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Given $Y\sim N(\mu, \sigma^2)$. I'm trying to find the moment generating function of $Z=\frac{Y-\mu}{\sigma}$ using the MGF transform method.

Here's what I've tried:
$$M_Y(t)=e^{\mu t+\frac{\sigma^2t^2}{2}}$$ $$M_Z(t) = E(e^{Zt}) = E(e^{t\frac{Y-\mu}{\sigma}})$$ and I'm stuck completely. I'm thinking that Z follows a standard normal distribution, hence the resulting MGF of it will be: $$M_Z(t) = e^{\frac{1}{2}t^2}$$ However, I do not know how to get there from where I'm stuck at. Can anyone help me with this? Thank you.

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$Ee^{t\frac {Y-\mu} {\sigma}}=Ee^{t\frac Y {\sigma}} e^{-\frac {t\mu} {\sigma}}$. Note that $e^{-\frac {t\mu} {\sigma}}$ is a constant and it can be pulled out of the expectation. Now $Ee^{t\frac Y {\sigma}} $ is nothing but $Ee^{s Y } $ where $s=\frac t {\sigma}$. Use the formula you have for $M_Y(s)$ to finish.

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  • $\begingroup$ Got it! Thank you so much! $\endgroup$
    – user672518
    Aug 5 '20 at 12:03

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