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Suppose the support of a distribution is $\{12 , 13 \}$ with $P(X = 12) = p$ and $P(X = 13) = 1-p$. Is this still a Bernoulli distribution even if the support is not $\{1, 0 \}$?

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As Bernoulli distribution is most often defined, no. Because it is such a close relative, results about Bernoulli distributions can be readily adapted to this one. – André Nicolas Apr 17 '12 at 23:32

1 Answer 1

As Bernoulli distribution is ordinarily defined, if $X$ is a random variable with Bernoulli distribution, and $Y=X+12$, then $Y$ does not have Bernoulli distribution. Even a random variable that takes on values $-1$ and $1$, each with probability $1/2$, has its own name (it has Rademacher distribution).But what's in a name? Certainly results about $X$ can be readily adapted to yield results about $Y$.

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