Find the Distribution of W-Z

Question

I'm currently stuck on the following question: We are told that $W,Z$~$N(0,1)$ (ie they are distributed by the standard normal curve) and are independent

Find the Distribution of Y=W-Z

This is my attempt:

$E(e^{t(w-z)}) =E(e^{tw}e^{-tz}) = E(e^{tw})E(e^{-tz})$ (from independence)

It can be shown that the mgf of the standard normal = $e^{\frac{t^2}{2}}$ so $m_{w-z}(w-z) =e^{\frac{t^2}{2}}e^{\frac{-t^2}{2}} =1 $

I can't figure out what type of distribution has an mgf of 1? Where have I made a mistake?

Answer 1

It can be shown that the mgf of the standard normal = $e^{\frac{t^2}{2}}$ so $m_{w-z}(w-z) =e^{\frac{-t^2}{2}}e^{\frac{-t^2}{2}} =1 $

It can be shown that the mgf of the standard normal random variable, such as $Z$ is: $\mathsf M_Z(t) = e^{\frac{t^2}{2}}$ so $$\mathsf M_{W-Z}(t) ~{=\mathsf M_W(t)\,\mathsf M_{-Z}(t) \\=e^{\frac{t^2}{2}}e^{\frac{t^2}{2}} \\= e^{t^2} \\= \mathsf M_{Z}(\pm t\surd 2) \\ = \mathsf E (e^{\pm tZ\surd 2})}$$

Now... What does that mean?