Finding the density function of sum of random variables The question asks given that $X_1,X_2,\ldots,X_N$ are independent normal variables with mean $\mu$ and $\sigma^2$. Find the $$\frac{1}{n}\sum\limits_{i=1}^n X_i$$
So I know that the normal distribution is
$\dfrac{1}{\sigma \sqrt{2\pi}} e^{\left(-\dfrac{(x-\mu)^2}{2 \sigma^2}\right)}$.
Since they are independently and identically distributed then to get the joint would just be
$$
  f(x) = \left(\dfrac{1}{\sigma \sqrt{2\pi}} e^{\left(-\dfrac{(x-\mu)^2}{2 \sigma^2}\right)}\right)^n $$
And then you would multiply by $\dfrac{1}{n}$.
I'm not sure if this is right.
 A: Th sum of independent normals is normal, and a constant times a normal is normal. Let $\overline{X}=\frac{1}{n}(X_1+X_2+\cdots +X_n)$. Then $\overline{X}$ is normally distributed, mean $\mu$, and variance $\frac{\sigma^2}{n}$.
The value of the mean follows from the linearity of expectation. As to the variance of $\overline{X}$, it is $\frac{1}{n^2}$ times the variance of $X_1+X_2+\cdots+X_n$. And the variance of a sum of independent random variables is the sum of their variances.
Thus the density function of $\overline{X}$ is equal to
$$\frac{1}{\sqrt{2\pi\tau^2}}e^{-(x-\mu)^2/(2\tau^2)},$$ 
where $\tau^2=\frac{\sigma^2}{n}$. 
A: I think you are confusing the joint density with the density of the sum of the $n$ random variables divided by $n$.
Given that the random variables are independently and identically distributed with mean $\mu$ and variance $\sigma^2$, the joint density of the $n$ variables is indeed the product of the individual densities. You do not multiply this result by $\frac{1}{n}$.
On the other hand, the density of the sum of the $n$ normal variables will be a single normal density with mean $n\mu$ and variance $n\sigma^2$. 
The mean of the sum of random variable is the sum of the means (linearity property of expectation):-
$$E\{\sum_{i=1}^NX_i\}=\sum_{i=1}^NE\{X_i\}=n\mu$$ 
As the random variables are independent, the variance of the sum is equal to the sum of the variances:- 
$$Var\{\sum_{i=1}^NX_i\}=\sum_{i=1}^NVar\{X_i\}=n\sigma^2$$ 
If you were to divide the sum by $n$, the result would have mean $\mu$ but variance $\large\frac{\sigma^2}{n}$ (multiplying a random variable by factor $k$ results in the mean being scaled by $k$ but the variance being scaled by $k^2$). The density of the scaled sum will have mean $\mu$ and variance $\large\frac{\sigma^2}{n}$.
Thus the density of $\large\frac{1}{n}\sum_{i=1}^nX_i$ will be $$\large f(\frac{1}{n}\sum_{i=1}^nX_i)=\frac{1}{\sigma}\sqrt{\frac{n}{2\pi}}e^{-\frac{n(x-\mu)^2}{2\sigma^2}}$$ 
