Given: For some X , V a r ( X ) = 9 , E ( X ) = 2

tenes6x

tenes6x

Answered question

2022-05-20

Given: For some X, V a r ( X ) = 9, E ( X ) = 2, E ( X 2 ) = 13
Problem: P r [ X = 2 ] > 0
The solution in my book says to construct a r.v. to satisfy the above conditions and confirm or deny the statement from there. For simplicity, assume X can take on two values P r [ a ] = 1 2 and P r [ b ] = 1 2 . Finding that a and b are not 2 is enough to disprove the statement. We can apply the constraints:
1 2 a + 1 2 b = 2
1 2 a 2 + 1 2 b 2 = 13
So I did this:
a = 4 b from the first equation
b 2 4 b 5 = 0 by substituting a
The above quadratic has two solutions, b = 1 and b = 5 which happen to be the same solutions the book got for a and b respectively.
I'm not sure if this is the right way to solve this problem since I never explicitly solved for a and the two solutions for b may coincidentally be the same.

Answer & Explanation

Maximo Sweeney

Maximo Sweeney

Beginner2022-05-21Added 7 answers

You haven't made a mistake.Once you solve for b, you have two options, 1 and 5. Using the first equation now, you'll get the corresponding two possibilities for a, since
a = 4 b
This will give you
(1) ( a , b ) = ( 4 ( 1 ) , 1 ) = ( 5 , 1 )
or
(2) ( a , b ) = ( 4 5 , 5 ) = ( 1 , 5 )
Clearly, both are valid (by your choice of parameterization, the roles of a and b are symmetric: if ( x , y ) is solution, so is ( y , x ).
Janiyah Huffman

Janiyah Huffman

Beginner2022-05-22Added 1 answers

Small comments to expert answer: the statement of the exercise is redundant, as if you have E [ X 2 ] and E [ X ], you already know the variance as V a r X = E [ X 2 ] E [ X ] 2 .
Further, for simplicity I would have personally solved it with a random variable taking two values, 0 and α, (respectively with probability 1 p and p), and solved for ( α , p ). This gives a different counterexample — try it out.

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