Show that the Gumbel distribution is max stable, i.e. if $X$ and $Y$ are two independent Gumbel distributed random variables with the same scale parameter, then $\max(X,Y)$ is Gumbel with the same scale parameter as $X$ and $Y$ . Recall that the distribution function in the Gumbel distribution is
$$F(x) = \exp(−e^{−(x−b)/a}),$$
where $a$ is the scale parameter.
Solution: If $X$ and $Y$ are independent and Gumbel distributed with scale parameter $a$ and location parameters $b_1$ and $b_2$ respectively and $Z=\max(X,Y)$ then
$$P(Z\le x)=P(X\le x)P(Y\le x)=\exp\left(-e^{-(x-b_1)/a}-e^{-(x-b_2)/a}\right)= \\ = \exp\left(-e^{-x/a}(e^{b_1/a}+e^{b_2/a})\right).$$
Since $e^{b_1/a}+e^{b_2/a}$ is positive and $e^{c/a}$ is continuous on $[0,\infty)$ there has to exist a $c:e^{b_1/a}+e^{b_2/a}=e^{c/a}.$
Questions: I don't understand how they treat the $\max$-function here. How does this apply:
$$P(Z\le x)=P(X\le x)P(Y\le x)?$$
This has to mean that $$P(Z\le x)=P(\max(X,Y)\leq x)=P(X\le x)P(Y\le x),$$
but then how is $$P(\max(X,Y)\leq x)=P(X\le x)P(Y\le x)?$$
I understand that the independency is responible for the multiplication, but what happens with the $\max(X,Y)?$
Let's assume that $X>Y,$ then it follows that
$$P(\max(X,Y)\leq x)=P(Y\le x),$$
If the reveres is true, i.e if $Y > X$ then it follows that
$$P(\max(X,Y)\leq x)=P(X\le x),$$
I don't see how these end upp being multiplied. I suppose that what I want is that
$$P(\max(X,Y)\leq x) = P(X\le x \ \cap \ Y\le x)=P(X\le x)P(Y\le x).$$
But why is it an intersection and not a union? Only one of the two cases can happen, so either $X>Y$ OR $X<Y$, and in probability, OR = $\cup$ and not $\cap.$ Can anyone explain why this is?
The statement $$P(\max(X,Y)\leq x) = P(X\le x \ \cap \ Y\le x)\tag1 $$ involves an intersection because the event in the LHS is the same as the event in the RHS: If $x$ and $y$ are numbers, then $\max(x,y)\le c$ exactly when $x\le c$ and $y\le c$. (Convince yourself that $\max(x,y)\le c$ is not the same as $x\le c$ or $y\le c$.)
Your line of reasoning fails because the statement $$P(\max(X,Y)\leq x)=P(Y\le x)\tag2 $$ isn't true; the event in the LHS is not the same as the event in the RHS. I know you are arguing by cases, but the probability function $P(\cdot)$ is unaware of that. When you are arguing by cases, you need to insert the case into the probability computation, so that the events on LHS and RHS remain equal. The proper statement for case $X\le Y$ is: $$P(\max(X,Y)\leq x\cap X\le Y)=P(Y\le x\cap X\le Y),\tag3 $$ because the events are now equal. Argue as follows: "If the max is less than or equal to $x$, and $X\le Y$, then $Y\le x$ and $X\le Y$. Conversely, if $X\le Y$ and $Y\le x$, then the max is less than or equal to $x$, and $X\le Y$." Yes, it's clunky, but without the qualifier $X\le Y$, you're back at (2) and the two sides of (2) are not equal.
The problem now is that (3) is a true statement but you haven't got yourself any closer to (1). You can rescue the computation, by taking it one step further: $$P(\max(X,Y)\leq x\cap X\le Y)=P(Y\le x\cap X\le Y)=P(X\le x\cap Y\le x\cap X\le Y),\tag4 $$ again because the events are equal. Now write the version for case $X>Y$:
$$P(\max(X,Y)\le x\cap X> Y)=P(X\le x\cap X> Y)=P(X\le x\cap Y\le x\cap X> Y),\tag5 $$ and now add (4) and (5) together: $$P(\max(X,Y)\le x)=P(X\le x\cap Y\le x).\tag6$$ In this last step (adding the two previous equalities), we are using the fact: $$P(A\cap B)+P(A\cap B^c)=P(A).$$