Is there an equation for the maximum of n random draws from a Gamma distribution

Suppose a random variable $X$ follows a Gamma distribution with parameters $\alpha$ and $\beta$ with the probability density function for $x>0$ as

$$f(x;\alpha,\beta)= \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} \exp(-\beta x)$$

where $\Gamma(\alpha)$ represents the Gamma function. Suppose we take n samples from this distribution and only record the maximum value. For a given n, is there a general way to describe the expectation of this maximum value that we record? Below is a plot of the expectation (circles) and also the standard deviation (lines) for 1000 trials for each n draws, which produces a log-like curve with $\alpha=0.02$ and $\beta=0.0025$. Yes, indeed there is! However, that limit is going to be rather trivial, since for iid random variables $X_1,\ldots, X_n$ with distribution function $F$, $$P(\max\{X_1,\ldots, X_n\}\leq x) = F^n(x)$$ and for $x_F = \sup\{x\in \mathbb R:F(x)<1\}$ then $$F^n(x) \to \begin{cases} 0, &x< x_F\\1&x\geq x_F\end{cases}$$ for $n\to \infty$. That is a rather boring result.

Instead, theory has been developed to investigate the behavior of normed maxima in what we today know as Extreme Value Theory, which origins from the theorem of Fisher and Tippet back in 1928: if there exists norming constants $c_n>0$ and $d_n\in\mathbb R$ and a non-degenerate distribution function $G$ for a sequence of iid random variables $X_1,X_2,\ldots,X_n$ such that $$c_n^{-1}(M_n - d_n) \stackrel d\to H$$ where $M_n = \max\{X_1,\ldots,X_n\}$ then $H$ is called an extreme value distribution and is limited to the following three distributions: \begin{align} \Phi_\alpha(x) &= \begin{cases} 0, &x\leq 0,\\ \exp\{-x^{-\alpha}\},& x>0 \end{cases}\quad \alpha>0\\\\ \Psi_\alpha(x) &= \begin{cases} \exp\{ -(-x)^\alpha\}, & x\leq 0, \\0,&x>0 \end{cases}\quad \alpha >0\\ \\ \Lambda(x) &= \exp\{-e^{-x}\}, \quad x\in \mathbb R. \end{align} We call these three distributions Fréchet, Weibull and Gumbel, respectively and if $X_i$ have distribution function $F$ we say that $F$ is in the maximum domain of attraction of $H$ and write $F\in \operatorname{MDA}(H)$.

For the Gamma distribution, it can be shown that it belongs to the $\operatorname{MDA}(\Lambda)$ by choosing $$d_n = F^{\leftarrow}(1-1/n) \quad \text{and} \quad c_n = a(d_n)$$ where $$a(u) = \beta^{-1} \left(1+\frac{\alpha-1}{\beta u} + o\left(\frac 1u\right)\right) \quad \text{and}\quad F^{\leftarrow}(q) = \inf\{x\in\mathbb R:F(x)\geq q\}$$ is the auxiliary function, or equivalently the mean excess function for the Gamma distribution (see Embrechts et al. (1997) referred to below) and the generalized inverse respectively. This means that properly scaled, then $$P(c_n^{-1}(M_n - d_n)\leq x) \stackrel d\to \Lambda(x), \quad \forall x, \quad n\to \infty$$ The expected value of the Gumbel distribution can then be calculated. With $\Lambda(x) = \exp\{-e^{-x}\}$, the density function exists everywhere and is given as $\Lambda'(x) = e^{-x}\exp\{-e^{-x}\}$. For $Z\sim \Lambda$ then \begin{align} E[Z] &= \int_{-\infty}^{\infty}x e^{-x} e^{-e^{-x}}\, \mathrm d x \\&= - \int_{0}^{\infty}{\log y}e^{-y}\, \mathrm dy \qquad \qquad \text{(subs } y = e^{-x} \text{)} \\&= -\int_0^\infty \frac{\mathrm d}{\mathrm d\alpha} \bigl[y^\alpha e^{-y}\bigr]_{\alpha = 0} \, \mathrm dy \\&= -\frac{\mathrm d}{\mathrm d\alpha}\biggl[\int_0^\infty y^\alpha e^{-y}\, \mathrm dy\biggr]_{\alpha=0} \\&= -\frac{\mathrm d}{\mathrm d\alpha}\bigl[\Gamma(\alpha+1)\bigr]_{\alpha=0} \\&= -\Gamma'(1) \\&= \gamma \approx 0.577. \end{align} With $c_n^{-1}(E[M_n]-d_n) \sim E[Z]$ for $n\to \infty$ this means that

properly scaled, the expectation is the Euler-Mascheroni constant $\approx 0.577$.

I have run 100 simulations of the max of $n = \{10^1, 10^2,10^3,10^4,10^5,5\cdot 10^5, 10^6, 2\cdot 10^6 \}$ and taken the expectation for each run, then plotted this against $n$ and added the limiting theoretical value. For the auxiliary function, the little-o part has been ignored. I have used the same parameters as in the question, i.e. $\alpha = 0.02$ and $\beta = 0.0025$. For further reading, may I suggest either the book Modelling Extremal Events by Embrechts, Klüppelberg and Mikosch (1997) or Extreme Value Theory: An Introduction by de Haan (2006).

• Any idea then as to the expected value of the sample maximum, which is what the OP asks? – wolfies Dec 18 '16 at 15:29
• Yes, thank you @wolfies for pointing out that. In my eager I forgot to actually answer the question. – Therkel Dec 18 '16 at 16:46
• Thank you for the answer and providing further reading. A few further questions, how come $E[M_n]$ is independent of $n$? Is it to do with norming constants? I.e should $E[M_n]= c_n \gamma + d_n$? If so, this does not look like the plot I have above even for large n ($10^5$). Also, what about if I want to know the distribution of $M_n$, is it just a matter of multiplying/dividing the norming constants as well so that the distribution of $M_n$ is $c_n \Lambda(x) + d_n$ or am I missing something? Also, I think you left out a negative in the first definition of the Gumbel distribution. – Free_Apples Dec 19 '16 at 2:59
• @Free_Apples I have corrected some mistakes. The expectation is not independent of $n$ as you noted and it is only an asymptotic result. Such independence would only happen if $X_i$ were from the max-stable distribution themselves, i.e. $F = \Lambda$. Also, it seems like there is a mistake in the table I got the norming constants from in Embrechts et al. (1997). I got some misspecification, so I changed to the general norming constants for distributions in the $\operatorname{MDA}(\Lambda)$, which is a result on Von Mises functions. Yes, I forgot a minus in the definition of the Gumbel! Thanks! – Therkel Dec 19 '16 at 12:51
• If you want to investigate the distribution of $M_n$ generally, you could perhaps look at the distribution of order statistics close to the maxima. This is discussed in the book Statistics of Extremes by Beirlant, Goegebeur, Segers and Teugels (2004) in chapter 3. I will only mention this in a comment since it may be relevant, but I have not attempted their results on the Gamma distribution so I cannot say much more. – Therkel Dec 19 '16 at 12:58