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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!
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!