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Moments of the exponential law on the half-line

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (ACω)), the choice hypothesis used by the change-of-variables lemma cited below. For every real λ>0 and every integer k≥0 the function r↦rke−λr is Lebesgue integrable on (0,∞) and

∫0∞rke−λr dr=k!λk+1.

The assertion is an identity for the Lebesgue integral of a nonnegative Borel function; no probabilistic interpretation and no normalised density are asserted. In particular no statement is made about any other exponent, about complex λ, or about negative powers.

Facts & Assumptions

Given: A real λ>0, an integer k≥0, and the Axiom of Countable Choice.

[F1]

Gamma integral. For every s>0 one has Γ(s)=∫0∞ts−1e−t dt, the Euler integral being finite and positive for exactly these s (The real Gamma function by Euler's integral).

[F2]

Integer values. Γ(n+1)=n! for every n∈N0 (Γ(n+1)=n! for every natural number n).

[F3]

Change of variables. Let m≥1 and let T:U→V be a C1 diffeomorphism between open subsets of Rm. For every nonnegative Borel h:V→[0,∞] one has ∫Vh(y) dy=∫Uh(T(x))∣det⁡DT(x)∣ dx, with equality in [0,∞] (Borel change of variables from the compact-support formula and Radon uniqueness).

Proof

technique · direct; change variables $r\mapsto\lambda r$ on the open half-line and evaluate the resulting gamma integral at an integer argument

Given: The data of the statement and [F1]-[F3].

1.1givenF3algebra

[Change of variables] Let T:(0,∞)→(0,∞) be T(x):=x/λ. Then T is a C1 diffeomorphism of open subsets of R with T′(x)=1/λ and det⁡DT(x)=1/λ nonzero and constant, and its inverse is y↦λy. The function h(r):=rke−λr is continuous, hence Borel, and nonnegative on (0,∞). With y=T(x) one has h(T(x))⋅∣det⁡DT(x)∣=(x/λ)ke−x⋅(1/λ)=λ−(k+1)xke−x, a nonnegative Borel function of x.

1.2F1F2

[Gamma value] Taking s:=k+1>0 in [F1], the integral ∫0∞xke−x dx=Γ(k+1) is finite, and [F2] gives Γ(k+1)=k!.

2.1step 1.1step 1.2F3algebra∎

[Assembly] By [F3] applied to the diffeomorphism of step 1.1 and the function h of that step, ∫0∞rke−λr dr=∫0∞λ−(k+1)xke−x dx=λ−(k+1)∫0∞xke−x dx=k!λk+1, where the middle equality is the scalar rule for the nonnegative integral and the last equality is step 1.2; the value is finite, so r↦rke−λr is integrable on (0,∞).

Remarks

  • Only the half-line is used. The change of variables is applied to the open half-line (0,∞), which is an open subset of R; no endpoint value of the integrand at 0 or at +∞ is used, and the equality is an equality in [0,∞].
  • Scaling convention. The exponent k+1 in the denominator and the factorial k! in the numerator are fixed by the convention that the unscaled integral is Γ(k+1)=k!; in particular the mean of the normalised density r↦λe−λr is 1/λ.

Depends on

Dependency tree · 2 levels

Sources