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ISI MStat PSB 2014 Problem 2 | Properties of a Function

This is a very beautiful sample problem from ISI MStat PSB 2014 Problem 2 based on the use and properties of a function . Let's give it a try !!

Problem- ISI MStat PSB 2014 Problem 2


Let a_{1}<a_{2}<\cdots<a_{m} and b_{1}<b_{2}<\cdots<b_{n} be real numbers such
that \sum_{i=1}^{m}\left|a_{i}-x\right|=\sum_{j=1}^{n}\left|b_{j}-x\right| \text { for all } x \in \mathbb{R}
Show that m=n and a_{j}=b_{j} for 1 \leq j \leq n

Prerequisites


Differentiability

Mod function

continuity

Solution :

Let , \sum_{i=1}^{m}\left|a_{i}-x\right|=\sum_{j=1}^{n}\left|b_{j}-x\right|=f(x)  \text { for all } x \in \mathbb{R}

Then , f(x)=\sum_{i=1}^{m}\left|a_{i}-x\right| is not differentiable at x=a_1,a_2, \cdots , a_m ---(1)

As we know the function |x-a_i| is not differentiable at x=a_i .

Again we have , f(x) = \sum_{j=1}^{n}\left|b_{j}-x\right| it also not differentiable at x= b_1,b_2, \cdots , b_n ----(2)

Hence from (1) we get f has m non-differentiable points and from (2) we get f has n non-differentiable points , which is possible only when m and n are equal .

And also the points where f is not differentiable must be same in both (1) and (2) .

As we have the restriction that a_{1}<a_{2}<\cdots<a_{m} and b_{1}<b_{2}<\cdots<b_{n} .

So , we have a_{j}=b_{j} for 1 \leq j \leq n .


Food For Thought

a<b \in \mathbb{R} . Let f:[a, b] \rightarrow[a, b] be a continuous and differentiable on (a,b) . Suppose that \left|f^{\prime}(x)\right| \leq \alpha<1 for all x \in(a, b) for some \alpha . Then prove that there exists unique x \in[a, b] such that f(x)=x


Similar Problems and Solutions



ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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This is a very beautiful sample problem from ISI MStat PSB 2014 Problem 2 based on the use and properties of a function . Let's give it a try !!

Problem- ISI MStat PSB 2014 Problem 2


Let a_{1}<a_{2}<\cdots<a_{m} and b_{1}<b_{2}<\cdots<b_{n} be real numbers such
that \sum_{i=1}^{m}\left|a_{i}-x\right|=\sum_{j=1}^{n}\left|b_{j}-x\right| \text { for all } x \in \mathbb{R}
Show that m=n and a_{j}=b_{j} for 1 \leq j \leq n

Prerequisites


Differentiability

Mod function

continuity

Solution :

Let , \sum_{i=1}^{m}\left|a_{i}-x\right|=\sum_{j=1}^{n}\left|b_{j}-x\right|=f(x)  \text { for all } x \in \mathbb{R}

Then , f(x)=\sum_{i=1}^{m}\left|a_{i}-x\right| is not differentiable at x=a_1,a_2, \cdots , a_m ---(1)

As we know the function |x-a_i| is not differentiable at x=a_i .

Again we have , f(x) = \sum_{j=1}^{n}\left|b_{j}-x\right| it also not differentiable at x= b_1,b_2, \cdots , b_n ----(2)

Hence from (1) we get f has m non-differentiable points and from (2) we get f has n non-differentiable points , which is possible only when m and n are equal .

And also the points where f is not differentiable must be same in both (1) and (2) .

As we have the restriction that a_{1}<a_{2}<\cdots<a_{m} and b_{1}<b_{2}<\cdots<b_{n} .

So , we have a_{j}=b_{j} for 1 \leq j \leq n .


Food For Thought

a<b \in \mathbb{R} . Let f:[a, b] \rightarrow[a, b] be a continuous and differentiable on (a,b) . Suppose that \left|f^{\prime}(x)\right| \leq \alpha<1 for all x \in(a, b) for some \alpha . Then prove that there exists unique x \in[a, b] such that f(x)=x


Similar Problems and Solutions



ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


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One comment on “ISI MStat PSB 2014 Problem 2 | Properties of a Function”

  1. This problem has an absolutely elementary solution. For any x < \min (a_1, b_1), the given condition implies that:

        \[\sum_{i=1}^m a_i - mx= \sum_{j=1}^n  b_j - nx.\]

    i.e.,

        \[(m-n)x= 	\sum_{i=1}^m a_i  - \sum_{j=1}^n  b_j.\]

    If m\neq n, this linear equation has only one solution. But the above equation should be valid for any of the infinitely many x<\min (a_1, b_1). Hence we must have m=n and \sum_{i=1}^na_i =\sum_{i=1}^n b_i.\\
    Next, if a_1 \neq b_1, assume a_1 < b_1, and choose x such that a_1<x< \min(a_2, b_1). The given condition implies that

        \[x-a_1 + \sum_{i=2}^m a_i - (m-1)x = \sum_{i=1}^nb_i -nx.\]

    This, coupled with the previous result, implies that x=a_1, which is a contradiction. Hence we have a_1=b_1. Proceeding similarly, it follows that a_i=b_i, \forall 1\leq i \leq n.

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