Signal, noise, and, unequal variances

  • Thread starter Thread starter Math Is Hard
  • Start date Start date
  • Tags Tags
    Noise Signal
Math Is Hard
Staff Emeritus
Science Advisor
Gold Member
Messages
4,650
Reaction score
39
I have two overlapping Gaussian distributions. One for probability of a noise event, one for probability of a signal event.
My noise distribution has mean = 0, variance = 1, and standard deviation = 1.
My signal distribution has mean = .80, variance = 3, and standard deviation = √3.

My criterion (λ ) is at 0.5 standard deviations above the mean on the noise distribution.

I need to calculate probability of a hit (PH) and probability of a false alarm (PF).

To get PF , I use 1- PCorrect Rejection, or 1 – Ф(0.5), which, using a Z to % conversion table, gives me 1- 0.691 = 0.31. (In my book, Ф(Z) just means converting a Z-score to % of area under the curve.)

To get PH, I first need to standardize the Z score for λ on the signal distribution since it has a different variance:

λ = (λ - μsignal)/σ signal = (0.5-0.8) / √3 = - 0.173

PH = 1 - Ф(-0.173) or by symmetry, PH = Ф(0.173) = 0.57. (again, using a Z to % conversion table to look up the area.)

So.. PF = = 0.31 and PH = 0.57.

I'm not sure if I did this right. I am shaky with unequal variances. If someone could check I'd appreciate it.

I need to sketch the overlapping distributions. Since I have standardized the z-score on the signal distribution, do I make it look identical to the noise curve (with a standard deviation of 1), or do I draw it short and wide as it is originally described?

Thanks!
 
Physics news on Phys.org
It looks like you did the calculations correctly! To sketch the overlapping distributions, draw the noise distribution with a standard deviation of 1 and then draw the signal distribution with the specified standard deviation (√3). You can make the shapes the same and just adjust the size to represent the different standard deviations. Good luck!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top