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

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