Using the definition of covariance, cov(x, y) = e[(x – my)(y – my)], prove the followings. a. cov(x, y) = e(xy) - mx hy b. cov(x, y) = cov(y, x) c. cov(x, x) = var(x) d. cov(x + z, y) = cov(x, y) + cov(z, y) e. cov(ex, y)= xcov(x, y) f. var(x + y) = var(x) + var(y) + 2cov(x, y) g. var(ex) = var(x) + ej cov(x, x; )

Step-by-step explanation:

We have by definition

where Mx and My are means of X and Y respectively

a)

b) Since right side inside expectation is commutative, we get cov(x,y) = cov (y,x)

c)

d)

f) Var(x+y) =

Expanding inside we get

=

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option

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step-by-step explanation:

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