Question

Prove that the standard deviation of first n natural numbers is equal to
$\sigma =\sqrt{\dfrac{{{n}^{2}}-1}{12}}$

Answer 1

Hint: Use the definition of standard deviation as the square root of the variance and $\operatorname{var}\left( X \right)=E\left( {{\left( X-\mu \right)}^{2}} \right)$ where $\mu =E\left( X \right)$. Use linearity of expression to prove that $\operatorname{var}\left( X \right)=E\left( {{X}^{2}} \right)-{{\left[ E\left( X \right) \right]}^{2}}$.

Using formula form the sum of squares of first n natural numbers and the sum of first n natural number to find the individual terms in the expression for var(X) , $\sum\limits_{r=1}^{n}{{{r}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ and $\sum\limits_{r=1}^{n}{r}=\dfrac{n\left( n+1 \right)}{2}$

Complete step-by-step answer:

We know that
$\operatorname{var}\left( X \right)=E\left( {{\left( X-\mu \right)}^{2}} \right)$
Using ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ we get
$\operatorname{var}(X)=E\left( {{X}^{2}}+{{\mu }^{2}}-2X\mu \right)$

Now we know that $E\left( X+Y \right)=E\left( X+Y \right)$

Using the above formula, we get
$\operatorname{var}(X)=E\left( {{X}^{2}} \right)+E\left( {{\mu }^{2}} \right)+E\left( -2X\mu \right)$

We know that $E\left( aX \right)=aE\left( X \right)$ and $E\left( a \right)=a$, we get
$\operatorname{var}(X)=E\left( {{X}^{2}} \right)+{{\mu }^{2}}-2\mu E\left( X \right)$
Using $\mu =E\left( X \right)$
$\begin{align}
  & \operatorname{var}(X)=E\left( {{X}^{2}} \right)+{{\left( E\left( X \right) \right)}^{2}}-2E\left( X \right)E\left( X \right) \\
 & \Rightarrow \operatorname{var}\left( X \right)=E\left( {{X}^{2}} \right)+{{\left( E\left( X \right) \right)}^{2}}-2{{\left( E\left( X \right) \right)}^{2}} \\
 & \Rightarrow \operatorname{var}\left( X \right)=E\left( {{X}^{2}} \right)-{{\left( E\left( X \right) \right)}^{2}} \\
\end{align}$

The standard deviation of discrete statistic data D is equal to the standard deviation of random variable X having the following properties,
$P\left( X=x \right)=\dfrac{1}{\left| D \right|},\text{ if }x\in D$and $P\left( X=x \right)=0,\text{ if }x\notin D$

So we choose X to be the random variable such that $P\left( X=r \right)=\dfrac{1}{n},1\le r\le n$ and $P\left( X \right)=0\text{ otherwise}$.

The standard deviation of this random variable will be equal to the standard deviation of first n natural numbers.

We know that \[E\left( f\left( X \right) \right)=\sum\limits_{r\in S}{P\left( X=r \right)f\left( r \right)}\]

Using the above formula taking f(x) = x, we get
$\begin{align}
  & E\left( X \right)=\sum\limits_{r=1}^{n}{P\left( X=r \right)r} \\
 & =\sum\limits_{r=1}^{n}{\dfrac{1}{n}r} \\
 & =\dfrac{1}{n}\sum\limits_{r=1}^{n}{r} \\
\end{align}$

Using $\sum\limits_{r=1}^{n}{r}=\dfrac{n\left( n+1 \right)}{2}$, we get
$\begin{align}
  & E\left( X \right)=\dfrac{n\left( n+1 \right)}{2n} \\
 & \Rightarrow E\left( X \right)=\dfrac{n+1}{2} \\
\end{align}$

We know that \[E\left( f\left( X \right) \right)=\sum\limits_{r\in S}{P\left( X=r \right)f\left( r \right)}\]

Using the above formula taking $f\left( x \right)={{x}^{2}}$, we get
$\begin{align}
  & E\left( {{X}^{2}} \right)=\sum\limits_{r=1}^{n}{P\left( X=r \right){{r}^{2}}} \\
 & =\dfrac{1}{n}\sum\limits_{r=1}^{n}{{{r}^{2}}} \\
\end{align}$

Using $\sum\limits_{r=1}^{n}{{{r}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$, we get
$\begin{align}
  & E\left( {{X}^{2}} \right)=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6n} \\
 & =\dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6} \\
\end{align}$

Using $\operatorname{var}\left( X \right)=E\left( {{X}^{2}} \right)-{{\left( E\left( X \right) \right)}^{2}}$, we get
$\begin{align}
  & \operatorname{var}\left( X \right)=\dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}-{{\left( \dfrac{n+1}{2} \right)}^{2}} \\
 & =\dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}-\dfrac{{{\left( n+1 \right)}^{2}}}{4} \\
 & =\dfrac{n+1}{12}\left( 4n+2-3n-3 \right) \\
 & =\dfrac{\left( n+1 \right)\left( n-1 \right)}{12} \\
 & =\dfrac{{{n}^{2}}-1}{12} \\
\end{align}$

We know that $\sigma \left( X \right)=\sqrt{\operatorname{var}\left( X \right)}$
Using the above formula, we get
$\sigma =\sqrt{\dfrac{{{n}^{2}}-1}{12}}$

Note: [1] Instead of using probability for finding the standard deviation, you can use the formula
\[E\left( D \right)=\dfrac{\sum\limits_{x\in D}{x}}{\left| D \right|}\] and $E\left( {{D}^{2}} \right)=\dfrac{\sum\limits_{x\in D}{{{x}^{2}}}}{\left| D \right|}$
Here we have $D=\left\{ 1,2,3,..,n \right\}$
So \[E\left( D \right)=\dfrac{\sum\limits_{x\in D}{x}}{\left| D \right|}=\dfrac{\sum\limits_{r=1}^{n}{r}}{n}=\dfrac{n+1}{2}\]
and $E\left( {{D}^{2}} \right)=\dfrac{\sum\limits_{r=1}^{n}{{{r}^{2}}}}{n}=\dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}$
which is the same as obtained above.