There are two service lines. The random variables X and Y are the proportions of time that line 1 and line 2 are in use, respectively. The joint probability density function for (X, Y ) is given by

In this exercise we must use the function given to calculate the variance and covariance that will be:

A) The random variables X and Y are not independent.      

B)

C)

D)

A) First of all, let's confirm the independece of the random variables X and Y:

Where G and H are the marginal distributions of X and Y respectively.

Now, let's find the functions G and H:

Also, the marginal distribution of Y:

So, let's confirm if eq(1) is true:

So, we can see that g(x)h(y) is not equal to the function in the question:

Thus, we cam say that the random variables X and Y are not independet.

B) To find out the expected value of 2, we can use:

Thus:

Hence, we obtain:

Now,

For E(XY), We can't use E(XY)=E(X)E(Y) became X and Y are not independent random variables. Thus:

C) Variances of X and Y, first of all let's find :

Now covariance:

D)

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The question is not complete and i have attached. It's the first image.

A) The random variables X and Y are not independent

B) E(X) = 5/8 ; E(Y) = 5/8 ; E(X+Y) = 5/4 ; E(XY) = 3/8

C) E(X²) = 7/15 ; E(Y²) = 7/15 ; Var (X) = 73/960 ; Var(Y) = 73/960 ; Cov(X, Y) = - 1/64

D) Var(X+Y) = 29/240

Step-by-step explanation:

I have attached all explanations.

no he doesnt because "-16' is negatiive

step-by-step explanation:

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