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For each one of the following statements, indicate whether it is true or false.
(a) If X = Y (i.e., the two random variables always take the same values), then Van X | Y = 0.
(b) If X = Y (the two random variables always take the same values), then Var (X | Y) = Var (X).
(c) If Y takes on the value y, then the random variable Var (X | Y) takes the value E[(X – E[X | Y = y])2 |Y = y].
(d) If Y takes on the value y, then the random variable Var (X | Y) takes the value E[(X - E[X | Y])2 | Y = y].
(e) If Y takes on the value y, then the random variable Var ( X | Y) takes the value E[(X – E[X])2 | Y = y].

Sagot :

Solution :

a). [tex]$\text{Var} (X|Y) =E ((X-E(X|Y))^2 |Y)$[/tex]

  Now, if X = Y, then :

  [tex]P(X|Y)=\left\{\begin{matrix} 1,& \text{if } x=y \\ 0, & \text{otherwise }\end{matrix}\right.[/tex]

Then, E[X|Y] = x = y

So, [tex]$\text{Var} (X|Y) =E((X-X)^2 |Y)$[/tex]

                      [tex]$=E(0|Y)$[/tex]

                      = 0

Therefore, this statement is TRUE.

b). If X = Y , then Var (X) = Var (Y)

And as Var (X|Y) = 0, so Var (X|Y) ≠ Var (X), except when all the elements of Y are same.

So this statement is FALSE.

c). As defined earlier,

  [tex]$\text{Var} (X|Y) =E ((X-E(X|Y))^2 |Y=y)$[/tex]

  So, this statement is also TRUE.

d). The statement is TRUE because [tex]$\text{Var} (X|Y) =E ((X-E(X|Y))^2 |Y=y)$[/tex].

e). FALSE

   Because, [tex]$\text{Var} (X|Y) =E ((X-E(X|Y=y))^2 |Y=y)$[/tex]

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