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 !!
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 \( \lim_{k \rightarrow \infty} a_{k} \) .
Integration
Gamma function
Definite integral as the limit of the Riemann sum
\( a_{k}=\lim {n \rightarrow \infty} \frac{1}{n} \sum_{m=1}^{kn} \exp \left(-\frac{1}{2} \frac{m^{2}}{n^{2}}\right) = \int_{0}^{k} e^{\frac{-y^2}{2}} dy \) , 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 ) .
Find the same when \( a_{k}=\lim {n \rightarrow \infty} \frac{1}{n} \sum_{m=1}^{kn} {(\frac{m}{n})}^{5} \exp \left(-\frac{1}{2} \frac{m}{n}\right) \).
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 !!
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 \( \lim_{k \rightarrow \infty} a_{k} \) .
Integration
Gamma function
Definite integral as the limit of the Riemann sum
\( a_{k}=\lim {n \rightarrow \infty} \frac{1}{n} \sum_{m=1}^{kn} \exp \left(-\frac{1}{2} \frac{m^{2}}{n^{2}}\right) = \int_{0}^{k} e^{\frac{-y^2}{2}} dy \) , 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 ) .
Find the same when \( a_{k}=\lim {n \rightarrow \infty} \frac{1}{n} \sum_{m=1}^{kn} {(\frac{m}{n})}^{5} \exp \left(-\frac{1}{2} \frac{m}{n}\right) \).
