Hey Jame.
The answer is yes. You can prove this by showing that if Z = kX, then the PDF P(Z = z, Y = y) = P(Z=z)*P(Y=y). As long as you can show this separable property, you will always be able to show independence.
Hey Jame.
The answer is yes. You can prove this by showing that if Z = kX, then the PDF P(Z = z, Y = y) = P(Z=z)*P(Y=y). As long as you can show this separable property, you will always be able to show independence.