My Math Forum Probabilities in the normal distribution

 May 14th, 2012, 11:45 AM #1 Member   Joined: Aug 2009 From: Copenhagen, Denmark Posts: 45 Thanks: 0 Probabilities in the normal distribution Hello all I am writing an assignment where I, among other things, have to write about simple properties of the normal distributions. I have to describe why the form of the normal distribution changes as the variance changes, e.g. why the normal distribution flattens out. My argument is as follows: Consider a normal distribution with mean $\mu$ and variance $\sigma ^2$. Now we look at the probability of getting a value in the interval $[a;b]$, where $a. When we use the probability density function, we get that: $P(X=x)=\int _a^b\frac{1}{\sqrt{2 \pi \sigma ^2}}e^{-\frac{1}{2}\left(\frac{x-\mu }{\sigma }\right)^2}dx$ Now, let $\sigma ^2\to \infty$. We get that $\underset{\sigma ^2\to \infty }{\text{limit}}\int_a^b e^{-\frac{1}{2}\left(\frac{x-\mu }{\sigma }\right)^2} \, dx=a-b$ and $\underset{\sigma ^2\to \infty }{\text{limit}}\int_a^b \frac{1}{\sqrt{2\pi \sigma ^2 }} \, dx=0$ By using some rule of limits (?) we conclude that the limit of the normal distribution on an interval $[a;b]$ approaches 0 as $\sigma ^2\to \infty$. Does this statement hold? Don't pay much attention to the formulations - but does the mathematical part make sense?
 May 14th, 2012, 01:36 PM #2 Global Moderator   Joined: May 2007 Posts: 6,854 Thanks: 744 Re: Probabilities in the normal distribution Effectively you are showing that the normal distribution flattens out as ? gets larger. There are some errors (typos?) in your analysis. You have P(X=x) when you should have P(a < X < b) The integral limit should be b - a, not a - b.
May 14th, 2012, 01:54 PM   #3
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Re: Probabilities in the normal distribution

Quote:
 Originally Posted by mathman There are some errors (typos?) in your analysis.
Thanks for the quick reply! Yes, the integral limit was a typo, while the P(X=x) was a misunderstanding.Thanks for clearing that up

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