2024.Q3 – Stable Limits, Characteristic Functions, and Densities
2024 Probability Prelim Exam (PDF)
Problem Statement (verbatim)
Let ${X, X_n, n \ge 1}$ be a sequence of i.i.d. random variables whose characteristic function satisfies \(\phi(t) = e^{-|t|^\alpha(1+|t|)}, \quad \text{for } -1 < t < 1\) where $\alpha \in (0, 2]$ is a constant.
For $n \ge 1$, let $S_n = \sum_{i=1}^n X_i$ and let $\phi_n(t)$ be the characteristic function of $n^{-1/\alpha}S_n$.
(a) Find $\displaystyle \lim_{n\to\infty} \phi_n(t)$.
(b) Prove that $n^{-1/\alpha}S_n$ converges in distribution to a random variable $Y$.
(c) Prove that the random variable $Y$ in (ii) has a continuous and bounded probability density function.
Part (a)
Claim.
For each fixed $t \in \mathbb{R}$, \(\phi_n(t) \;=\; E\big[e^{it n^{-1/\alpha} S_n}\big] \;=\; \big(\phi(n^{-1/\alpha}t)\big)^n \;\longrightarrow\; e^{-|t|^\alpha} \quad \text{as } n\to\infty.\)
Proof.
By independence and identical distribution of ${X_i}$, \(\phi_n(t) = E\big[e^{it n^{-1/\alpha} S_n}\big] = E\left[\exp\!\left(it n^{-1/\alpha}\sum_{i=1}^n X_i\right)\right] = \prod_{i=1}^n E\big[e^{it n^{-1/\alpha} X_i}\big] = \big(\phi(n^{-1/\alpha}t)\big)^n.\)
For large enough $n$, we have $|n^{-1/\alpha}t|<1$, so the given formula for $\phi$ applies: \(\phi(n^{-1/\alpha}t) = \exp\Big(-|n^{-1/\alpha}t|^\alpha\big(1+|n^{-1/\alpha}t|\big)\Big) = \exp\Big(-n^{-1}|t|^\alpha\big(1+n^{-1/\alpha}|t|\big)\Big).\)
Therefore \(\phi_n(t) = \Big(\phi(n^{-1/\alpha}t)\Big)^n = \exp\Big( n\cdot\big[-n^{-1}|t|^\alpha(1+n^{-1/\alpha}|t|)\big] \Big) = \exp\Big(-|t|^\alpha\big(1+n^{-1/\alpha}|t|\big)\Big).\)
As $n\to\infty$, we have $n^{-1/\alpha}|t|\to 0$, hence \(\phi_n(t) \longrightarrow \exp(-|t|^\alpha) \equiv \phi_Y(t).\)
Conclusion.
The pointwise limit of the characteristic functions of $n^{-1/\alpha}S_n$ is \(\lim_{n\to\infty}\phi_n(t) = e^{-|t|^\alpha}.\)
Key Takeaways
- Scaling property of characteristic functions:
For $Y=cX$, $\phi_Y(t) = \phi_X(ct)$. - For sums of i.i.d. variables,
$\phi_{S_n}(t)=\big(\phi_X(t)\big)^n$. - Combine both:
$\phi_{n^{-1/\alpha}S_n}(t) = \left[\phi_X(n^{-1/\alpha}t)\right]^n$. - Then plug in the explicit form and let $n\to\infty$.
Part (b)
Claim.
There exists a random variable $Y$ with characteristic function \(\phi_Y(t) = e^{-|t|^\alpha}, \quad t\in\mathbb{R},\) such that \(n^{-1/\alpha}S_n \xrightarrow{d} Y.\)
Proof.
From part (a), for every fixed $t\in\mathbb{R}$, \(\phi_n(t) \longrightarrow \phi_Y(t) := e^{-|t|^\alpha}.\)
We now apply Lévy’s Continuity Theorem:
If $\phi_n(t)$ are characteristic functions of random variables $Z_n$ and
$\phi_n(t) \to \phi(t)$ pointwise, where $\phi$ is continuous at $t=0$ and $\phi(0)=1$,
then $\phi$ is a characteristic function of some $Y$, and $Z_n \xrightarrow{d} Y$.
Here:
- $\phi_n(t)$ is the c.f. of $Z_n := n^{-1/\alpha}S_n$.
-
The pointwise limit is $\phi_Y(t) = e^{- t ^\alpha}$. - $\phi_Y(0)=e^0=1$.
- $\phi_Y(t)$ is continuous at $0$ (in fact everywhere).
So by Lévy’s Continuity Theorem, there exists a random variable $Y$ with c.f. $e^{-|t|^\alpha}$, and \(n^{-1/\alpha}S_n \xrightarrow{d} Y.\)
Conclusion.
The normalized sums $n^{-1/\alpha}S_n$ converge in distribution to a random variable $Y$ whose characteristic function is $e^{-|t|^\alpha}$.
Key Takeaways
- Lévy Continuity Theorem is the standard tool to go from pointwise c.f. convergence to distributional convergence.
- To apply it, you must check:
- pointwise limit exists,
- the limit function satisfies $\phi(0)=1$,
- the limit is continuous at 0.
- Once those conditions hold, no further tightness argument is needed; Lévy gives both existence of $Y$ and convergence in distribution.
Part (c)
Claim.
The limiting random variable $Y$ with c.f. $\phi_Y(t)=e^{-|t|^\alpha}$ has a continuous and bounded probability density function.
Proof.
We use a standard Fourier inversion / density existence criterion:
If a characteristic function $\phi$ is integrable, \(\int_{-\infty}^{\infty} |\phi(t)|\,dt < \infty,\) then the corresponding distribution has a bounded continuous density given by \(f(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-itx}\phi(t)\,dt.\)
So we need to show $\phi_Y \in L^1(\mathbb{R})$.
We compute: \(\int_{-\infty}^{\infty} |\phi_Y(t)|\,dt = \int_{-\infty}^{\infty} e^{-|t|^\alpha} dt = 2\int_{0}^{\infty} e^{-t^\alpha} dt.\)
Use the change of variable $u=t^\alpha$, so $t = u^{1/\alpha}$, $dt = \frac{1}{\alpha}u^{1/\alpha-1}du$. Then \(2\int_0^{\infty} e^{-t^\alpha} dt = 2\int_0^{\infty} e^{-u}\frac{1}{\alpha}u^{1/\alpha-1}du = \frac{2}{\alpha} \int_0^{\infty} e^{-u} u^{1/\alpha-1}du = \frac{2}{\alpha}\,\Gamma\!\left(\frac{1}{\alpha}\right) < \infty.\)
Thus $\phi_Y\in L^1(\mathbb{R})$. By the inversion theorem, $Y$ has a density \(f_Y(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-itx}\,e^{-|t|^\alpha}\,dt,\) which is continuous and bounded (since it is the Fourier transform of an $L^1$ function).
Conclusion.
The limit distribution $Y$ has a continuous and bounded density $f_Y$, obtained as the inverse Fourier transform of $\phi_Y(t)=e^{-|t|^\alpha}$.
Key Takeaways
- Integrability of a characteristic function implies the existence of a bounded, continuous density via the Fourier inversion theorem.
-
To show integrability for $\phi_Y(t)=e^{- t ^\alpha}$, use: - symmetry to reduce to $\int_0^\infty e^{-t^\alpha}dt$,
- change of variables $u=t^\alpha$,
- recognition of the Gamma function: $\displaystyle \int_0^\infty e^{-u}u^{p-1}du = \Gamma(p)$.
-
This problem is a prototype for stable-like limits: the limit c.f. is of the form $e^{- t ^\alpha}$, and such laws are absolutely continuous with nice densities.
Global Key Takeaways for Question 3
- Sums of i.i.d. random variables with an explicit c.f. can often be normalized to converge to a stable-type limit law.
- The workflow:
- Compute the c.f. of the normalized sum using scaling + independence.
- Take the limit of c.f.s pointwise.
- Use Lévy’s continuity theorem to get the limit in distribution.
- Use integrability of the limit c.f. to deduce a bounded continuous density.
- Techniques reinforced:
- Characteristic function manipulations (scaling, powers, limits).
- Lévy continuity theorem.
- Fourier inversion + $L^1$ criterion.
- Gamma-function substitution $u = t^\alpha$.
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