I know that the pmf is $$p(x) = p^{k}(1-p)^{1-k}$$ for $k \in \{0,1\}$. Suppose we want to find the joint pdf for i.i.d $X_1, \dots X_n$. Then would if be $$p_{n}(x) = p^{kn} (1-p)^{n(1-k)}$$
How do you write the pmf of a sum of Bernoulli random variables?
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probability
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0I think you mean $p(k) = p^k (1-p)^{1-k}$, i.e. $p(0) = 1-p$ and $p(1) = p$. Please note that it is a pmf (probability mass function), not a pdf (probability density function). – 2012-05-29
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0$f(k) = p^{nk} (1-p^n)^{1-k}$ for $k \in \{0,1\}$ would be the pmf for the product $X_1 \ldots X_n$. The joint pmf for $X_1,\ldots,X_n$ has to be a function of $n$ variables: $p_n(k_1,\ldots,k_n) = p(k_1) p(k_2) \ldots p(k_n)$ – 2012-05-29