1
$\begingroup$

I have two normally distributed random variables that add together to form a third normally distributed variable. The means in all cases are 0. Thus, the standard deviations of the first 2 variables add in quadrature:

sigma_c^2=sigma_a^2+sigma_b^2

I want to measure directly sigma_a, but the instrument used adds sigma_b noise. Currently I measure (separately) sigma_b and sigma_c, then solve for sigma_a.

The problem is when the instrument noise floor dominates the signal of interest. In the extreme case when this occurs, sigma_c (the signal of interest plus noise) approximately equals sigma_b (the noise), and my estimate of sigma_a (the signal of interest) returns 0, which is incorrect (e.g. I've reached the noise floor of the instrument).

Is it possible to somehow quantify the error in sigma_a as sigma_b approaches sigma_c?

Ultimately I need to determine whether sigma_a is below a compliance limit L with some amount of certainty. Is this possible given the above data?

UPDATE 1

Is the following true?

"If n samples were taken to compute each of sigma_c and sigma_b, then the standard error for the signal of interest is sigma_a/sqrt(n). Thus, one can be 95% confident that the sample mean of distribution a is within plus or minus 1.96*(standard error) from the true population mean."

$\endgroup$

1 Answer 1

0
$\begingroup$

Yes, averaging reduces the noise. As you already stated $$Std[\bar X_n] = Std[X_1]/\sqrt{n}$$ where $Std[X_1]$ is the standard deviation and the mean is defined as $$\bar X_n = \frac{1}{n} \sum_{i=1}^n X_i$$ However, you should be careful, because shifts in the mean could broaden your averaged signal.

Another approach is an experimental one: If you are interested in a periodic signal, maybe you could use a lock-in amplifier to reduce the noise.

$\endgroup$
2
  • $\begingroup$ Assuming the $X_i$ variables are IID, shouldn't the variance be reduced by a factor $n$ (instead of $\sqrt{n}$)? $\endgroup$
    – Luca Citi
    Jul 3, 2017 at 21:53
  • $\begingroup$ I'm sorry. As stated, user46688s initial formula is correct, I corrected mine. Thank you, Luca. $\endgroup$
    – Semoi
    Jul 3, 2017 at 22:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.