contestada

The pair of random variables (X,Y) is equally likely to take any of the four pairs of values (0,1), (1,0), (−1,0), (0,−1). Note that X and Y each have zero mean.

a) Find E[XY].
E[XY]=
b) YES or NO: For this pair of random variables (X,Y), is it true that Var(X+Y)=Var(X)+Var(Y)?

Select an option Yes No
c) YES or NO: We know that if X and Y are independent, then Var(X+Y)=Var(X)+Var(Y). Is the converse true? That is, does the condition Var(X+Y)=Var(X)+Var(Y) imply independence?

Select an option Yes No

Respuesta :

mirianmoses

Answer and Step-by-step explanation:

(a)

E[XY] = 1/4*0*1 + 1/4*1*0 + 1/4*-1*0 +1/4*0*-1 = 0

(b)

E[X] = 0, E[Y] = 0

Thus, Cov(X,Y) = E[XY] - E[X]E[Y] = 0

So, Var(X + Y) = Var(X) + Var(Y) is True

The answer is Yes

(c) No, the converse is not true

(a) The value of E[XY] is zero.

(b) Yes  For this pair of random variables (X, Y) is true for the given relation.

(c) No, the converse is not true.

What is a variance?

Variance is the value of the squared variation of the random variable from its mean value, in probability and statistics.

The given data in the problem will be;

(X, Y)  is a random variable

pairs of values is given as  (0,1), (1,0), (−1,0), (0,−1).

E[XY] =?

(a) The value of E[XY] is zero.

[tex]E[XY] = \frac{1}{4}\times 0\times1 +\frac{1}{4}\times 1\times 0 + \frac{1}{4} \times 1\times 0 +\frac{1}{4}\times0\times1 = 0[/tex]

Hence the value of E[XY] is zero.

(b) Yes or this pair of random variables (X, Y) is true for the given relation.

[tex]E[X] = 0, E[Y] = 0[/tex]

[tex]\rm Cov(X,Y) = E[X,Y] - E[X]E[Y] = 0Var(X + Y) = Var(X) + Var(Y)[/tex]

Hence ) Yes  For this pair of random variables (X, Y) is true for the given relation.

(c) No, the converse is not true

X and Y are independent

[tex]Var(X+Y)=Var(X)+Var(Y).[/tex]

Hence ) No, the converse is not true

To learn more about the variance refer to the link;

https://brainly.com/question/7635845

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