Find the density of their average

Question

If $f_{X,Y,Z}(x,y,z)=e^{-(x+y+z)}I_{[0,\infty]}(x)I_{[0,\infty]}(y)I_{[0,\infty]}(z)$ find the density of their average $\frac{X+Y+Z}{3}$

I'm a little lost on how to solve this exercise, $f_{X,Y,Z}(x,y,z)$ It looks like the product of three exponential random variables $X\sim exp(1),Y\sim exp(1),Z\sim exp(1)$.

This would be the case of making the transformation with Jacobian calling $$A=\frac{X+Y+Z}{3},B=Y,C=Z$$ then finding $$f_{A,B,C}(a,b,c)$$ and finally $$f_A(a)=\int f_{A,B,C}(a,b,c)dbdc$$ this seems to me somewhat complicated and rather laborious

Answer 1

1. $X,Y,Z\sim_{iid}\text{Exp}(1)$ because $F_{X,Y,Z}(x,y,z)=F_X(x)F_Y(y)F_Z(z)$ for all $(x,y,z)\in\mathbb{R}^3$.
2. Denote $S_n=\sum_{i=1}^nX_i$ where $X_i\sim_{iid}\text{Exp}(\lambda)$. Distribution of $S_n/n$ follows $\text{Gamma}(n,n\lambda)$ because

$$\varphi_{S_n/n}(t)=\left[\varphi_{X_1}\left(\frac{t}{n}\right)\right]^n=\left[1-\frac{it}{n\lambda}\right]^{-n}$$

so that

$$f_{S_n/n}(s)=(n\lambda)^n e^{-n\lambda s}\frac{s^{n-1}}{(n-1)!}$$

Answer 2

Consider $F(a) = \Bbb{P}(\frac{X+Y + Z}{3} \leq a) =\Bbb{P}(X+Y + Z \leq 3a)$.

Now calculate

$$F(a) = \int_0^{3a} \int_0^{3a-x} \int_0^{3a - x - y} e^{-z}e^{-y}e^{-x} \, dz\,dy\,dx = \\ = \int_0^{3a} \int_0^{3a-x} (1 - e^{-3a + x + y})e^{-y}e^{-x} \,dy\,dx \\ = \int_0^{3a} \int_0^{3a-x} (e^{-y} - e^{-3a + x} )e^{-x} \,dy\,dx \\ = \int_0^{3a} (1 - e^{-3a + x})e^{-x} - e^{-3a + x} (3a - x)e^{-x} \,dx \\ = 1-e^{-3a} - 3a e^{-3a} - (e^{-3a}3a)3a + (e^{-3a})\frac{1}{2}(3a)^2\\ = 1 - e^{-3a}(1 + 3a +\frac{9a^2}{2})$$

To obtain the density you just need to derivate:

$$f(a) = F'(a) = 3 e^{-3a}(1 + 3a +\frac{9a^2}{2}) - e^{-3a}( 3 + 9a) \\ =3 e^{-3a}(\frac{9a^2}{2}) $$

As you said it is troublesome, maybe one can get a better way to get this result