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  "body": "\n## Proof\n\nWe first express $E[X]$ as a series, then shift the index and recognise the exponential series.\n\n1. Let $\\lambda \u003e 0$, $S = \\set{0, 1, 2, \\dots}$ and $p(k) = e^{-\\lambda} \\lambda^k / k!$ for $k \\in S$.\n2. Let $\\mu$ be the [[law|0TST4MHA3A]] of $X$. Since $X$ has the [[Poisson distribution|0VEFSSKX18]], $\\mu(B) = \\sum_{k \\in B \\cap S} p(k)$ 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 $\\int_\\R h \\, d\\mu = \\sum_{k \\in S} h(k) \\, p(k)$ in $[0, \\infty]$, 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 $k \\in A_i$. All terms are nonnegative, 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. For each $N$, the function $s_N = \\sum_{k=0}^{N} h(k) \\mathbf{1}_{\\set{k}}$ is simple with $0 \\le s_N \\le h$. So $\\sum_{k=0}^{N} h(k) \\, p(k) = \\int s_N \\, d\\mu \\le \\int h \\, d\\mu$.\n8. Letting $N \\to \\infty$ in step 7 gives $\\sum_{k \\in S} h(k) \\, p(k) \\le \\int h \\, d\\mu$. With step 6, equality holds.\n9. We compute $\\sum_{k \\in S} k \\, p(k)$. The term with $k = 0$ is zero. For $k \\ge 1$, $k \\, \\lambda^k / k! = \\lambda \\cdot \\lambda^{k-1} / (k-1)!$.\n10. So for each $N \\ge 1$, $\\sum_{k=0}^{N} k \\, p(k) = \\lambda e^{-\\lambda} \\sum_{j=0}^{N-1} \\lambda^j / j!$, with $j = k - 1$.\n11. By the [[power series of the exponential function|06JZGMYTYD]], the right side tends to $\\lambda e^{-\\lambda} e^{\\lambda}$ as $N \\to \\infty$. By the [[exponential of a sum|0E8HNVBCSP]], $e^{-\\lambda} e^{\\lambda} = e^0 = 1$. So $\\sum_{k \\in S} k \\, p(k) = \\lambda$.\n12. Step 3 with $h(x) = \\abs{x}$ and step 11 give $\\int \\abs{x} \\, \\mu(dx) = \\lambda \u003c \\infty$.\n13. By the [[law of the unconscious statistician|01ZWX875AT]] with $g(x) = x$, $X$ is integrable and $E[X] = \\int_\\R x \\, \\mu(dx)$.\n14. 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 $x^- = \\max(-x, 0)$.\n15. On $S$ we have $x^+ = x$ and $x^- = 0$. Step 3 gives $\\int x^+ \\, d\\mu = \\sum_{k \\in S} k \\, p(k)$ and $\\int x^- \\, d\\mu = 0$.\n16. Hence $E[X] = \\sum_{k=0}^{\\infty} k \\, e^{-\\lambda} \\lambda^k / k! = \\lambda$, by step 11.\n\n## Idea\n\nThe expectation of a variable on $\\set{0, 1, 2, \\dots}$ is the weighted sum $\\sum_k k \\, p(k)$. The factor $k$ cancels against $k!$ and leaves $(k-1)!$. After a shift of index, the sum is $\\lambda e^{-\\lambda}$ times the exponential series for $e^{\\lambda}$.\n",
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