this one, sorry.

[D H]

It depends on what you mean by "random function". If, on one hand, you mean taking a forming a variable Z by adding random variable X to a strictly deterministic variable Y, then Z will indeed be a random variable.

 

If, on the other hand, you mean taking a function that represents a probability distribution such as [math]f(x)=\frac 1 {\sqrt{2\pi}}\exp\left(-\,\frac {x^2} 2\right)[/math] and forming some other function [math]h(x)=f(x)+g(x)[/math] where g(x) is some other function, then no, the new function h(x) most likely is not a "random function". It is nonsense, kind of like asking "what is ten meters plus five seconds".