Sum of exponential random variables follows Gamma, confused by the parameters

Question

I've learned sum of exponential random variables follows Gamma distribution.

But everywhere I read the parametrization is different. For instance, Wiki describes the relationship, but don't say what their parameters actually mean? Shape, scale, rate, 1/rate?

Exponential distribution: $x$~$exp(\lambda)$ $$f(x|\lambda )=\lambda {{e}^{-\lambda x}}$$ $$E[x]=1/ \lambda$$ $$var(x)=1/{{\lambda}^2}$$

Gamma distribution: $\Gamma(\text{shape}=\alpha, \text{scale}=\beta)$ $$f(x|\alpha ,\beta )=\frac{1}{{{\beta }^{\alpha }}}\frac{1}{\Gamma (\alpha )}{{x}^{\alpha -1}}{{e}^{-\frac{x}{\beta }}}$$ $$E[x]=\alpha\beta$$ $$var[x]=\alpha{\beta}^{2}$$

In this setting, what is $\sum\limits_{i=1}^{n}{{{x}_{i}}}$? What would the correct parametrization be? How about extending this to chi-square?

Answer 1

The sum of $n$ independent Gamma random variables $\sim \Gamma(t_i, \lambda)$ is a Gamma random variable $\sim \Gamma\left(\sum_i t_i, \lambda\right)$. It does not matter what the second parameter means (scale or inverse of scale) as long as all $n$ random variable have the same second parameter. This idea extends readily to $\chi^2$ random variables which are a special case of Gamma random variables.

Answer 2

The sum of $n$ iid exponential distributions with scale $\theta$ (rate $\theta^{-1}$) is gamma-distributed with shape $n$ and scale $\theta$ (rate $\theta^{-1}$).

Answer 3

gamma distribution is made of exponential distribution that is exponential distribution is base for gamma distribution. then if $f(x|\lambda)=\lambda e^{−\lambda x}$ we have $\sum_n x_i \sim \text{Gamma}(n,\lambda)$, as long as all $X_i$ are independent.

$$f(x|\alpha,\beta)=\frac{\beta^α}{\Gamma(\alpha)} \cdot x^{\alpha−1} \cdot e^{−x\beta} $$

Answer 4

1. Exponential distribution is a special case of Gamma distribution. If $X$ is a r.v. and

   $X \sim Exp(\lambda) \implies f(x) = \lambda e^{-\lambda x} = \frac{\lambda^1}{\Gamma(1)} x^{1-1}e^{-\lambda x}=\frac{\lambda^1}{\Gamma(\alpha)} x^{\alpha-1}e^{-\lambda x}$, where $\alpha=1$

   $\implies X \sim \Gamma(1,\lambda)$

   i.e., $X$ follows a Gamma distribution with shape parameter $=\alpha=1$ and rate paramater $=\lambda$

2. Sum of two independent Gamma r.v.s sharing the same rate parameter value, is a Gamma r.v. again. From here and here,

   $X_1\sim\Gamma(\alpha_1, \beta)$ and $X_2\sim\Gamma(\alpha_2, \beta)$ with $X_1\perp X_2 \implies X_1 + X_2 \sim \Gamma(\alpha_1+\alpha_2,\beta)$.

   By Induction then we have,

   $X_i\sim\Gamma(\alpha_i, \beta)$, for $i=1\ldots n$ and mutually independent $X_i$ r.v.s, $\implies\sum\limits_{i=1}^{n}X_i\sim \Gamma\left(\sum\limits_{i=1}^{n}\alpha_i, \beta\right)$

Now, combining (1) and (2),

$X_i\sim Exp(\lambda)$, for $i=1\ldots n$ and i.i.d. r.v.s $X_i$, we have $X_i\sim \Gamma(1,\lambda)\;,\forall{i} \implies \sum\limits_{i=1}^{n}X_i\sim \Gamma(n,\lambda)$.