Given: For some X, V a r ( X ) = 9, E ( X ) = 2, E ( X 2 ) = 13Problem: P r [ X = 2 ] &gt; 0The 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 = 13So I did this: a = 4 − b from the first equation b 2 − 4 b − 5 = 0 by substituting aThe 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.

Question

Given: For some   X,   V  a  r  (  X  )  =  9,   E  (  X  )  =  2,   E  (      X    2    )  =  13Problem:   P  r  [  X  =  2  ]  &amp;gt;  0The 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    =  13So I did this:  a  =  4  −  b from the first equation      b    2    −  4  b  −  5  =  0 by substituting   aThe 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&#039;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.

Maximo Sweeney · Accepted Answer

You haven&#039;t made a mistake.Once you solve for   b, you have two options,   −  1 and   5. Using the first equation now, you&#039;ll get the corresponding two possibilities for   a, since  a  =  4  −  bThis 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 · Accepted Answer

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.

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

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