Joint Density for Independent Normal Variables

\(X\) is a normally distributed variable with mean, \(\mu\), and known variance, \(\sigma^2\):
\( X_1,..., X_n \overset{iid}{\sim} N(\mu, \sigma^2) \)

The joint density is:
\begin{align}
f_{\theta}(x_1,...,x_n) &= \prod_{i=1}^{n} \frac{1}{\sqrt{2 \pi \sigma^2}}e^{-\frac{1}{2\sigma^2}(x_i-\mu)^2} \\
&= (2 \pi \sigma^2)^\frac{n}{2} \prod_{i=1}^{n} e^{-\frac{1}{2\sigma^2}(x_i-\mu)^2} \\
&= (2 \pi \sigma^2)^\frac{n}{2} e^{-\frac{1}{2\sigma^2} \sum_{i=1}^n (x_i-\mu)^2} \\ 
\end{align}

In the exponent, use the following substitution \((x_i-\mu)^2 = x_i^2 - 2x_i\mu + \mu^2\):

\begin{align}
f_{\theta}(x_1,...,x_n) &= (2 \pi \sigma^2)^\frac{n}{2} e^{-\frac{1}{2\sigma^2} \sum_{i=1}^n (x_i-\mu)^2} \\ 
&= (2 \pi \sigma^2)^\frac{n}{2} e^{-\frac{1}{2\sigma^2} \sum_{i=1}^n x_i^2 - 2x_i\mu + \mu^2} \\
&= (2 \pi \sigma^2)^\frac{n}{2} e^{-\frac{1}{2\sigma^2} \sum_{i=1}^n x_i^2 + \frac{1}{2\sigma^2} \sum_{i=1}^n 2x_i\mu -\frac{1}{2\sigma^2} \sum_{i=1}^n \mu^2} \\ 
&= (2 \pi \sigma^2)^\frac{n}{2} e^{-\frac{1}{2\sigma^2} \sum_{i=1}^n x_i^2 + \frac{\mu}{\sigma^2} \sum_{i=1}^n x_i -\frac{n\mu^2}{2\sigma^2}} \\ 

&= (2 \pi \sigma^2)^\frac{n}{2} e^{-\frac{1}{2\sigma^2} \sum_{i=1}^n x_i^2} e^{\frac{\mu}{\sigma^2} \sum_{i=1}^n x_i } e^{-\frac{n\mu^2}{2\sigma^2}} \\ 

\end{align}

Now set \( T= \sum_{i=1}^n x_i \), then re-arrange the expression to get:

\begin{align}
f_{\theta}(x_1,...,x_n) &= (2 \pi \sigma^2)^\frac{n}{2} e^{-\frac{1}{2\sigma^2} \sum_{i=1}^n x_i^2} e^{\frac{\mu}{\sigma^2} \sum_{i=1}^n x_i } e^{-\frac{n\mu^2}{2\sigma^2}} \\
&= (2 \pi \sigma^2)^\frac{n}{2} e^{-\frac{1}{2\sigma^2} \sum_{i=1}^n x_i^2} e^{\frac{\mu}{\sigma^2} T } e^{-\frac{n\mu^2}{2\sigma^2}} \\
&= (2 \pi \sigma^2)^\frac{n}{2} e^{-\frac{n\mu^2}{2\sigma^2}} e^{\frac{\mu}{\sigma^2} T } e^{-\frac{1}{2\sigma^2} \sum_{i=1}^n x_i^2}   \\
\end{align}

This expression can be factorized into \( g_{\mu}(t) = (2 \pi \sigma^2)^\frac{n}{2} e^{-\frac{n\mu^2}{2\sigma^2}} e^{\frac{\mu}{\sigma^2} T } \) and \( h(x_1,...,x_n) = e^{-\frac{1}{2\sigma^2} \sum_{i=1}^n x_i^2} \), so \(T\) is a sufficient statistic for \( \mu \).