Question

Statistics,

https://www.dropbox.com/s/znr8vg1hadk9nt6/Screenshot%202015-05-25%2015.37.00.jpg?dl=0

How would you do the Var one

Answer 1

Basically what I am asking is... How would you split Var(5X-4Y)

Answer 2

Var(5X) + Var(4Y) or Var(5X) - Var(4Y)

Answer 3

variance of a sum is the sum of the variances

Answer 4

I think you add them

Answer 5

so then Var(5X-4Y) = Var(5X) + Var(4Y)

Answer 6

What if I have Var(X-Y-Z) ? Var(X-Y-Z) = Var(X) + Var(Y) + Var(Z) ?

Var(X-Y+Z) = Var(X) + Var(Y) + Var(Z) ?
Var(X+Y-Z) = Var(X) + Var(Y) + Var(Z) ?

Answer 7

yes , because you would write var(-4y) as (-4)^2 var(y)= 16 var(y)

Answer 8

so
Var(5X-4Y) is
25 var(x) + 16 var(y)
25*1.5 +16*2

Answer 9

E(X-2Y) = 5-2*7 ? why it comes up negative

Answer 10

Expected values can be negative ?

Answer 11

for a normal distribution, the expected value is the mean (peak of the bell curve)
if you "shift" the curve by multiplying by -2, then yes, the mean can be negative

Answer 12

Derivation true only if X and Y are independent...

Var[X-Y]
(by definition of variance)
= E[(X-Y)^2] - E[X-Y]^2
(by linearity of expectation)
= E[X^2] - 2E[XY] + E[Y^2] - E[X]^2 + 2E[X]E[Y] - E[Y]^2
(by independence of X and Y)
= E[X^2] - E[X]^2 + E[Y^2] - E[Y]^2
(by definition of variance)
Var[X] + Var[Y]