

Subject:
The normal distribution in probability theory
Category: Science > Math Asked by: iaindewittga List Price: $2.00 
Posted:
13 May 2006 01:46 PDT
Expires: 12 Jun 2006 01:46 PDT Question ID: 728354 
Why is it that the normal distribution contains a factor of 1/2 in the exponent? (Alternately, why does the coefficient contain a factor of sq. rt. 1/2?) The two factors, 1/2 in the exponent and sq. rt. 1/2 in the coeffiecient, come as a set (i.e., once 1/2 is put into the exponent, a correction of sq. rt. 1/2 is needed in the coeffiecient to preserve unity in the distribution's integral). Each is necessary when the other is present but, why include either in the distribution at all? What functionality do they add? It seems that all the essential properties of the distribution would be present absent these factors. My two speculations are: One, these factors help the distribution better fit 'normal' data (i.e., the distribution is designed to accommodate a pattern seen in nature). This implies there is a hitherto undiscovered explanation that rests on a principle of natural phenomena. Two, the normal distribution was originally derived from other distributions (as an approximation to them) and the factor of 1/2 in the exponent is a vestigial remnant, an emblem of lineage. What's the real reason? 

There is no answer at this time. 

Subject:
Re: The normal distribution in probability theory
From: kottekoega on 13 May 2006 07:56 PDT 
The 1/2 in the exponent is used to give the distribution a variance of 1. This is an arbitrary convention, but a convenient one. 
Subject:
Re: The normal distribution in probability theory
From: berkeleychocolatega on 18 May 2006 16:23 PDT 
The sqrt of 2 in the denominator and the 1/2 in the numerator are not arbitrary. The central limit theorem of statistics says that averages of simple random samples converge to the normal distribution. That's where the numbers come from altnough it isn't trivial to show this is true. 
Subject:
Re: The normal distribution in probability theory
From: kottekoega on 18 May 2006 19:27 PDT 
Chocolate: It is still a normal distribution with any number replacing the 2. The general form is: (1/sqrt(2*pi*s))exp(x^2/(2s)) Where s is the variance of the distribution. The variance just measures the width of the peak of the normal distribution. Clearly this width will be different if you were talking about the average size of a mosquito or the average size of a star, measuring both in meters. The Central Limit Theorem just tells us that the distribution of the sum of a large number of independent random variable will obey the general form given above. It doesn't tell us what the variance is. 
Subject:
Re: The normal distribution in probability theory
From: iaindewittga on 19 May 2006 17:47 PDT 
Thank you, Kottekoe (and Berkeleychocolate). To be clear, the 1/2 in the exponent is to ensure that the variance can be succinctly described by the parameter 's'. Without the 1/2, finding the variance from the mgf would result in a product of 's' and some constant, presumably 2*sI haven't completed the algebra myself. That said, the value of 's' itself is not necessarily 1. 's' may take any value. Correct? 
Subject:
Re: The normal distribution in probability theory
From: kottekoega on 21 May 2006 20:57 PDT 
Yes, that is correct. 
Subject:
Re: The normal distribution in probability theory
From: activealexaokiga on 11 Jun 2006 18:00 PDT 
I think everyone missed the mathematical background. I even think everyone is using analogies to explain that the 1/2 is conventionally inserted  which is TOTALLY WRONG. They are suggested due to the nature of probability  never can be over 1. Normal distribution curves or general distribution curves are Probability Density Function (PDF). The horizontal axis covers all the possible outcomes X. Suppose each outcome is x. PDF of a particular outcome is thus defined by PDF(x), when x is continuous varialbe Statisticians say P(X=x)=PDF(x) but the fact is P(X=x)=PDF(x)*dx. By calculus, dx is the smallest possible interval thus concluding the probability to be approximately 0, which is good enough for them. Anyway, the probability that is always 1 is the cumulative probability of all the outcome. We call it Cumulative Density Function (CDF). As PDf suggests, CDF is the summation of the probability of all possible outcomes. Geometrically it is the area under the distribution curve. Therefore the area under distribution curve needs to be ALWAYS 1 to be reasonable. By the way, the general form using normal scores is: 1/{sigma*sqrt(2*pi)} * exp{(1/2)*z^2} where z=(xaverage)/sigma, The general form of any level of variables is: 1/{sigma*sqrt(2*pi)} * exp{(1/2)*(x/s)^2}, then the distribution crosses vertical axis (average=0) And the function we know very well satisfies both property of symmetric bell curve (that outcomes nearer to the average occur more likely than extremities) and the cumulative probability is always 1. In conclusion, it is no sort of inserting 1/2 conveniently together with sqrt(1/2) 
Subject:
Re: The normal distribution in probability theory
From: kottekoega on 12 Jun 2006 18:18 PDT 
Active Alex: I beg to differ. In his question, the original poster made it very clear that he understood that the factor in the exponent and in the normalizing factor were connected by the fact that the integral of the PDF must be unity. Your general form and mine are identical and, as you can plainly see from either one, the factor in the exponent can take on any positive value. 
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