# PDF of $Y = \frac{X_1 + X_2 + ... + X_n}{n}$

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.

• What does $*$ denote here? Product would not be okay. On LHS if find argument $z$ but on RHS I find argument $x$, so things are wrong. Commented Apr 23, 2021 at 10:32
• * is convolution not product Commented Apr 23, 2021 at 10:38
• Okay, but where is argument $z$ on RHS? If you can find the characteristic function then look here for finding PDF. Commented Apr 23, 2021 at 10:39
• The PDF of $Z$ is right. Now you just need to answer this question: $A$ is an RV with PDF $f(x)$, what is the PDF of the random variable $B = \lambda A$ where $\lambda$ is a constant. Commented Apr 23, 2021 at 10:43
• thanks drhab. I edited it Commented Apr 23, 2021 at 10:43

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.
• thanks. but as I said I want to use characteristic functions to find $f_Y(y)$ Commented Apr 23, 2021 at 10:49
• Given your characteristic function, don't you just have to calc. the Fouriertransform of it? $$f_Y(y)=\frac{1}{2\pi} \int_{-\infty}^\infty \Phi_Y(w) e^{-iyw} \, {\rm d}w$$ Commented Apr 23, 2021 at 11:55