Let $0 . Let $X_n=\binom {n} {k}p^k(1-p)^{n-k}$. Based on the definition of $X_n$, can I conclude that the $0 Finally, can I conclude that $\mathbb{E}[|X_n|]<1$ for all $n$?
Does $X_n=\binom {n} {k}p^k(1-p)^{n-k}$ have finite expectation?
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probability-theory
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1$X_n$ as defined is not quite a random variable (OK, it is a random variable which happens to be constant). Indeed the constant r.v. $X_n$ has expectation $\le 1$, since it is a constant $\le 1$. But I expect there is some confusion here, with a mistranscription of the real question. – 2012-05-28
1 Answers
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Notice that $X_n(k) \geq 0$ for any $0 \leq k \leq n$ and $p \in [0,1]$. Now by the binomial theorem, $(a + b)^n = \sum_{k=0}^n {n \choose k} a^k b^{n-k}.$ Take $a = p$ and $b = 1-p$ to get $1 = \sum_{k=0}^n {n \choose k} p^k (1-p)^{n-k} = \sum_{k=0}^n X_n(k).$ So $X_n(k) \geq 0$ and they add up to $1$, hence they must all be between $0$ and $1$. And from your earlier question you already know that if $X_n \in [0,1]$ then $E[|X_n|] \leq 1$.