Decomposition of exponential random variable

Question

I know that sum of independent Exponential random variables follows Gamma distribution.

But

Is it possible to decompose exponential random variate into independent and identically gamma random variates?

Answer 1

The exponential distribution with parameter $\lambda$ is the gamma $(1,\lambda)$ distribution. The convolution of a gamma $(a,\lambda)$ distribution and a gamma $(b,\lambda)$ distribution is a gamma $(a+b,\lambda)$ distribution. Hence, summing $n$ i.i.d. random variables with gamma $(\frac1n,\lambda)$ distribution yields an exponential random variable with parameter $\lambda$

Answer 2

The exponential distribution of mean $\lambda^{-1} > 0$ is actually $\Gamma(1,\lambda)$. So, if you take $X_1,\dots,X_n$ i.i.d. with $\Gamma(n^{-1},\lambda)$, then $$ X_1 + X_2 + \dots + X_n \sim \mathcal{E}(\lambda). $$