joint density formula of i.i.d. exponential random variables

Question

Let $X_1 , X_2 , ...., X_n$ be a set of independent and identically distributed exponential random variables. Find the joint density function formula $f (x_1, x_2, ...., x_n)$, simplifying as much as possible.

Why is the answer $$\lambda^n(e^{-\lambda\sum_{i=0}^n X_i})$$ instead of just $$\lambda^n(e^{-n\lambda X})$$

Answer 1

The random variables are iid; therefore,

\begin{align*} f_{X_1 \dots X_n}(x_1,\dots,x_n | \lambda) = \prod_{i=1}^n f_{X_i}(x_i | \lambda) = \prod_{i=1}^n \lambda e^{-\lambda x} = \lambda^n \prod_{i=1}^n e^{-\lambda x} = \lambda^n e^{-\lambda \sum_{i=1}^n x_i}\\ \end{align*}