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) [tex]E(X) = 5/8 ; \\E(Y) = 5/8 ; \\E(X+Y) = 5/4 ;\\ E(XY) = 3/8[/tex]
C)[tex]E(X^2) = 7/15 ; \\E(Y^2) = 7/15 ; \\Var (X) = 73/960 ; \\Var(Y) = 73/960 ; \\Cov(X, Y) = - 1/64[/tex]
D) [tex]Var(X+Y) = 29/240[/tex]
A) First of all, let's confirm the independece of the random variables X and Y:
[tex]f(x,y)= g(x)h(y), for all 0\leq x, y\leq 1 (1)[/tex]
Where G and H are the marginal distributions of X and Y respectively.
Now, let's find the functions G and H:
[tex]g(x)= \int\limits^ {y} f(x,y)= \int\limits^1_0 {3/2(X^2+Y^2} \, dy= 3/2X^2+1/2[/tex]
Also, the marginal distribution of Y:
[tex]h(y)= \int\limits^x {f(x, y)} \, dx = \int\limits^1_0 {3/2(X^2+Y^2)} \, dx = 3/2Y^2+1/2[/tex]
So, let's confirm if eq(1) is true:
[tex]g(x)h(y)= (3/2x^2+1/2)(3/2y^2+1/2)=1/4(3x^2+1)(3y^2+1)[/tex]
So, we can see that g(x)h(y) is not equal to the function in the question:
[tex]f(x,y) = 3/2(x^2+y^2)[/tex]
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:
[tex]E(X+Y)= E(x)+E(Y)[/tex]
Thus:
[tex]E(x)= \int\limits^x {x(3/2X^2+1/2)} \, dx = (1/2)\int\limits^1_0 {3x^2+x} \, dx = 5/8[/tex]
Hence, we obtain:
[tex]E(x)= 5/8\\E(y)= \int\limits^1_0 {3y^3+y} \, dy= 5/8[/tex]
Now,
[tex]E(z)= E(X+Y)= E(X)+E(Y)= (5/8)+(5/8)= 5/4[/tex]
For E(XY), We can't use E(XY)=E(X)E(Y) became X and Y are not independent random variables. Thus:
[tex]E(XY)= \int\limits^x \int\limits^y {xyf(x,y)} \, dx dy= 3/8[/tex]
C) Variances of X and Y, first of all let's find :
[tex]E(x^2)=\int\limits^x \int\limits^y {x^2f(x,y)} \, dx = 7/15\\E(y^2)=\int\limits^x \int\limits^y {y^2f(x,y)} \, dx = 7/15\\\\\\Var(x)=E(x^2)-(E(x))^2=Var(y)= 73/960[/tex]
Now covariance:
[tex]Caos(X, Y)= E(x, Y)-E(x)E(Y)= 3/8-(5/8*5/8)= -1/64[/tex]
D) [tex]Var(X+Y)= var(X)+var(Y)+2cov(X,Y)\\= 29/240[/tex]
Learn more: brainly.com/question/13487072
