Confidence interval for the expectation of two stochastic processes Let's say I have one stochastic process $X$ which represents the production of something. I make some improvements over the production line and I hopefully end up with a different stochastic process, $Y$, with mean greater than $X$ and less variance. Let's suppose each process was run for the same time.
If I want to quantify the improvement (I) using their expectations ratio ($I = \frac{\mathrm{E}[Y]}{\mathrm{E}[X]}$), what would be the 95% confidence interval of I?
Can I just treat them as independent Gaussian variables and calculate the improvement ($I$) variance with the variance product formula? Since $X$ and $Y$ are variations of the same production line, I don't know if they can be considered independent. Also, I'm not sure if they can be considered Gaussian.
 A: As @whuber points out, independence is a probabilistic concept and you must try to "map" your real-life situation to it, to see whether it is reasonable to assume it or not. In your case, many production factors (eg people, machines) will be the same before and after your intervention. But this alone does not imply dependence between $X$ and $Y$. Does your production process (stochastic process) has a "memory"? Does how people perform today affect how they will perform tomorrow? Does an unexpected breakdown of a machine today, will somehow affect its capacity and output after it has been repaired and re-introduced in the process (namely, does stochastic disturbances exhibit persistence)? If the answers to such questions are "yes" then your process possesses memory, hence tomorrow is not independent of today, irrespective of your intervention, hence you should model some dependence, like a Markov chain for example.
If the answers to such questions are "no", or "rather not", or even "no, most of the times", then you can accept as a reasonable approximation the independence assumption, since it simplifies greatly the technical setup. 
Continuing with the "mapping exercise", your intervention can be thought of as a "structural break" - if you think "production function", it means some coefficients shifted due to your intervention. Note that this does not mean that you can write something like $Y = a + bX$ even though it may emerge that the new mean and variance can be mathematically related to the old mean and variance through such a representation) -because there is one critical dimension of $Y$ and $X$ that is different: the time dimension (and this is why it is advisable to write stochastic processes always with a time index : so we have $X(t), \; t=1,...T$ but $Y(s), s=T+1,..., 2T$).
But if independence applies, you can "view" $X$ and $Y$ as two distinct processes, irrespective of the fact that it seems they share "common characteristics" (same people etc) - well, they don't: under independence, they change of the time index alone is sufficient to make things distinct (although possibly identical) from a probabilistic, stochastic, statistical point of view.
Then, @whuber has given you the answer.
