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The standard Gaussian law is determined by its moments

Statement

Assume AC. Let $X$ be a real random variable with $E ∣ X ∣^{k} < \infty$ for every $k \geq 0$ such that $E [X^{k}] = E [Z^{k}]$ for every $k \geq 0$ , where $Z \sim N (0, 1)$ (Standard normal and normal laws). Then $X \sim N (0, 1)$ . More generally, let $Z$ be a real random variable with moment generating function finite on a neighbourhood of $0$ , and set $m_{k} : = E [Z^{k}]$ . If $X$ has all absolute moments finite and $E [X^{k}] = m_{k}$ for every $k \geq 0$ , then $X$ has the same law as $Z$ .

Facts & Assumptions

Given: AC; real random variables $X$ and $Z$ with $E ∣ X ∣^{k} < \infty$ and $E ∣ Z ∣^{k} < \infty$ for every $k \geq 0$ , and $E [X^{k}] = E [Z^{k}]$ for every $k \geq 0$ . In the Gaussian case $Z \sim N (0, 1)$ . In the general case there is a real $r > 0$ with $E [e^{t Z}] < \infty$ for every real $t$ with $∣ t ∣ < r$ ; an "analytic moment generating function on a neighbourhood of $0$ " is read as exactly this finiteness assertion, which analyticity on an interval implies. Write $φ_{X}, φ_{Z}$ for the characteristic functions (Characteristic function of a real random variable) and $h : = φ_{X} - φ_{Z}$ .

[F1]

For every $j$ , if $E ∣ Y ∣^{j} < \infty$ then $φ_{Y} \in C^{j} (R)$ , $φ_{Y}^{(l)} (t) = E [(i Y)^{l} e^{i t Y}]$ for $0 \leq l \leq j$ , hence $∣ φ_{Y}^{(l)} (t) ∣ \leq E ∣ Y ∣^{l}$ and $φ_{Y}^{(l)} (0) = i^{l} E [Y^{l}]$ (Moments give derivatives of the characteristic function); the moments are those of Moments, variance, and covariance on a probability space.

[F2]

If $X, Y \in L^{2} (P)$ are real random variables then $E [∣ X Y ∣] \leq (E [X^{2}])^{1 / 2} (E [Y^{2}])^{1 / 2}$ (Cauchy-Schwarz for random variables).

[F3]

For $Z \sim N (0, 1)$ one has $E [Z^{2 m}] = (2 m - 1)!! = 1 \cdot 3 \dots (2 m - 1)$ for every integer $m \geq 1$ , and $E [Z^{2 m + 1}] = 0$ for every $m \geq 0$ by symmetry of the density $e^{- x^{2} / 2} / \sqrt{2 π}$ (Gaussian even moments for Brownian increments, Standard normal and normal laws).

[F4]

For $v \geq 0$ , $e^{v} = \sum_{j \geq 0} v^{j} / j!$ , so $e^{v} \geq v^{k} / k!$ for every integer $k \geq 0$ (The power-series, product-limit, IVP, functional-equation, and Picard definitions agree).

[F5]

Taylor remainder bound: if $f$ has derivatives through order $n + 1$ on the closed interval between $a$ and $x$ , with $∣ f^{(n + 1)} ∣ \leq M$ there, then $∣ R_{n, a} f (x) ∣ \leq M ∣ x - a ∣^{n + 1} / (n + 1)!$ , where $R_{n, a} f (x) = f (x) - T_{n, a} f (x)$ and $T_{n, a} f$ is the Taylor polynomial of degree at most $n$ (Taylor polynomials and their remainders, A uniform derivative bound gives a uniform Taylor remainder bound).

[F6]

Two Borel probability laws on $R$ with equal characteristic functions are equal (Uniqueness of a law from its characteristic function).

[F7]

For every real $x$ there is a natural number $n \geq 1$ with $x < n$ (Every complete ordered field is Archimedean).

[F8]

Expectations of integrable variables are linear, monotone for real variables, and satisfy $∣ E U ∣ \leq E ∣ U ∣$ (Linearity, monotonicity, and the modulus bound for expectation).

Proof

technique · direct

1.1givenF1algebraF8

Setup: by [F1], $φ_{X}$ and $φ_{Z}$ are $C^{\infty}$ on $R$ with $∣ φ_{Y}^{(l)} (t) ∣ \leq E ∣ Y ∣^{l}$ for every $l \geq 0$ and every real $t$ , and $φ_{Y}^{(l)} (0) = i^{l} E [Y^{l}]$ ; consequently $h = φ_{X} - φ_{Z}$ is $C^{\infty}$ with $h^{(l)} (0) = i^{l} (E [X^{l}] - E [Z^{l}]) = 0$ for every $l \geq 0$ , and $∣ h^{(l)} (t) ∣ \leq E ∣ X ∣^{l} + E ∣ Z ∣^{l}$ for all $t$ . Also, by the standing hypothesis, in the general case $A_{x} : = E [e^{x Z}] + E [e^{- x Z}] < \infty$ for every real $x$ with $0 < x < r$ .

1.2givenF2F3algebra

Gaussian moment bounds: let $m \geq 0$ . The arithmetic inequality $(2 m)! \leq 4^{m} (m!)^{2}$ holds for $m = 0$ and is preserved by passing from $m$ to $m + 1$ , since $(2 m + 2) (2 m + 1) \leq 4 (m + 1)^{2}$ ; hence for $m \geq 1$ , using [F3], $E [Z^{2 m}] = (2 m - 1)!! = (2 m)! / (2^{m} m!) \leq 4^{m} (m!)^{2} / (2^{m} m!) = 2^{m} m! \leq (\sqrt{2})^{2 m} (2 m)!$ , and for $m \geq 1$ the Cauchy-Schwarz bound [F2] gives $E ∣ Z ∣^{2 m - 1} \leq (E [Z^{4 m - 2}])^{1 / 2} = ((4 m - 3)!!)^{1 / 2} \leq (2^{2 m - 1} (2 m - 1)!)^{1 / 2} \leq (\sqrt{2})^{2 m - 1} (2 m - 1)!$ , while $E ∣ Z ∣^{0} = 1$ . Therefore $E ∣ Z ∣^{k} \leq (\sqrt{2})^{k} k!$ for every $k \geq 0$ ; and in the Gaussian case $E ∣ X ∣^{k} \leq (E [X^{2 k}])^{1 / 2} = (E [Z^{2 k}])^{1 / 2} = ((2 k - 1)!!)^{1 / 2} \leq 2^{k / 2} (k!)^{1 / 2} \leq (\sqrt{2})^{k} k!$ for $k \geq 1$ , with equality $E ∣ X ∣^{0} = 1$ for $k = 0$ . Thus $E ∣ Y ∣^{k} \leq 4^{k} k!$ for $Y \in {X, Z}$ and every $k \geq 0$ , in the Gaussian case.

1.3givenF5algebra

Local vanishing: let $g$ be a real-valued $C^{\infty}$ function on $R$ and suppose there are $M > 0$ , $τ > 0$ with $∣ g^{(l)} (t) ∣ \leq M l! τ^{- l}$ for all $l \geq 0$ and all real $t$ , and let $t_{0}$ be a point with $g^{(l)} (t_{0}) = 0$ for every $l \geq 0$ . Then for every real $t$ with $∣ t - t_{0} ∣ < τ$ and every $j \geq 0$ , the Taylor polynomial satisfies $T_{j, t_{0}} g (t) = 0$ , so [F5] applied with $n = j$ and the bound on the $(j + 1)$ -th derivative gives $∣ g (t) ∣ = ∣ R_{j, t_{0}} g (t) ∣ \leq M (j + 1)! τ^{- (j + 1)} ∣ t - t_{0} ∣^{j + 1} / (j + 1)! = M (∣ t - t_{0} ∣ / τ)^{j + 1}$ ; letting $j \to \infty$ gives $g (t) = 0$ . Hence $g$ vanishes on $(t_{0} - τ, t_{0} + τ)$ .

2.1givenF2F4step 1.2algebraF8

MGF moment bounds: fix $0 < x < r$ and put $A : = E [e^{x Z}] + E [e^{- x Z}] \geq 2$ . Since $e^{x ∣ Z ∣} \leq e^{x Z} + e^{- x Z}$ , [F4] gives $E ∣ Z ∣^{k} \leq A k! x^{- k}$ for every $k \geq 0$ , including $k = 0$ . By [F2] and moment equality, $E ∣ X ∣^{k} \leq (E [X^{2 k}])^{1 / 2} = (E [Z^{2 k}])^{1 / 2} \leq A^{1 / 2} \sqrt{(2 k)!} x^{- k} \leq A^{1 / 2} 2^{k} k! x^{- k}$ , using the factorial inequality proved in step 1.2. Hence $E ∣ Y ∣^{k} \leq C k! (x / 2)^{- k}$ for $Y \in {X, Z}$ , with $C : = \max {A, A^{1 / 2}}$ . No integral over a zero power is used.

2.2givenF7step 1.3algebra

Global vanishing: let $g$ be as in step 1.3 and suppose in addition that $g^{(l)} (0) = 0$ for every $l \geq 0$ . Then $g \equiv 0$ on $R$ : step 1.3 with $t_{0} = 0$ gives $g \equiv 0$ on $I_{0} : = (- τ, τ)$ ; suppose $g \equiv 0$ on $I_{j} : = (- (1 + j / 2) τ, (1 + j / 2) τ)$ for some $j \geq 0$ and put $t_{\pm} : = \pm (j + 1) τ / 2$ , so $∣ t_{\pm} ∣ = (1 + j / 2) τ - τ / 2$ lies in the interior of $I_{j}$ and all derivatives of $g$ vanish at $t_{\pm}$ . Step 1.3 at $t_{\pm}$ gives $g \equiv 0$ on $((j - 1) τ / 2, (j + 3) τ / 2)$ and on $(- (j + 3) τ / 2, - (j - 1) τ / 2)$ ; since $(j + 3) τ / 2 = (1 + (j + 1) / 2) τ$ and $(j - 1) τ / 2 < (1 + j / 2) τ$ , the union of these intervals with $I_{j}$ contains $I_{j + 1} : = (- (1 + (j + 1) / 2) τ, (1 + (j + 1) / 2) τ)$ . By induction $g \equiv 0$ on $I_{j}$ for every $j \geq 0$ ; for an arbitrary real $T$ , [F7] supplies a natural number $n \geq 1$ with $n > 2 ∣ T ∣ / τ$ , hence $∣ T ∣ < n τ / 2 < (1 + (n - 1) / 2) τ$ and $T \in I_{n - 1}$ , so $g (T) = 0$ .

3.1givenF6step 1.1step 1.2step 2.2algebra

Gaussian case: by step 1.1, $h^{(l)} (0) = 0$ for every $l$ , and by steps 1.2 and 1.1, $∣ h^{(l)} (t) ∣ \leq E ∣ X ∣^{l} + E ∣ Z ∣^{l} \leq 2 \cdot 4^{l} l! = 2 l! (1 / 4)^{- l}$ for all $t, l$ . Thus both $Re h$ and $Im h$ satisfy the real Taylor hypotheses of step 2.2 with $M = 2$ and $τ = 1 / 4$ ; applying it separately to the two components gives $h \equiv 0$ , that is $φ_{X} = φ_{Z}$ . By [F6] the laws of $X$ and $Z$ are equal, so $X \sim N (0, 1)$ .

3.2givenF6step 1.1step 2.1step 2.2algebra

General case: fix $0 < x < r$ and let $C, τ : = x / 2$ be as in step 2.1, so $E ∣ Y ∣^{k} \leq C k! τ^{- k}$ for $Y \in {X, Z}$ and every $k \geq 0$ . By steps 1.1 and 2.1, $h^{(l)} (0) = 0$ for every $l$ and $∣ h^{(l)} (t) ∣ \leq 2 C l! τ^{- l}$ for all $t, l$ ; step 2.2 with $M = 2 C$ , applied separately to $Re h$ and $Im h$ , gives $h \equiv 0$ , hence $φ_{X} = φ_{Z}$ , and [F6] gives that $X$ and $Z$ have the same law.

4.1givenstep 3.1step 3.2∎

Conclusion: if $X$ has all moments and $E [X^{k}] = E [Z^{k}]$ for all $k$ with $Z \sim N (0, 1)$ , step 3.1 shows $X \sim N (0, 1)$ , which is the first assertion; if $Z$ has an analytic moment generating function on a neighbourhood of $0$ (so that the finiteness hypothesis of step 2.1 holds) and $X$ has all moments with $E [X^{k}] = m_{k}$ , step 3.2 shows that $X$ has the same law as $Z$ , which is the general assertion.

Depends on

Used by

The multivariate method of moments for a determinate limit Theorem

Dependency tree · two levels

64 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Vladimir Ivanov and Grigori Olshanski, Kerov's central limit theorem for the Plancherel measure on Young diagrams, arXiv:math/0304010; survey-paper version in Symmetric Functions 2001, NATO Science Series II 74 (2002), 93-151 (standard reference, not scraped)