ISI MStat PSB 2008 Problem 2 | Definite integral as the limit of the Riemann sum

This is a very beautiful sample problem from ISI MStat PSB 2008 Problem 2 based on definite integral as the limit of the Riemann sum . Let's give it a try !!

Problem- ISI MStat PSB 2008 Problem 2

For $k \geq 1,$ let $a_{k}=\lim {n \rightarrow \infty} \frac{1}{n} \sum_{m=1}^{kn} \exp \left(-\frac{1}{2} \frac{m^{2}}{n^{2}}\right)$

Find $\lim_{k \rightarrow \infty} a_{k}$ .

Prerequisites

Integration

Gamma function

Definite integral as the limit of the Riemann sum

Solution :

$a_{k}=\lim {n \rightarrow \infty} \frac{1}{n} \sum_{m=1}^{kn} \exp \left(-\frac{1}{2} \frac{m^{2}}{n^{2}}\right) = \int_{0}^{k} e^{\frac{-y^2}{2}} dy$ , this can be written you may see in details Definite integral as the limit of the Riemann sum .

Therefore , $lim_{k \to \infty} a_{k}= \int_{0}^{ \infty} e^{\frac{-y^2}{2}} dy$ ----(1) , let $\frac{y^2}{2}=z \Rightarrow dy= \frac{dz}{\sqrt{2z}}$

Substituting we get , $\int_{0}^{ \infty} z^{\frac{1}{2} -1} e^{z} \frac{1}{\sqrt{2}} dz =\frac{ \gamma(\frac{1}{2}) }{\sqrt{2}} = \sqrt{\frac{\pi}{2}}$

Statistical Insight

Let $X \sim N(0,1)$ i.e X is a standard normal random variable then,

$Y=|X|$ called folded Normal has pdf $f_{Y}(y)= \begin{cases} \frac{2}{\sqrt{2 \pi }} e^{\frac{-x^2}{2}} & , y>0 \\ 0 &, otherwise \end{cases}$ . (Verify!)

So, from (1) we can say that $\int_{0}^{ \infty} e^{\frac{-y^2}{2}} dy = \frac{\sqrt{2 \pi }}{2} \int_{0}^{ \infty}\frac{2}{\sqrt{2 \pi }} f_{Y}(y) dy$

$=\frac{\sqrt{2 \pi }}{2} \times 1$ ( As that a PDF of folded Normal distribution ) .

Food For Thought

Find the same when $a_{k}=\lim {n \rightarrow \infty} \frac{1}{n} \sum_{m=1}^{kn} {(\frac{m}{n})}^{5} \exp \left(-\frac{1}{2} \frac{m}{n}\right)$ .

Outstanding Statistics Program with Applications