Thread: Show Y has normal distribution

If X~Normal(u,v^2), show that Y = aX + b is Normal(au+b, (a^2)(v^2)) by using moment generating functions to complete the proof. In order to show that Y is normal with mean au + b and variance (a^2)(v^2), it suffices to show that the m.g.f. of Y is the same as the m.g.f. of a normal with that mean and variance. So find some way to derive the m.g.f. of Y using the information that you are given.


I'm really not sure where to start with this. I know the mgf of X should be M(t) = exp(ut + (v^2)(t^2)/2) and the mgf of Normal(au+b, (a^2)(v^2)) should be exp((au+b)t + (a^2)(v^2)(t^2)/2). Should I substitute ax+b in for x in the pdf of X and then find the mgf, or is there a simpler way?

Originally Posted by uberbandgeek6

If X~Normal(u,v^2), show that Y = aX + b is Normal(au+b, (a^2)(v^2)) by using moment generating functions to complete the proof. In order to show that Y is normal with mean au + b and variance (a^2)(v^2), it suffices to show that the m.g.f. of Y is the same as the m.g.f. of a normal with that mean and variance. So find some way to derive the m.g.f. of Y using the information that you are given.


I'm really not sure where to start with this. I know the mgf of X should be M(t) = exp(ut + (v^2)(t^2)/2) and the mgf of Normal(au+b, (a^2)(v^2)) should be exp((au+b)t + (a^2)(v^2)(t^2)/2). Should I substitute ax+b in for x in the pdf of X and then find the mgf, or is there a simpler way?

There is a theorem regarding the mgf of aX + b when you know the mgf of X. Look it up and use it.