In CLT we have $$Y = \frac{X_1 + X_2 + \cdots + X_n}{n}$$ where $X_1, X_2, \dots, X_n$ are statistically independent and identically distributed (i.i.d.) random variables. Is there a way to find the PDF of $Y$ for any $n$?
I tried to calculate the PDF of $Z = X_1 + X_2 + \cdots + X_n$ and I reached
$$f_Z(z) = f_{x_1}(z) * f_{x_2}(z) * \cdots * f_{x_n}(z)$$
but I couldn't find the PDF where the sum of $X_i$'s are divided by $n$ and that's why I asked this question.
I calculated the characteristic function of $Y$:
$$\Phi_Y(w) = \Phi_{X_1} \left(\frac{w}{n}\right) \cdot \Phi_{X_2}\left(\frac{w}{n}\right) \cdots \Phi_{X_n}\left(\frac{w}{n}\right)$$
but I don't know how to use this to find its PDF.
You can always think of the probability (density) to measure $Y$ as all the possible probabilities summed up for $X_1,...,X_n$ under the constraint that $Y=g(X_1,...,X_n)$ (Here you have $g(X_1,...,X_n)=\frac{X_1+...+X_n}{n}$). Mathematically speaking you are looking for a PDF for Y s.t. the probability to measure $y$ is given by $$f_Y(y) \, {\rm d}y = \int {\rm d}x_1 \dots \int {\rm d}x_n \left(\delta(y-g(x_1,...,x_n)) \, {\rm d}y \right)\, f_{X_1}(x_1)\dots f_{X_n}(x_n)$$ where $\delta$ is the delta function. You can think of $\delta(y-g(x_1,...,x_n)) \, {\rm d}y$ being $1$ if the constraint is fulfilled and $0$ (giving no contribution) if not. In your case you can eliminate one x-integration using the $\delta$-function and end up with some similar sort of convolution.