Variance of Uniform Normal Distribution

Posted on February 21, 2017 by Sumant Sumant

Let the corresponding probabilities are \frac{1}{n+1} for all at the points \{0,1,2,\cdots,n+1\}
The general formula is \sigma^2 = E[x^2]-(E[X])^2
Let’s first calculate E[X]=0\cdot \frac{1}{n+1}+1\cdot \frac{1}{n+1}+\cdots+n\cdot \frac{1}{n+1}=\frac{0+1+2+\cdots+n}{n+1}=\frac{n(n+1)}{2(n+1)}=\frac{n}{2}
Now lets calculate E[X]=0^2\cdot \frac{1}{n+1}+1^2\cdot \frac{1}{n+1}+\cdots+n^2\cdot \frac{1}{n+1}=\frac{0^2+1^2+2^2+\cdots+n^2}{n+1}=\frac{n(n+1)(2n+1)}{6(n+1)}=\frac{n(2n+1)}{6}
Therefore \sigma^2 = \frac{n(2n+1)}{6}-\left ( \frac{n}{2}\right )^2
\Rightarrow \frac{n(2n+1)}{6}-\frac{n^2}{4}
\Rightarrow \frac{2n(2n+1)-3n^2}{12}
\Rightarrow \frac{4n^2+2n-3n^2}{12}
\Rightarrow \frac{n^2+2n}{12}=\frac{n(n+2)}{12}

Now suppose we have same n+1 terms shifted from a to b in that case the variance becomes \frac{(b-a)(b-a+2)}{12}

About Sumant Sumant

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