Question

Sean X y Y variables aleatorias, en donde E(X) = 2k, E(X2) = 5k, E(Y) = c y E(Y2) = 2c, siendo k y c, constantes, deducir: 1. E(kX2 + Y + c) y E(kX2 + cY2) 2. V(Y + c) y V(kX)

Answer 1

E(kx2+y+c ) = E(kx2) +E (y) +E(c) 
E (kx2) = KE (X2) = k ( 5k)   = 5k2
E (y) = c
E(c) = c
E ( kx2+y+c) = 5k2+2c
E( KX2+CY2) = E(kx2) + E ( CY2)
E ( KX2) = KE(x2) = K (5k) = 5 K2
E ( cy2) = cE ( Y2) = c( 2C)  = 2C2



2.  v ( Y+C)  y v ( kx)
V ( y+c) = v (y)
v ( y) = E ( y2) = 2c
E² ( y) = c²  
v( y+c) = c(2-c)
V ( kx) =  k² ( E(x²)- E²(x))
E(x²)= 5k
E²(x) = 4k²
k²v(x) = K²( 5-4k²)
V(kx)= k²(5-4K²)