Let $X$,$Y$,$Z$ be independent random variables with exponential distribution of parameter $\lambda$, then $X,Y,Z$ ~ $\xi(\lambda)$.
Is it true that $f_{X, Y, Z}(x,y,z)=\lambda^{3}e^{-\lambda (x+y+z)}1_{[0,+\infty) \times [0,+\infty) \times [0,+\infty)}(x,y,z)$ ?
where$f_{X, Y, Z}$ is the joint density of $X,Y,Z$ and $f_{X},f_{Y},f_{Z}$ the density of $X,Y$ and $Z$.
Edit: If it's true, why are they mutually independent?
If you change $e^{-\lambda x+y+z}$ to $e^{-\lambda(x+y+z)}$, then yes, that is correct.
Why? Well, for an exponentially distributed random variable $W$, the PDF is given by $$f_W(w)=\lambda e^{-\lambda w}1_{[0,+\infty)}(w)$$ And for three independently distributed RVs $X,Y,Z$, their joint density function is given by the product of their density functions: $$f_{XYZ}(x,y,z)=f_X(x)f_Y(y)f_Z(z)$$ Since the PDFS of $X,Y,Z$ are all equivalent to to $f_W$, you have $$f_{XYZ}(x,y,z)=f_W(x)f_W(y)f_W(z)$$ Which, with some substitution and algebraic simplification, is exactly what you have.