Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Discrete mathematics, continuous mathematics, and probability / Probability / 9

The cumulative distribution function is $F_X(x)=\mathbb{P}(X\le x)$. In the absolutely continuous setting of a probability density function, the probability density function is a nonnegative integrable function $f_X$ satisfyingThus $\mathbb{P}(a<X\le b)={\int }_a^b{f}_X(t)dt$ and $F_X^{\prime }=f_X$ almost everywhere. A merely continuous cumulative distribution function need not possess a probability density function; the density assumption means absolute continuity.

Every cumulative distribution function is nondecreasing, right-continuous, valued in $[0,1]$, and has limits $0$ at $-\infty$ and $1$ at $+\infty$. For a continuous random variable it is continuous. When a probability density function exists, it is in fact absolutely continuous, with nonnegative derivative almost everywhere.

The proposed cumulative distribution function satisfies $0<F(x)<1$ and is smooth. Its derivative isso it is increasing and continuous, hence right-continuous. As $x\to -\infty$, $e^{-x}\to \infty$ and $F(x)\to 0$; as $x\to \infty$, $e^{-x}\to 0$ and $F(x)\to 1$. Thus it has all the required cumulative distribution function properties, and defines the standard Gumbel distribution.

Because $Z=e^{-Y}>0$, its cumulative distribution function is zero for $z\le 0$. For $z>0$, the transformation is decreasing, and $Y$ has a continuous cumulative distribution function, soThus the exponential transform of a standard Gumbel variable givesThis is the exponential distribution with rate $1$.

Write $M_n={\max }_i{E}_i$. Each exponential distribution has cumulative distribution function $1-e^{-x}$ for $x\ge 0$. Independence therefore gives the maximum of independent exponential random variables formulaFor any fixed $y$, the threshold $y+\ln n$ is nonnegative for all sufficiently large $n$. ThenThis is convergence in distribution to the standard Gumbel distribution.

First $0\le M_n\le {\sum }_i{E}_i$, so linearity of expectation gives $\mathbb{E}{M}_n\le n<\infty$. The tail integral formula for expectation, followed by the finite binomial theorem expansion, givesThe exchange involves only a finite sum. As a useful check, substitution $t=e^{-x}$ rewrites the integral as ${\int }_0^1[1-(1-t{)}^n]/tdt={\sum }_{j=0}^{n-1}{\int }_0^1(1-t{)}^{j}dt={\sum }_{k=1}^{n}1/k$, the harmonic number $H_n$.