Cheenta
Get inspired by the success stories of our students in IIT JAM MS, ISI  MStat, CMI MSc DS.  Learn More

ISI MStat PSB 2009 Problem 2 | Linear Difference Equation

This is a very beautiful sample problem from ISI MStat PSB 2009 Problem 2 based on Convergence of a sequence. Let's give it a try !!

Problem- ISI MStat PSB 2009 Problem 2

Let $$\{x_{n}: n \geq 0\}$$ be a sequence of real numbers such that
$$x_{n+1}=\lambda x_{n}+(1-\lambda) x_{n-1}, n \geq 1,$$ for some $$0<\lambda<1$$
(a) Show that $$x_{n}=x_{0}+(x_{1}-x_{0}) \sum_{k=0}^{n-1}(\lambda-1)^{k}$$
(b) Hence, or otherwise, show that $$x_{n}$$ converges and find the limit.

Prerequisites

Limit

Sequence

Linear Difference Equation

Solution :

(a) We are given that $$x_{n+1}=\lambda x_{n}+(1-\lambda) x_{n-1}, n \geq 1,$$ for some $$0<\lambda<1$$

So, $$x_{n+1} - x_{n} = -(1- \lambda)( x_n-x_{n-1})$$ ---- (1)

Again using (1) we have $$( x_n-x_{n-1})= -(1- \lambda)( x_{n-1}-x_{n-2})$$ .

Now putting this in (1) we have , $$x_{n+1} - x_{n} = {(-(1- \lambda))}^2 ( x_{n-1}-x_{n-2})$$ .

So, proceeding like this we have $$x_{n+1} - x_{n} = {(-(1- \lambda))}^n ( x_{1}-x_{0})$$ for all $$n \geq 1$$ and for some $$0<\lambda<1$$---- (2)

So, from (2) we have $$x_{n} - x_{n-1} = {(-(1- \lambda))}^{n-1} ( x_{1}-x_{0})$$ , $$\cdots , (x_2-x_1)=-(\lambda-1)(x_1-x_{0})$$ and $$x_1-x_{0}=x_{1}-x_{0}$$

Adding all the above n equation we have $$x_{n}-x_{0}=(x_{1}-x_{0}) \sum_{k=0}^{n-1} {(\lambda-1)}^{k}$$

Hence , $$x_{n}=x_{0}+(x_{1}-x_{0}) \sum_{k=0}^{n-1}(\lambda-1)^{k}$$ (proved ) .

(b) As we now have an explicit form of $$x_{n}=x_{0}+(x_{1}-x_{0}) \times \frac{1-{( \lambda -1)}^n}{1-(\lambda -1)}$$ ----(3)

Hence from (3) we can say $$x_{n}$$ is bounded and monotonic ( verify ) so , it's convergent .

Now let's take $$\lim_{n\to\infty}$$ both side of (3) we get , $$\lim_{x\to\infty} x_{n} = x_{0}+(x_{1}-x_{0}) \times \frac{1}{2 - \lambda}$$ .

Since , $$\lim_{x\to\infty} {( \lambda - 1)}^{n} = 0$$ as , $$-1 < \lambda -1 < 0$$ .

Food For Thought

$$\mathrm{m}$$ and $$\mathrm{k}$$ are two natural number and $$a_{1}, a_{2}, \ldots, a_{m}$$ and $$b_{1}, b_{2}, \ldots, b_{k}$$ are two sets of positive real numbers such that $$a_{1}^{\frac{1}{n}}+a_{2}^{\frac{1}{n}}+\cdots+a_{m}^{\frac{1}{n}}$$ = $$b_{1}^{\frac{1}{n}}+\cdots+b_{k}^{\frac{1}{n}}$$

for all natural number $$\mathrm{n} .$$ Then prove that $$\mathrm{m}=\mathrm{k}$$ and $$a_{1} a_{2} \ldots a_{m}=b_{1} b_{2} . . b_{k}$$ .

Outstanding Statistics Program with Applications

This is a very beautiful sample problem from ISI MStat PSB 2009 Problem 2 based on Convergence of a sequence. Let's give it a try !!

Problem- ISI MStat PSB 2009 Problem 2

Let $$\{x_{n}: n \geq 0\}$$ be a sequence of real numbers such that
$$x_{n+1}=\lambda x_{n}+(1-\lambda) x_{n-1}, n \geq 1,$$ for some $$0<\lambda<1$$
(a) Show that $$x_{n}=x_{0}+(x_{1}-x_{0}) \sum_{k=0}^{n-1}(\lambda-1)^{k}$$
(b) Hence, or otherwise, show that $$x_{n}$$ converges and find the limit.

Prerequisites

Limit

Sequence

Linear Difference Equation

Solution :

(a) We are given that $$x_{n+1}=\lambda x_{n}+(1-\lambda) x_{n-1}, n \geq 1,$$ for some $$0<\lambda<1$$

So, $$x_{n+1} - x_{n} = -(1- \lambda)( x_n-x_{n-1})$$ ---- (1)

Again using (1) we have $$( x_n-x_{n-1})= -(1- \lambda)( x_{n-1}-x_{n-2})$$ .

Now putting this in (1) we have , $$x_{n+1} - x_{n} = {(-(1- \lambda))}^2 ( x_{n-1}-x_{n-2})$$ .

So, proceeding like this we have $$x_{n+1} - x_{n} = {(-(1- \lambda))}^n ( x_{1}-x_{0})$$ for all $$n \geq 1$$ and for some $$0<\lambda<1$$---- (2)

So, from (2) we have $$x_{n} - x_{n-1} = {(-(1- \lambda))}^{n-1} ( x_{1}-x_{0})$$ , $$\cdots , (x_2-x_1)=-(\lambda-1)(x_1-x_{0})$$ and $$x_1-x_{0}=x_{1}-x_{0}$$

Adding all the above n equation we have $$x_{n}-x_{0}=(x_{1}-x_{0}) \sum_{k=0}^{n-1} {(\lambda-1)}^{k}$$

Hence , $$x_{n}=x_{0}+(x_{1}-x_{0}) \sum_{k=0}^{n-1}(\lambda-1)^{k}$$ (proved ) .

(b) As we now have an explicit form of $$x_{n}=x_{0}+(x_{1}-x_{0}) \times \frac{1-{( \lambda -1)}^n}{1-(\lambda -1)}$$ ----(3)

Hence from (3) we can say $$x_{n}$$ is bounded and monotonic ( verify ) so , it's convergent .

Now let's take $$\lim_{n\to\infty}$$ both side of (3) we get , $$\lim_{x\to\infty} x_{n} = x_{0}+(x_{1}-x_{0}) \times \frac{1}{2 - \lambda}$$ .

Since , $$\lim_{x\to\infty} {( \lambda - 1)}^{n} = 0$$ as , $$-1 < \lambda -1 < 0$$ .

Food For Thought

$$\mathrm{m}$$ and $$\mathrm{k}$$ are two natural number and $$a_{1}, a_{2}, \ldots, a_{m}$$ and $$b_{1}, b_{2}, \ldots, b_{k}$$ are two sets of positive real numbers such that $$a_{1}^{\frac{1}{n}}+a_{2}^{\frac{1}{n}}+\cdots+a_{m}^{\frac{1}{n}}$$ = $$b_{1}^{\frac{1}{n}}+\cdots+b_{k}^{\frac{1}{n}}$$

for all natural number $$\mathrm{n} .$$ Then prove that $$\mathrm{m}=\mathrm{k}$$ and $$a_{1} a_{2} \ldots a_{m}=b_{1} b_{2} . . b_{k}$$ .

Outstanding Statistics Program with Applications

This site uses Akismet to reduce spam. Learn how your comment data is processed.