Question

Consider the two data sets

          X_{1} = \begin{bmatrix} 1 &2 &4 \\ 5 &3 &5 \\ \end{bmatrix} and \: X_{2} =\begin{bmatrix} 3 & 2 & 4\\ 9 & 5& 7\\ \end{bmatrix}

for which 

          _{x_{1}}^{-}= \begin{bmatrix} 2\\ 4 \end{bmatrix} ,=_{x_{2}}^{-} =\begin{bmatrix} 3\\ 5 \end{bmatrix}

and

     S \: \: \: _{pooled} =\begin{bmatrix} 1 & 1\\ 1& 2 \end{bmatrix}

i) Calculate the linear discriminant function.

ii) Classify the observation  X_{0}^{{}'} = [ 2 , 7] as population  \pi_{1} or population  \pi_{2} with equal priors and equal costs.

      

07 Feb 2021
Answer :
Word Count : 371
To solve this problem, we will calculate the linear discriminant function (LDF) and classify the observation \( X_0' = [2, 7] \) as belonging to either population \( \pi_1 \) or \( \pi_2 \) based on the given information. Here’s how to proceed step by step: ### Step 1: Calculate the linear discriminant function The linear discriminant function is given by: _____ _________ ______ _________ ________ ________.
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Related Questions

A single repairperson looks after the two machines 1 and 2. Each time it is repaired, machine i stays up for an exponential time with rate \lambda , where i = 1, 2  . When machine i fails, it requires an exponentially distributed amount of work with rate \mu to complete its repair. The repairperson will always service machine 1 when it is down. For instance, if machine 1 fails while 2 is being repaired, then the repairperson will immediately stop work on machine 2 and start on

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 p( 1,1)= \frac{1}{4} ,\: \: \: \: p(2,1)= \frac{1}{4} ,\: \: \: \: p( 3,1)= \frac{1}{16}

p( 1,2)= \frac{1}{16} ,\: \: \: \: p(2,1)= 0 ,\: \: \: \: p( 3,2)= \frac{1}{16}

p( 1,3)= 0 ,\: \: \: \: p(2,3)=\frac{1}{16} ,\: \: \: \: p( 3,3)= \frac{1}{4}

i)  Computer E [ X \mid Y = i ] for\: \: i = 1,2,3

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                                                    \sum _{x} = \begin{bmatrix} \sigma _{11} &\sigma_{12} \\ \sigma_{12} & \sigma _{22} \end{bmatrix}

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                                                          Z_{1} = X_{1} - X_{2}

                                                          Z_{2} = X_{1} + X_{2}

or  

           Z = \begin{bmatrix} Z_{1} \\Z _{2} \end{bmatrix} =\begin{bmatrix} 1 &-1 \\ 1 &1 \end{bmatrix}\begin{bmatrix} X_{1}\\ X_{2} \end{bmatrix} = CX

       in terms of \mu _{x} and\sum _{X}

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  i ) \: \: P_{o} = \frac{1}{5},\: \: P_{2} = \frac{4}{5}

i i) \: \: P_{o} = \frac{1}{5},\: \: P_{1} = \frac{3}{5} P_{2}= \frac{1}{5}

ii i) \: P_{o} = \frac{1}{6},\: \: P_{1} = \frac{1}{2} P_{3}= \frac{1}{3}

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\sum =\begin{bmatrix} 16 &-2 &4 \\ -2& 9&1 \\ 4&1 & 25 \end{bmatrix}

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ii) Multiply your matrices to check the relation  V 1/2 P V 1/2

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c) The relation of accessibility in states is transitive.

d) For any pair of discrete random variables X and  y \sum P [y/y\mid X/x ] < 1.

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i) What is E [ X | N (t)] ?

ii) Argue that Var  [ x \mid N (t) ] = N (t) t^{2} /12

iii) What is Var ( X ) ?

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