Olympiad Maths Prep

Track / Stage 6 / 132 of 400 #1132 of 2000

Problem 1132

National olympiad, first round
Algebra Difficulty 6.2 Prove it

49. Let T=T1T=T_{1}, where TaT_{a} is defined in problem II.13.48. Show that

T=dN−2, T \stackrel{d}{=} N^{-2},

where N∼N(0,1)N \sim \mathscr{N}(0,1) is a standard Gaussian random variable. Show also that the Laplace transform is given by

Ee−λ2T2=Ee−λ22N2=e−λ,λ⩾0 \mathrm{E} e^{-\frac{\lambda^{2} T}{2}}=\mathrm{E} e^{-\frac{\lambda^{2}}{2 N^{2}}}=e^{-\lambda}, \quad \lambda \geqslant 0

and the Fourier transform is given by

EeitT=EeitN2=exp⁡{−∣t∣1/2(1−it∣t∣)},t∈R \mathrm{E} e^{i t T}=\mathrm{E} e^{\frac{i t}{N^{2}}}=\exp \left\{-|t|^{1 / 2}\left(1-i \frac{t}{|t|}\right)\right\}, \quad t \in \mathbb{R}

Thus, the random variable N−2N^{-2} can be considered as a constructively defined random variable with a stable distribution (with parameters α=1/2,β=0,θ=−1,d=1\alpha=1 / 2, \beta=0, \theta=-1, d=1).

Remark. The results of II.13.48 and II.13.49 imply Lévy's formula

∫0∞ae−a22t2πt3e−λtdt=e−a2λ \int_{0}^{\infty} \frac{a e^{-\frac{a^{2}}{2 t}}}{\sqrt{2 \pi t^{3}}} e^{-\lambda t} d t=e^{-a \sqrt{2 \lambda}}

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Solution. Reasoning as in the solution of II.13.48, we find that P(T1⩽t)=2P(Bt⩾1)=2P(tN⩾1)=P(tN2⩾1)=P(N−2⩽t)\mathrm{P}\left(T_{1} \leqslant t\right)=2 \mathrm{P}\left(B_{t} \geqslant 1\right)=2 \mathrm{P}(\sqrt{t} N \geqslant 1)=\mathrm{P}\left(t N^{2} \geqslant 1\right)=\mathrm{P}\left(N^{-2} \leqslant t\right). Hence,

Ee−λ2T2=Ee−λ22N2=∫Re−λ22x2−x222πdx==e−λ∫Re−12(x−λx)22πdx=e−λ∫Re−x222πdx=e−λ, \begin{aligned} \mathrm{E} e^{-\frac{\lambda^{2} T}{2}}=\mathrm{E} e^{-\frac{\lambda^{2}}{2 N^{2}}}= & \int_{\mathbb{R}} \frac{e^{-\frac{\lambda^{2}}{2 x^{2}}-\frac{x^{2}}{2}}}{\sqrt{2 \pi}} d x= \\ & =e^{-\lambda} \int_{\mathbb{R}} \frac{e^{-\frac{1}{2}\left(x-\frac{\lambda}{x}\right)^{2}}}{\sqrt{2 \pi}} d x=e^{-\lambda} \int_{\mathbb{R}} \frac{e^{-\frac{x^{2}}{2}}}{\sqrt{2 \pi}} d x=e^{-\lambda}, \end{aligned}

where the last equality follows from the fact that the Boolean transformation preserves the Lebesgue measure on R\mathbb{R} (see problem II.6.103).

Consider the function of a complex variable f(z)=Ee−z2T2f(z)=\mathrm{E} e^{-\frac{z^{2} T}{2}}, defined in the domain D={z∈C:Re⁡z2⩾0}D=\left\{z \in \mathbb{C}: \operatorname{Re} z^{2} \geqslant 0\right\}, where, in particular, ∣e−z2T/2∣⩽1\left|e^{-z^{2} T / 2}\right| \leqslant 1. It is now sufficient to show that the function ff is holomorphic inside this domain and continuous on the boundary. Indeed, in this case, by the uniqueness theorem, f(z)f(z) necessarily equals e−ze^{-z} in the domain DD, since f(z)=e−zf(z)=e^{-z} for z∈Rz \in \mathbb{R}. The continuity of the function ff in DD follows from the Lebesgue dominated convergence theorem. We will prove that ff is differentiable with respect to zz inside DD. Since

∣e−z2T2−e−z02T2z−z0∣⩽e−Re⁡z02T2∣e−(z2−z02)T2−1∣∣z−z0∣⩽⩽e−Re⁡z02T2e∣z2−z02∣T2−1∣z−z0∣⩽⩽sup⁡x⩾0[ex−1xexp⁡{−xRe⁡z022∣z2−z02∣}]∣z+z0∣e−Re⁡z02T4⋅T2⩽⩽sup⁡x⩾0[ex−1xe−x]3∣z0∣e−Re⁡z02T4⋅T2⩽3∣z0∣e−Re⁡z02T4⋅T2 \begin{aligned} & \left|\frac{e^{-z^{2} \frac{T}{2}}-e^{-z_{0}^{2} \frac{T}{2}}}{z-z_{0}}\right| \leqslant e^{-\operatorname{Re} z_{0}^{2} \frac{T}{2}} \frac{\left|e^{-\left(z^{2}-z_{0}^{2}\right) \frac{T}{2}}-1\right|}{\left|z-z_{0}\right|} \leqslant \\ & \leqslant e^{-\operatorname{Re} z_{0}^{2} \frac{T}{2}} \frac{e^{\left|z^{2}-z_{0}^{2}\right| \frac{T}{2}}-1}{\left|z-z_{0}\right|} \leqslant \\ & \leqslant \sup _{x \geqslant 0}\left[\frac{e^{x}-1}{x} \exp \left\{-\frac{x \operatorname{Re} z_{0}^{2}}{2\left|z^{2}-z_{0}^{2}\right|}\right\}\right]\left|z+z_{0}\right| e^{-\operatorname{Re} z_{0}^{2} \frac{T}{4}} \cdot \frac{T}{2} \leqslant \\ & \leqslant \sup _{x \geqslant 0}\left[\frac{e^{x}-1}{x} e^{-x}\right] 3\left|z_{0}\right| e^{-\operatorname{Re} z_{0}^{2} \frac{T}{4}} \cdot \frac{T}{2} \leqslant 3\left|z_{0}\right| e^{-\operatorname{Re} z_{0}^{2} \frac{T}{4}} \cdot \frac{T}{2} \end{aligned}

for all z0z_{0} inside D,z∈DD, z \in D and ∣z−z0∣⩽∣z0∣,∣z2−z02∣⩽Re⁡z022\left|z-z_{0}\right| \leqslant\left|z_{0}\right|,\left|z^{2}-z_{0}^{2}\right| \leqslant \frac{\operatorname{Re} z_{0}^{2}}{2}, applying the Lebesgue dominated convergence theorem again, we get

lim⁡z→z0f(z)−f(z0)z−z0=−Ez0Te−z02T2=f′(z0) \lim _{z \rightarrow z_{0}} \frac{f(z)-f\left(z_{0}\right)}{z-z_{0}}=-\mathrm{E} z_{0} T e^{-\frac{z_{0}^{2} T}{2}}=f^{\prime}\left(z_{0}\right)

i.e., the function ff is holomorphic inside DD. Thus,

EeitT=Ee−z2T2=e−z,z=∣t∣(1−it∣t∣) \mathrm{E} e^{i t T}=\mathrm{E} e^{-\frac{z^{2} T}{2}}=e^{-z}, \quad z=\sqrt{|t|}\left(1-i \frac{t}{|t|}\right)

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.