The sum of multiple [[independent and identically distributed]] [[random variable|random variables]] in some cases has the same or a new distribution. You can use [[moment generating function]] to determine which distribution the sum has, but here is a list ## Bernoulli The sum of $n$ *iid* [[Bernoulli distribution]] random variables $X\sim bernoulli(p)$ has the [[Binomial distribution]] $X\sim bin(n,p)$ ## Binomial The sum of $m$ *iid* [[Binomial distribution]] random variables $X\sim bin(n,p)$ has the binomial distribution $X\sim bin(mn,p)$ ## Exponential The sum of $n$ *iid* [[exponential distribution]] random variables $X\sim exp(rate=\lambda)$ has the [[gamma distribution]] $X\sim \Gamma(n,\lambda)$ ## Gamma The sum of $n$ *iid* [[gamma distribution]] random variables $X\sim \gamma(\alpha, \beta)$ has the gamma distribution $X\sim \Gamma(n\alpha, \beta)$ ## Poisson The sum of $n$ *iid* [[Poisson distribution]] random variables $X\sim Pois(\lambda)$ has the Poisson distribution $X\sim Pois(n\lambda)$ ## Normal (Gaussian) The sum of $n$ *iid* [[normal distribution]] random variables $X\sim N(\mu, \sigma^2)$ has the normal distribution $X\sim N(n\mu, n\sigma^2)$ In the case of the normal distribution, the random variables do not need to be independent for this result to hold. See [[linear combination of normal random variables]] for more detail. ## Chi-squared The sum of $k$ *iid* [[chi-squared distribution]] random variables also has a chi-squared distribution with $kn$ degrees of freedom. $X \sim \chi^2(kn)$ [[base/Statistics/convolution]] > [!Tip]- Additional Resources > - [Convolutions | Why X+Y in probability is a beautiful mess](https://youtu.be/IaSGqQa5O-M?si=dgNqXIsREx-hinQR) | 3Blue1Brown