Question:
Assume that the $X_1,\dots,X_n$ are continuous i.i.d random variables. Calculate the probability density function of:
$$Y_n:=\frac{1}{n}\sum_{i=1}^n X_i$$
My try:
I thought maybe finding the CDF is easier. So I wrote:
$$\mathbb{P}(Y_n\le y)=\mathbb{P}(\frac{1}{n}\sum_{i=1}^n X_i\le y)$$
However, this gives me nothing. I do not know how to go further.
Let's $\varphi_X(t) = E(e^{itX})$ the characteristic function of $X$ (if the density function of $X$ is known, then $\varphi_X(t) = E(e^{itX}) = \int_{\Bbb R}e^{itx}f_X(x)dx$). Then
$$\varphi_Y(t)=E\left(e^{it\frac{1}{n}\sum X_i}\right)=\left(E\left(e^{i\frac{t}{n}X} \right) \right)^{n} = \varphi^n_X\left(\frac{t}{n} \right)$$
Use the inverse formulae, we obtain the density function of $Y$ $$f_Y(y) = \frac{1}{2\pi}\int_{\Bbb R}e^{-ity}\varphi_Y(t)dt = \frac{1}{2\pi}\int_{\Bbb R}e^{-ity}\varphi^n_X\left(\frac{t}{n} \right)dt$$
Another representation without using characteristic function is
$$f_Y(y) = \frac{1}{2\pi}\int_{\Bbb R}e^{-ity}\varphi_Y(t)dt = \frac{1}{2\pi}\int_{\Bbb R}e^{-ity}\left( \int_{\Bbb R}e^{itx}f_X(x)dx\right)^ndt$$