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  "body": "\n## Proof\n\nWe first express $E[X]$ as a finite sum, then evaluate the sum with the binomial theorem.\n\n1. Let $n \\ge 1$, $p \\in [0, 1]$, $q = 1 - p$ and $S = \\set{0, 1, \\dots, n}$. Let $p(k) = \\binom{n}{k} p^k q^{n-k}$ for $k \\in S$.\n2. Let $\\mu$ be the [[law|0TST4MHA3A]] of $X$. Since $X$ has the [[binomial distribution|0VRWZDWW6B]], $\\mu(B) = \\sum_{k \\in B \\cap S} p(k)$ for each Borel set $B \\subseteq \\R$. In particular $\\mu(\\set{k}) = p(k)$ for $k \\in S$.\n3. Let $h \\colon \\R \\to [0, \\infty)$ be Borel measurable. We show $\\int_\\R h \\, d\\mu = \\sum_{k \\in S} h(k) \\, p(k)$, using the [[integral of a non-negative function|GEA8H3228S]].\n4. Let $s = \\sum_{i=1}^{m} a_i \\mathbf{1}_{A_i}$ be a [[simple function|7WMBX7M4P6]] with $0 \\le s \\le h$, $a_i \\ge 0$ and pairwise disjoint Borel sets $A_i$. By step 2, $\\int s \\, d\\mu = \\sum_{i} a_i \\sum_{k \\in A_i \\cap S} p(k)$.\n5. Each $k \\in S$ lies in at most one $A_i$, and $s(k) = a_i$ when $k \\in A_i$. So the sum in step 4 equals $\\sum_{k \\in S} s(k) \\, p(k) \\le \\sum_{k \\in S} h(k) \\, p(k)$.\n6. Taking the supremum over all such $s$ gives $\\int h \\, d\\mu \\le \\sum_{k \\in S} h(k) \\, p(k)$.\n7. The function $s_0 = \\sum_{k \\in S} h(k) \\mathbf{1}_{\\set{k}}$ is simple, with disjoint Borel sets $\\set{k}$, and $0 \\le s_0 \\le h$. Its integral is $\\sum_{k \\in S} h(k) \\, p(k)$. So equality holds in step 6.\n8. Apply step 3 to $h(x) = \\abs{x}$. This gives $\\int \\abs{x} \\, \\mu(dx) = \\sum_{k=0}^{n} k \\, p(k) \u003c \\infty$.\n9. By the [[law of the unconscious statistician|01ZWX875AT]] with $g(x) = x$, $X$ is integrable and $E[X] = \\int_\\R x \\, \\mu(dx)$.\n10. By the definition of the [[Lebesgue integral|XVEWR93EMV]], $\\int x \\, \\mu(dx) = \\int x^+ \\, \\mu(dx) - \\int x^- \\, \\mu(dx)$, with $x^+ = \\max(x, 0)$ and $x^- = \\max(-x, 0)$.\n11. On $S$ we have $x^+ = x$ and $x^- = 0$. Step 3 gives $\\int x^+ \\, d\\mu = \\sum_{k=0}^{n} k \\, p(k)$ and $\\int x^- \\, d\\mu = 0$. So $E[X] = \\sum_{k=0}^{n} k \\binom{n}{k} p^k q^{n-k}$, the first equality.\n12. The term with $k = 0$ is zero. Let $1 \\le k \\le n$. By the [[factorial formula|033KTKKRV6]] for [[binomial coefficients|092GEF38YD]], $k \\binom{n}{k} = \\dfrac{n!}{(k-1)! \\, (n-k)!} = n \\binom{n-1}{k-1}$, since $(n-1) - (k-1) = n - k$.\n13. So $E[X] = \\sum_{k=1}^{n} n \\binom{n-1}{k-1} p^k q^{n-k}$. Put $j = k - 1$, so that $n - k = (n - 1) - j$. Then $E[X] = np \\sum_{j=0}^{n-1} \\binom{n-1}{j} p^j q^{(n-1)-j}$.\n14. By the [[binomial theorem|0EHWPWGHB2]] with exponent $n - 1 \\ge 0$, the last sum is $(p + q)^{n-1} = 1^{n-1} = 1$.\n15. Hence $E[X] = np$.\n\n## Idea\n\nFor a distribution on finitely many points, the expectation is the weighted sum $\\sum_k k \\, p(k)$. The identity $k \\binom{n}{k} = n \\binom{n-1}{k-1}$ pulls out a factor $np$. What is left is the full sum of the $\\operatorname{Bin}(n - 1, p)$ probabilities, which is $1$ by the binomial theorem.\n",
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