I am looking at estimating the output of two machines. However, the only data I have is the combined output of these machines. These machines are pretty much identical and interchangeable.

The data I have is the weekly production rate (in linear feet / hr) of the WORKSTATION. The workstation has two identical machines which are both constantly fed from the same queue. So the data I have is basically (Machine1LF + Machine2LF) and (Machine1WorkHr + Machine2WorkHr) for the last 40 weeks. I need to take this aggregate data and estimate the weekly average and standard deviation of LF/hr for each machine.

To estimate the standard deviation of each machine's output (LF/hr) I intuitively arrived at:

$$ \sigma^{2} _{X} + \sigma^{2} _{Y} = \sigma^{2} _{(X+Y)} $$

Where $$ \sigma^{2}_X = \sigma^{2}_Y $$ yielding

$$ \sigma_{x} = \sigma_{y} = \frac{\sqrt{\sigma^{2} _{(X+Y)}}}{\sqrt{2}} $$

This seems right to me and looks to test correctly when I simulate the results, but searching the internet for a formal definition the only thing I can find is:

$$\sigma_{(X+Y)} = \sqrt{\sigma^{2}_X + \sigma^{2}_Y + 2cov(X,Y)} $$

which has the added covariance term, and understanding that the covariance of a number with itself is not 0, it is the variance of itself which leads me to:

$$\sigma_{(X+Y)} = \sqrt{\sigma^{2}_X + \sigma^{2}_Y + 2\sigma^{2}_X} $$ yielding

$$ \sigma_{x} = \sigma_{y} = \frac{\sqrt{\sigma^{2} _{(X+Y)}}}{2} $$

I would be fine going with this since it is the formal definition, but the problem is that it does not test out when I simulate the results. Could someone please advise where my assumptions or calculations went wrong?

  • $\begingroup$ Your problem is not well defined: Is the output of each machine a random variable? If not - the standard deviation is meaningless. Also - what do you mean by "estimating the output of two machines" - the average? Another remark - what does "have the same output" mean - the same distribution? Last thing - you can probably assume cov(X,Y)=0. You need to rewrite your question to make it mathematically sound. $\endgroup$
    – Zahava Kor
    Commented Feb 7, 2018 at 18:49
  • $\begingroup$ @ZahavaKor Noted, I will update my question for clarity $\endgroup$ Commented Feb 7, 2018 at 18:52
  • $\begingroup$ Your evaluation of the covariance is erroneous. You have no data with which to estimate it, either. In this circumstance one would expect it to be positive because common contributions from the input queue would tend to affect the two machines simultaneously in similar ways. You could use this assumption to provide bounds on the standard deviations. $\endgroup$
    – whuber
    Commented Feb 7, 2018 at 19:03

1 Answer 1


After @whuber and @ZahavaKor comments, especially noting since I actually have no data to base a covariance on I have arrived at the following:

Thinking on this further, I believe my assumption that the covariance was not zero is the one that is failing. Since the input of part complexity to the queue for the machines is random then the output of complexity is random. For the covariance to be nonzero, then the LF/hr of machine1 would need to somehow affect the LF/hr of machine2 which is not the case.

I was assuming that for each point in the dataset machine1 took 1/2 and machine2 took 1/2 which would show that for each datapoint X=Y=(X+Y)/2. This would lead to very high covariance. However, it is very unlikely that the weekly production rate of each machine was the same, however their long-term average production rates WOULD be the same.

TLDR: The covariance is 0 since the machines are independent and their inputs are random.


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