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Concentration and Exponential Bounds · Tutorial 932 of 994

Maxima of Sub-Gaussian Variables

Learn to bound the tail and expected value of a finite maximum, with explicit dependence on the number and sub-Gaussian parameters of the variables.

Advanced 11 min read

What You'll Learn

  • State the centered MGF definition of a sub-Gaussian variable and its scale parameter
  • Derive one-sided tail bounds for the maximum from individual sub-Gaussian tails
  • Calibrate a threshold using the number of variables and a target error probability
  • Handle unequal means and sub-Gaussian parameters without assuming independence
  • Bound the expected maximum using an exponential-moment argument
  • Distinguish bounds for a maximum from bounds for the maximum absolute deviation

Why Maxima Need a Separate Bound

A sub-Gaussian tail becomes less protective when we look across many variables: even if each one is unlikely to be unusually large, one among many may be. A useful bound must account for both the individual tail scales and the number of variables being considered. The union bound gives that adjustment directly, and it does not require the variables to be independent.

Throughout, let \(X_1,\ldots,X_m\) be real-valued random variables on the same probability space, with finite means \(\mu_i=E[X_i]\). They may be dependent. Write \(Z_i=X_i-\mu_i\) for the centered variables. This tutorial bounds the upper tail and expected value of \(\max_i X_i\); bounds for an absolute maximum require accounting for both tails.

Sub-Gaussian Scale and Individual Tails

Definition: A random variable \(X\) with finite mean \(\mu=E[X]\) is sub-Gaussian with parameter \(\sigma\), where \(\sigma\geq 0\), if $$ E\!\left[\exp\{\lambda(X-\mu)\}\right]\leq \exp\!\left(\frac{\sigma^2\lambda^2}{2}\right) \quad\text{for every }\lambda\in\mathbb{R}. $$ The parameter \(\sigma\) is an upper bound on the variable’s sub-Gaussian scale; it need not be the smallest possible such bound.

The MGF-to-tail argument gives, for \(\sigma>0\) and \(t>0\), $$ P(X-\mu\geq t)\leq \exp\!\left(-\frac{t^2}{2\sigma^2}\right), \qquad P(X-\mu\leq -t)\leq \exp\!\left(-\frac{t^2}{2\sigma^2}\right). $$

For the upper tail, for any \(\lambda>0\), Markov’s inequality and the MGF condition give $$ P(X-\mu\geq t) \leq e^{-\lambda t}E[e^{\lambda(X-\mu)}] \leq \exp\!\left(-\lambda t+\frac{\sigma^2\lambda^2}{2}\right). $$ The exponent is minimized at \(\lambda=t/\sigma^2\), giving the stated upper-tail bound. Apply the same argument to \(-(X-\mu)\) for the lower tail. If \(\sigma=0\), the MGF condition forces \(X=\mu\) almost surely, so the positive-deviation probabilities are zero.

These individual bounds are the input for a maximum bound. In particular, the scales and means need not be the same across variables.

Tail Bounds for a Finite Maximum

Theorem: Suppose \(X_i\) is sub-Gaussian with mean \(\mu_i\) and parameter \(\sigma_i>0\), for \(i=1,\ldots,m\). For \(t>0\), $$ P\!\left(\max_{1\leq i\leq m}(X_i-\mu_i)\geq t\right) \leq \sum_{i=1}^{m}\exp\!\left(-\frac{t^2}{2\sigma_i^2}\right). $$ In particular, if \(\sigma_i\leq\sigma\) for all \(i\), then $$ P\!\left(\max_{1\leq i\leq m}(X_i-\mu_i)\geq t\right) \leq m\exp\!\left(-\frac{t^2}{2\sigma^2}\right). $$ No independence assumption is needed.

Proof. The event that the maximum exceeds \(t\) is the union of the individual exceedance events: $$ \left\{\max_i(X_i-\mu_i)\geq t\right\} =\bigcup_{i=1}^{m}\{X_i-\mu_i\geq t\}. $$ The union bound and the individual sub-Gaussian tail bounds therefore give $$ P\!\left(\max_i(X_i-\mu_i)\geq t\right) \leq \sum_{i=1}^{m}P(X_i-\mu_i\geq t) \leq \sum_{i=1}^{m}\exp\!\left(-\frac{t^2}{2\sigma_i^2}\right). $$ If \(\sigma_i\leq\sigma\), then each exponential is at most \(\exp(-t^2/(2\sigma^2))\), yielding the common-scale bound. The argument uses no joint distribution calculation, so dependence does not affect its validity. \(\square\)

For a common upper bound \(\sigma>0\), choose a desired error probability \(\delta\in(0,1)\). Solving \(m\exp(-t^2/(2\sigma^2))=\delta\) gives the threshold $$ t=\sigma\sqrt{2\log(m/\delta)}. $$ Consequently, $$ P\!\left(\max_i(X_i-\mu_i)\geq \sigma\sqrt{2\log(m/\delta)}\right)\leq\delta. $$ The dependence on the number of variables is through \(\sqrt{\log m}\), rather than a factor proportional to \(m\) in the threshold. The factor \(m\) appears in the tail probability and becomes a logarithm when the bound is inverted.

Worked Example: A Common Scale Across 25 Variables

Suppose 25 random variables each have mean zero and are sub-Gaussian with parameter at most \(2\). They may be dependent. Find a threshold that the maximum exceeds with probability at most \(0.05\).

Here \(m=25\), \(\sigma=2\), and \(\delta=0.05\). The threshold formula gives $$ t=2\sqrt{2\log(25/0.05)} =2\sqrt{2\log(500)}. $$ Since \(\log(500)\approx6.214608\), this is approximately \(7.051\). Substitution verifies the probability bound: $$ P\!\left(\max_{1\leq i\leq25}X_i\geq2\sqrt{2\log(500)}\right) \leq25\exp\!\left(-\frac{[2\sqrt{2\log(500)}]^2}{2(2^2)}\right) =25e^{-\log(500)} =\frac{25}{500} =0.05. $$ The conclusion is an upper bound; the actual probability may be smaller, particularly if the variables are dependent in a way that makes large values occur together.

Unequal Scales and Unequal Means

If the centered variables have different parameters, retaining each \(\sigma_i\) can improve on replacing them all by their largest scale. The theorem already supplies this heterogeneous bound. If the quantities of interest are the uncentered \(X_i\), their means must also be taken into account. For any real \(x\), the event \(\{\max_i X_i\geq x\}\) is the union of \(\{X_i\geq x\}\). For indices with \(x>\mu_i\), apply the individual tail bound; indices with \(x\leq\mu_i\) can only be bounded trivially by 1. Thus $$ P(\max_i X_i\geq x) \leq \sum_{i:x>\mu_i}\exp\!\left(-\frac{(x-\mu_i)^2}{2\sigma_i^2}\right) +\#\{i:x\leq\mu_i\}. $$ This form is useful when means differ, though it may be uninformative if the threshold is below several means.

Worked Example: Three Different Sub-Gaussian Scales

Let \(X_1,X_2,X_3\) have mean zero and respective sub-Gaussian parameters \(1,2,3\). Bound the probability that at least one exceeds \(7.5\). No independence assumption is made.

Apply the heterogeneous union bound: $$ P(\max_i X_i\geq7.5) \leq e^{-7.5^2/(2\cdot1^2)} +e^{-7.5^2/(2\cdot2^2)} +e^{-7.5^2/(2\cdot3^2)}. $$ The three exponents are \(28.125\), \(7.03125\), and \(3.125\), respectively. Evaluating each term, $$ e^{-28.125}\approx0.00000000000061,\qquad e^{-7.03125}\approx0.0008838,\qquad e^{-3.125}\approx0.0439369. $$ Therefore the sum is approximately \(0.0448208\), and $$ P(\max_i X_i\geq7.5)\leq0.0448208. $$ For comparison, replacing all three parameters by the largest one, \(3\), gives the looser bound \(3e^{-7.5^2/18}=3e^{-3.125}\approx0.131811\). Keeping the individual scales substantially sharpens this particular union bound.

An Expected-Maximum Bound

The tail bound controls exceedance probabilities at a chosen threshold. A related argument gives a direct bound on the expected maximum. It also shows why the logarithm of the number of variables appears.

Theorem: If \(Z_1,\ldots,Z_m\) are centered sub-Gaussian random variables, each with parameter at most \(\sigma>0\), then $$ E\!\left[\max_{1\leq i\leq m}Z_i\right]\leq \sigma\sqrt{2\log m}. $$ For \(m=1\), the right-hand side is zero and \(E[Z_1]=0\). Independence is not required.

Proof. Each \(Z_i\) has finite exponential moments by its MGF bound and is integrable. For any \(\lambda>0\), the pointwise inequality $$ \max_i Z_i\leq\frac{1}{\lambda}\log\!\left(\sum_{i=1}^{m}e^{\lambda Z_i}\right) $$ holds because the sum includes \(e^{\lambda \max_i Z_i}\). The random variable on the right is integrable: it is bounded above by \(\lambda^{-1}\log m+\max_i Z_i\), and the finite maximum of integrable variables is integrable. Jensen’s inequality for the concave logarithm, followed by the MGF bounds, yields $$ E[\max_i Z_i] \leq\frac{1}{\lambda}\log\!\left(\sum_{i=1}^{m}E[e^{\lambda Z_i}]\right) \leq\frac{\log m}{\lambda}+\frac{\lambda\sigma^2}{2}. $$ For \(m>1\), minimize the last expression over \(\lambda>0\). Its derivative is \(-\log(m)/\lambda^2+\sigma^2/2\), so the minimizing value is \(\lambda=\sqrt{2\log(m)}/\sigma\). Substituting gives \(\sigma\sqrt{2\log m}\). For \(m=1\), the expected maximum is \(E[Z_1]=0\). \(\square\)

For variables with means \(\mu_i\) and individual parameters \(\sigma_i\), the same proof gives a more detailed form. For every \(\lambda>0\), $$ E[\max_i X_i] \leq\frac{1}{\lambda}\log\!\left( \sum_{i=1}^{m}\exp\!\left(\lambda\mu_i+\frac{\lambda^2\sigma_i^2}{2}\right) \right). $$ Indeed, apply the pointwise log-sum inequality to \(X_i\), take expectations using Jensen’s inequality, and use \(E[e^{\lambda X_i}]\leq\exp(\lambda\mu_i+\lambda^2\sigma_i^2/2)\). Taking the infimum over \(\lambda>0\) can yield a sharper bound when the means or scales differ.

Worked Example: Expected Maximum of Eight Variables

Suppose \(X_1,\ldots,X_8\) each have mean \(5\) and are sub-Gaussian with parameter at most \(3\). They need not be independent. Use the expected-maximum theorem to bound \(E[\max_i X_i]\).

Set \(Z_i=X_i-5\). Then the \(Z_i\) are centered and have parameters at most \(3\), and \(\max_i X_i=5+\max_i Z_i\). Hence $$ E[\max_i X_i]\leq5+3\sqrt{2\log 8}. $$ Since \(\log8\approx2.079442\), \(3\sqrt{2\log8}\approx6.1180\). Thus \(E[\max_i X_i]\leq11.1180\) (approximately). This is a one-sided upper bound on the mean of the maximum, not a claim that the expected maximum equals this value.

Absolute Maxima and Common Pitfalls

A bound for the largest positive deviation is not automatically a bound for the largest absolute deviation. For each \(i\), sub-Gaussianity bounds both \(P(Z_i\geq t)\) and \(P(Z_i\leq-t)\). Applying the union bound to all \(2m\) events gives $$ P\!\left(\max_i|X_i-\mu_i|\geq t\right) \leq 2\sum_{i=1}^{m}\exp\!\left(-\frac{t^2}{2\sigma_i^2}\right) \leq 2m\exp\!\left(-\frac{t^2}{2\sigma^2}\right), $$ when every \(\sigma_i\leq\sigma\). The corresponding common-scale threshold for probability at most \(\delta\) is \(\sigma\sqrt{2\log(2m/\delta)}\).

These results are useful whenever a statistic is selected as the largest among many candidates: the number of candidates affects the guarantee even if each candidate individually has a strong tail bound. The union-bound approach is simple and robust, while the expected-maximum argument supplies a complementary bound on average size. Both remain valid under dependence; any gains from a more detailed dependence structure require additional assumptions and analysis.

Check Your Understanding

Try these questions using the definitions and bounds above.

  1. Why does the union-bound tail estimate remain valid when the \(X_i\) are dependent?
  2. If \(m\) variables have a common sub-Gaussian parameter \(\sigma\), what threshold makes the upper-tail probability at most \(\delta\)?
  3. For three variables with parameters \(1,2,3\), write the union-bound expression for the probability that their centered maximum is at least \(t>0\).
  4. What changes when the target is \(\max_i|X_i-\mu_i|\) instead of the largest positive deviation?
  5. State the expected-maximum bound for \(m\) centered variables with common parameter at most \(\sigma\), and identify the condition under which it applies.
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