2018.Q2 — Gaussian Maxima, Extreme Value Limits, and Gumbel Convergence
2018 Probability Prelim Exam (PDF)
Let $Z, Z_1, Z_2, \dots$ be iid random variables where $Z \sim N(0,1)$ and denote by
$f_Z(x)$, $-\infty<x<\infty$ the density of $Z$.
Also let \(M_n = \max_{1\le k\le n} Z_k.\) (a) Recall the inequality, for $x \ge 0$, \(\left(\frac1x - \frac1{x^3}\right) f_Z(x) \le P(Z>x) \le \frac1x\, f_Z(x).\)(i) Prove that
\(P(Z>x) \sim \frac{1}{x} f_Z(x), \qquad\text{namely}\qquad \frac{P(Z>x)}{(1/x) f_Z(x)} \xrightarrow[x\to\infty]{} 1.\)(ii) Prove that for each $y\in\mathbb{R}$, \(\frac{P\!\left(Z>x + \frac{y}{x}\right)} {P(Z>x)} \xrightarrow[x\to\infty]{} e^{-y}.\) Hint: Use part (i).
(b) Let $a_n$ satisfy $P(Z > a_n)=1/n$ for $n\ge1$.
Let
\(q_n(y) = P\!\left(Z > a_n + \frac{y}{a_n}\right).\) (i) Prove that for each $y\in\mathbb{R}$, \((1 - q_n(y))^n \xrightarrow[n\to\infty]{} e^{-e^{-y}}.\) Hint: First show $n q_n(y) \to e^{-y}$.(ii) Prove that
\(a_n(M_n - a_n) \xRightarrow[n\to\infty]{} Y, \qquad F_Y(y)=e^{-e^{-y}}.\)
(c) Prove that
\(M_n - a_n \xrightarrow{P} 0 \qquad\text{and conclude}\qquad \frac{M_n}{a_n} \xrightarrow{P} 1.\)
(a)(i) Asymptotics of the Gaussian Tail
Assumptions / Notation
- $Z\sim N(0,1)$ with density $\phi(x)=f_Z(x)$.
- Given inequalities: \(\left(\frac1x - \frac1{x^3}\right)\phi(x) \le P(Z>x) \le \frac1x\phi(x).\)
Solution
Claim.
\(\frac{P(Z>x)}{(1/x)\phi(x)} \xrightarrow[x\to\infty]{} 1.\)
Proof.
Step 1. Divide the inequality by $\phi(x)$.
Since $\phi(x)>0$,
\(\frac1x - \frac1{x^3}
\;\le\;
\frac{P(Z>x)}{\phi(x)}
\;\le\;
\frac1x.\)
Step 2. Multiply through by $x$. \(1 - \frac1{x^2} \;\le\; \frac{P(Z>x)}{(1/x)\phi(x)} \;\le\; 1.\)
Step 3. Take limits.
As $x\to\infty$,
\(1 - \frac1{x^2} \to 1.\)
By the squeeze theorem: \(\frac{P(Z>x)}{(1/x)\phi(x)} \to 1.\)
Conclusion.
\(P(Z>x) \sim \frac{1}{x}\phi(x).\)
Key Takeaways
- This is a direct application of the squeeze theorem.
- Gaussian tails are asymptotically equivalent to $(1/x)\phi(x)$.
- This asymptotic expansion is the foundation of all extreme-value theory for Gaussians.
(a)(ii) Ratio of Shifted Tails
Assumptions
- Use result of (a)(i): \(P(Z>x) \sim \frac{1}{x}\phi(x).\)
Solution
Claim.
For each $y\in\mathbb{R}$, \(\frac{P(Z>x+y/x)}{P(Z>x)} \to e^{-y}.\)
Proof.
Step 1. Apply part (i) to numerator and denominator. \(\frac{P(Z>x+y/x)}{P(Z>x)} \sim \frac{(1/(x+y/x))\phi(x+y/x)}{(1/x)\phi(x)}.\)
Step 2. Simplify the prefactor. \(\frac{1/(x+y/x)}{1/x} = \frac{1}{1 + y/x^2} \xrightarrow[x\to\infty]{} 1.\)
Step 3. Compute the ratio of densities. \(\frac{\phi(x+y/x)}{\phi(x)} = \exp\!\left( -\tfrac12(x+y/x)^2 + \tfrac12 x^2 \right) = \exp\!\left( -y - \frac{y^2}{2x^2} \right) \to e^{-y}.\)
Conclusion.
\(\frac{P(Z>x+y/x)}{P(Z>x)} \to e^{-y}.\)
Key Takeaways
- Ratios of Gaussian tails reduce to exponent algebra in the density.
- The term $y/x^2$ vanishes, leaving $e^{-y}$.
(b)(i) Convergence of $(1 - q_n(y))^n$ to Gumbel Form
Assumptions
- $a_n$ defined by $P(Z>a_n)=1/n$.
- $q_n(y)=P(Z>a_n+y/a_n)$.
- From (a)(ii): \(\frac{q_n(y)}{1/n} \to e^{-y}.\)
Solution
Claim.
\((1 - q_n(y))^n \to e^{-e^{-y}}.\)
Proof.
Step 1. Use part (a)(ii) with $x=a_n$.
Since $P(Z>a_n)=1/n$,
\(\frac{q_n(y)}{1/n}
=\frac{P(Z>a_n+y/a_n)}{P(Z>a_n)}
\to e^{-y}.\)
Thus: \(n q_n(y) \to e^{-y}.\)
Step 2. Rewrite $(1 - q_n(y))^n$. \((1 - q_n(y))^n = \left(1 - \frac{n q_n(y)}{n}\right)^n \to e^{-\lim n q_n(y)} = e^{-e^{-y}}.\)
Conclusion.
\((1 - q_n(y))^n \to e^{-e^{-y}}.\)
Key Takeaways
- Classic limit: \((1 - u_n)^n \to e^{-c} \quad \text{if } nu_n\to c.\)
- This is the essential Gumbel limit mechanism.
- $nq_n(y)$ represents the expected number of exceedances.
(b)(ii) Convergence of the Maxima to the Gumbel Distribution
Assumptions
- $M_n = \max Z_k$.
- Independence implies
\(P(M_n \le z) = P(Z \le z)^n.\) - From (b)(i): $(1 - q_n(y))^n \to e^{-e^{-y}}$.
Solution
Claim.
\(a_n(M_n - a_n) \Rightarrow Y, \qquad F_Y(y)=e^{-e^{-y}}.\)
Proof.
Step 1. Rewrite the event. \(P(a_n(M_n - a_n) \le y) = P\left(M_n \le a_n + \frac{y}{a_n}\right).\)
Step 2. Use independence. \(P(M_n \le z) = P(Z\le z)^n = (1 - P(Z>z))^n.\)
Let $z = a_n + y/a_n$: \(P(a_n(M_n - a_n) \le y) = (1 - q_n(y))^n.\)
Step 3. Apply (b)(i): \((1 - q_n(y))^n \to e^{-e^{-y}}.\)
Conclusion.
\(a_n(M_n - a_n) \Rightarrow Y,
\qquad P(Y\le y)=e^{-e^{-y}}.\)
This is the Gumbel distribution.
Key Takeaways
- Maxima of Gaussians converge (after centering/scaling) to Gumbel.
- Independence yields a product structure, reducing everything to tail approximations.
(c) Show $M_n - a_n \to 0$ in probability and $M_n/a_n \to 1$
Assumptions
- From (b)(ii):
\(X_n := a_n(M_n - a_n) \Rightarrow Y,\) where $Y$ is nondegenerate. - $a_n \to \infty$.
Solution
Claim 1.
\(M_n - a_n \xrightarrow{P} 0.\)
Proof.
Step 1. Multiply and divide by $a_n$.
\(M_n - a_n
= \frac{1}{a_n}\, a_n(M_n - a_n)
= \frac{1}{a_n} X_n.\)
Step 2. Apply Slutsky’s theorem.
- $X_n \Rightarrow Y$ (from part (b)(ii))
- $1/a_n \to 0$
Thus: \(\frac{1}{a_n} X_n \Rightarrow 0.\)
A convergence in distribution to a constant implies convergence in probability.
Conclusion.
\(M_n - a_n \to 0 \quad \text{in probability}.\)
Claim 2.
\(\frac{M_n}{a_n} \xrightarrow{P} 1.\)
Proof.
For any $\varepsilon>0$, \(\left|\frac{M_n}{a_n} - 1\right| < \varepsilon \quad \Leftrightarrow \quad |M_n - a_n| < \varepsilon a_n.\)
Since $a_n\to\infty$, for large $n$, $\varepsilon a_n$ is large, and
\(P(|M_n - a_n| < \varepsilon a_n)
\ge
P(|M_n - a_n| < \varepsilon)
\to 1.\)
Conclusion.
\(\frac{M_n}{a_n} \to 1 \quad \text{in probability}.\)
Key Takeaways
- Use the “multiply by 1” trick:
\(M_n - a_n = \frac{1}{a_n}\, a_n(M_n - a_n).\) - Slutsky transforms extreme-value scaling into probability convergence.
- Convergence in distribution to a constant implies convergence in probability.
- Scaling arguments: multiplying $\varepsilon$ by $a_n\to\infty$ relaxes inequalities.
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