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Suppose $X_1,...,X_n$ is a random sample from a normal distribution with an unknown mean $\mu$ ,known standard deviation $\sigma$ and sample population $\bar{X}$. Show (using moment generating function) that $Z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$ has a standard normal distribution.

My attempt has been to take the MGT of a normal random variable and replace x with Z but I quickly run into integrals that I haven't learned how to evaluate yet

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Let $X \sim \operatorname{Normal}(\mu,\sigma)$. Then the MGF of $X$ is given by $$M_X(t) = e^{\mu t + (\sigma t)^2/2}.$$ The MGF of the sum $$n\bar X = \sum_{i=1}^n X_i$$ where each $X_i$ is IID as $X$, is simply $$M_{n \bar X}(t) = M_X(t)^n = e^{n (\mu t + (\sigma t)^2/2)}.$$ This demonstrates that $n \bar X$ is also normal, but with mean $\mu^* = n \mu$, and standard deviation $\sigma^* = \sigma \sqrt{n}$. Then let $$Z = \frac{\bar X - \mu}{\sigma/\sqrt{n}} = \frac{n \bar X - \mu^*}{\sigma^*}.$$ We now use the method of transformation, since we cannot presume that $Z$ so defined is itself normal. The transformation is clearly one-to-one and monotone, since $$g(x) = \frac{x - \mu^*}{\sigma^*}$$ is very clearly a linear transformation for $\sigma^* > 0$, and gives us $Z = g(n \bar X)$. Consequently, $$f_Z(z) = f_{n \bar X}(g^{-1}(z)) \left| \frac{dg^{-1}}{dz} \right| = f_{n \bar X}(\sigma^* z + \mu^*) \sigma^*.$$ Now recalling the normal density function, we have $$f_Z(z) = \frac{1}{\sqrt{2\pi}\sigma^*} e^{-(\sigma^* z + \mu^* - \mu^*)^2/(2 (\sigma^*)^2)} \sigma^* = \frac{1}{\sqrt{2\pi}} e^{-(\sigma^* z)^2/(2(\sigma^*)^2)} = \frac{1}{\sqrt{2\pi}} e^{-z^2/2}.$$ This concludes the proof.

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