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Let $Q_p$ be a class of probability distributions on non-negative reals parametrized by $p$, so that $$ Q_p([0,\infty)) = 1. $$ I wonder which known classes of distributions are closed under taking the maximum and, i.e. if $X_1\sim Q_{p_1}$ and $X_2\sim Q_{p_2}$ are independent then $\max(X_1,X_2)\sim Q_{p_3}$.

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    $\begingroup$ Are you looking for a mathematical characterization of such classes or are you inquiring about which of the generally known parametric families of distributions may have this property? $\endgroup$ – whuber May 16 '12 at 15:55
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    $\begingroup$ stat.lsa.umich.edu/~sstoev/talks/BU_talk_07.pdf, p. 3 $\endgroup$ – whuber May 16 '12 at 16:15
  • $\begingroup$ @whuber All three extreme value types work by the argument I have given below. I do not show that they are the only ones though. $\endgroup$ – Michael Chernick May 16 '12 at 16:50
  • $\begingroup$ The powerpoint of Stoev that whuber cites shows the result I have given for these distributions that llya decribed which are called maxi-stable and the theorem quoted in the presentation furthermore states that they are the only ones. $\endgroup$ – Michael Chernick May 16 '12 at 17:01
  • $\begingroup$ @Michael Did you notice the restriction to non-negative values in the question? That rules out the extreme-value distributions having positive support on the negative reals. $\endgroup$ – whuber May 16 '12 at 17:15
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Those would be the extreme value distributions. There are three of them, as they are usually presented, corresponding to three sets of conditions on the underlying distribution for which the limiting distribution of the maximum is being found. They are closed under finding the maximum, which is what you want.

More-or-less copying from an old version of Methods of Statistical Analysis for Reliability and Life Data (Mann, Schafer, Singpurwalla),

Type I: $F_{X(n)}(x) = \exp\left\{-\exp \left[-\frac{x-\gamma}{\alpha}\right] \right\},\space -\infty < x < \infty, \space \alpha > 0$

Type II: $F_{X(n)}(x) = \exp\left\{-\left(\frac{x-\gamma}{\alpha}\right)^{-\beta}\right\}, \space x \geq \gamma, \space \alpha,\beta > 0$

Type III: $F_{X(n)}(x) = \exp\left\{-\left[-\left(\frac{x-\gamma}{\alpha}\right)^\beta\right]\right\}. \space x \leq \gamma, \space \alpha,\beta > 0$

Edit: Read the comments, which extend this answer to make a greatly improved and more complete answer to this question!

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    $\begingroup$ +1 But Types I and III do not apply to the question. $\endgroup$ – whuber May 16 '12 at 16:16
  • $\begingroup$ Quite true (+1), I was answering a more general question without explaining the difference. Also, I should have described the normalization that has to happen in order to prevent degeneracy, as you did in your comment to MC's answer below. Teach me to write these answers when I'm about to head out the door! (well, maybe not...:) $\endgroup$ – jbowman May 16 '12 at 17:41
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    $\begingroup$ @whuber I am probably asking something obvious but, is it true that if $X_1\sim Frechet(\alpha_1,\beta_1)$ and $X_2\sim Frechet(\alpha_2,\beta_2)$ and they are independent, then $max(X_1,X_2)\sim Frechet(\alpha_3,\beta_3)$? $\endgroup$ – user10525 May 16 '12 at 18:42
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    $\begingroup$ That's a great question, @Procrastinator. I couldn't think of any reason why such a result should be true, so I simulated 1,000,000 iid values from Frechet$(3,1)$ and 1,000,000 iid values from Frechet$(10,1)$ and computed their pairwise maxima. The results cannot be fit--not even approximately--by any Frechet$(\alpha,\beta)$ distribution. You need all three parameters (including the location parameter) to close this family under maxima. Then--emulating an (incomplete) argument in Michael Chernick's reply--you can show $\max(X_1,X_2)$ must be shifted scaled Frechet. $\endgroup$ – whuber May 16 '12 at 19:18
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It seems to me that proposing extreme value distributions really answers a different question. I will demonstrate that by addressing this question directly and showing it leads to distributions that are not among the extreme value types.

Let's consider this from first principles. It is immediate, from the axioms of probability and definition of the CDF, that the distribution of the maximum of two independent random variables with CDFs $F_1$ and $F_2$ has $F_1 F_2$ for its CDF. Suppose a class of distributions $\Omega = \{F_{\theta}\}$ exists that is closed under pairwise maximum; that is,

$$F_\theta \in \Omega, \ F_\phi \in \Omega \text{ implies } F_\theta F_\phi \in \Omega.$$

It is convenient to take logarithms, extending (as in Rudin's advanced analysis texts) the real numbers to include $-\infty$ as the log of $0$. Logs of CDFs of random variables essentially supported on $[0,\infty)$ are (i) mononotically nonincreasing, (ii) equal to $-\infty$ on $(-\infty,0)$, (iii) have right limits of $0$, and (iv) are cadlag. From this point of view, $\Omega$ must be a convex subset of a cone within the space of cadlag functions on $\mathbb{R}$. For it to be finitely parameterized, that cone must generate a finite-dimensional vector subspace. That still leaves a lot of possibilities.

Some of these possibilities are well known. Consider, for example, the CDF of a uniform variable on $[0,1]$. Its CDF equals $0$ on $(-\infty,0]$, $x$ when $0 \le x \le 1$, and $1$ on $[1,\infty)$. The cone it generates is the set of CDFs of the form

$$F_\theta(x) = \exp(\theta \log(x)) = x^\theta,\quad 0 \lt x \lt 1$$

parameterized by $\theta \gt 0$. Clearly the maximum of any two independent random variables with distributions in this family has a distribution also in this family (their parameters simply add). We may, if we wish, restrict to a convex subset of the form $\{F_\theta | \theta \ge \theta_0\}$ and still have a maximum-closed family. Notice, please, that no member of this family is an extreme-value distribution.

This formulation includes discrete distributions (which obviously are not among the three types of extreme value distributions). For instance, consider the distributions supported on the natural numbers $0, 1, 2, \ldots, k, \ldots$ for which the probabilities are given by

$${\Pr}_\theta(k) = \theta^{1/(k+1)} - \theta^{1/k}$$

(taking $\theta^{1/k}=0$ when $k=0$), parameterized by $0 \lt \theta \lt 1$. By construction, the CDF $F_\theta(k) = \theta^{1/(k+1)}$, whence it follows

$$F_\theta(k) F_\phi(k) = \theta^{1/(k+1)}\phi^{1/(k+1)} = (\theta\phi)^{1/(k+1)},$$

and because the assumptions imply $0 \lt \theta\phi \lt 1$, this shows the family is closed under pairwise maxima.

I hope that this analysis and these two examples show that, contrary to an opinion expressed in a comment, the approach of starting with a finite number of well-chosen CDFs and closing them with respect to the pairwise maximum (that is, forming their cones in an appropriate related vector space) not only is constructive but yields interesting and potentially useful classes of distributions.

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    $\begingroup$ +1 for this analysis and checking the interpretation of extreme value distributions. $\endgroup$ – user10525 May 17 '12 at 22:06
  • $\begingroup$ @whuber: thank you very much for the attention paid to this problem, I really didn't expect so much of nice answers (and I'll greet everybody who replied). The cone (or, semigroup) construction you've given is indeed true: if $F_\theta$ is any family of distributions then its closure (w.r.t. $\max$) has every element of the form $(F_{\theta_1}^{\alpha_1}\times\dots\times F_{\theta_n}^{\alpha_n})$ where $\alpha_i\geq 0$ and $n\in\mathbb N$. Unfortunately, I realized that closure w.r.t. shift is also needed (i.e. if $F(x)\in \Omega$ then $F(x-a)\in \Omega$). Should I ask a new question for this? $\endgroup$ – Ilya May 18 '12 at 11:40
  • $\begingroup$ That is certainly a complication, Ilya. But before you change anything or post any new question, please consider how you would reconcile the shift closure requirement with the (apparently contradictory) requirement that all the variables have nonnegative support! (I guess you will need to restrict the possible values of $a$.) $\endgroup$ – whuber May 18 '12 at 14:20
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jbowman beat me to the answer. A explanation for why they work is that Gnedenko's Theorem states that if $X_1,...,X_n$ is a sequence of n independent identically distributed random variables $M_n=\max(X_1, X_2,...,X_n)$ under certain conditions on the tail of the distribution converges to 1 of the three type that jbowman listed in his answer. Now since any type I, type II or type III distribtion can be expressed as the limit of the max of an iid sequence then if $G_1$ is say type I and is the limit distribution of $M_n=\max(X_1, X_2,...,X_n)$ as n tends to infinity and $G_2$ is also type I and is the limit of $N_n=\max(Y_1,Y_2,...,Y_n)$ then say $V_n=\max(M_n,N_n)$ and $G_3$ is the distribution of the limit as n approaches infinity for Vn then $G_3$ will be type I and be the distribution for the maximum of a rv with distribution $G_1$ and another with distribution $G_2$ and hence type I is closed under maximization. The same argument works for type II and type III.

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    $\begingroup$ For the unbounded distributions, the maximum does not converge: it diverges with $n$. As with the CLT, an appropriate normalization is required. (This is why it's essential to include location and scale parameters in these families.) Gnedenko's classic paper on the subject begins (if I recall correctly) by asking whether a series of affine coefficients $a_n, b_n$ can be found such that $a M_n + b_n$ converges. After establishing this, he then obtains the possible forms of limiting distribution. $\endgroup$ – whuber May 16 '12 at 17:18
  • $\begingroup$ In all cases I should have said appropriately normalized. Thanks. Even in the bounded case you have to normalize to get the limit (I think, I should remember this, my dissertation was on extremes! But 34 years ago) $\endgroup$ – Michael Chernick May 16 '12 at 17:48
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    $\begingroup$ Notice also that the extreme value distributions do not exhaustively answer the question. (This is not criticism, it's just an observation.) For instance, restricting $p$ to the natural numbers we can define $Q_p$ to be the uniform distribution on $[p,p+1]$. This class is closed under the maximum ($\max(Q_p, Q_r) \sim Q_{\max(p,r)}$), but no member of it is an extreme value distribution. $\endgroup$ – whuber May 16 '12 at 18:28
  • $\begingroup$ @whuber all three types are unbounded cases but the short tailed type III include bounded cases like the uniform distribution. For U[0,1], P[Mn<= 1-x/n] converges to exp(-x) since P[Mn<=1-x/n] =(1-x/n)^n. $\endgroup$ – Michael Chernick May 16 '12 at 18:29
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    $\begingroup$ Your replies do not seem relevant in the example I gave, Michael. The distinction is that this question is not asking about countable sequences of iid variables or even countable sequences of anything; it's asking only about closure under pairs of variables that typically have different distributions. (But now I see there is a flaw in my example: the maximum when $p=r$ is no longer uniform, so I would have to enlarge the family appropriately to include maxima of arbitrarily many iid uniforms.) $\endgroup$ – whuber May 16 '12 at 18:40

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