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Is geometric distribution a type of Bernoulli distribution? But I remember geometric distribution has mean $1/p$ not matching the table here. enter image description here

Rodrigo de Azevedo
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spruce
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    "Is geometric distribution a type of Bernoulli distribution?" Not really. – callculus42 Mar 07 '20 at 17:47
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    Likely the “PMF” formula in that table was confusing you. A Bernoulli random variable $X$ can only take one of two values, 0 or 1, and $P[X=1]=p, P[X=0]=1-p$. – Michael Mar 07 '20 at 17:54
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    The distributions are related but not the same. The number of Bernoulli trials until there is a failure has geometric distribution, for example. – Math1000 Mar 07 '20 at 19:27

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A Bernoulli distribution can take only the values $0$ and $1$ while the geometric distribution takes values in the set of the positive integers hence cannot be a Bernoulli distribution.

However, there is a link between these two distributions: consider a sequence $(\xi_i)_{i\geqslant 1}$ of independent Bernouilli distributed random variables with parameter $p$. Let $X:=\inf\{i\geqslant 1, \xi_i=1\}$. Then $X$ has a geometric distribution (the number of attempts before one gets a first success).

Davide Giraudo
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