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Deriving the Gaussian density probability equation

by CuriousQuazim
Tags: density function, gaussian
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CuriousQuazim
#1
Feb29-12, 08:39 AM
P: 5
Hey ^^, new here but I already have a question haha

Does anyone here know how the coefficient (x-μ)^2 was derived in the following equation:

σ^3=(1/√2∏)∫(1/σ)*(x-μ)^2*exp((x-μ)^2)/(2σ^2))

I know the general equation for density probability is (1/σ)*exp((x-μ)^2)/(2σ^2))
but in this case I can't quite see how the coefficient came about... any help?

Thanks in advance!
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alan2
#2
Feb29-12, 01:17 PM
P: 199
Your expression looks wrong to me. Could you check it for accuracy?
mathman
#3
Feb29-12, 04:20 PM
Sci Advisor
P: 6,038
It looks like it should be σ2. The expression is essentially the definition of the variance, the second moment of the distribution centered at the mean.

CuriousQuazim
#4
Mar1-12, 04:46 AM
P: 5
Deriving the Gaussian density probability equation

Oh I'm sorry that was an error on my part, it is indeed σ^2

σ^2=(1/√2∏)∫(1/σ)*(x-μ)^2*exp((x-μ)^2)/(2σ^2))

Ah thank you so much mathman ^^, that's what I was looking for! I'm studying engineering so sometimes they just throw mathematical equations at us with no explanation _.


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