Is is possible to get characteristic function of maximum of i.i.d. random variable sequence? Such as $X_1, X_2$ are two i.i.d random variables, then what is characteristic function of $X=\max(X_1,X_2)$?
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2$\begingroup$ the cumulative distribution function of the maximum of $n$ iid random variables $x_i$ is just the $n$th power of the cumulative distribution of the $x_i$'s. $\endgroup$– Carlo BeenakkerCommented Jun 3, 2013 at 16:12

$\begingroup$ But $P(\max(X_1,X_2) \le t)= P(\{X_1 \le t\} \cup \{X_2 \le t\}).$ $\endgroup$– MarkCommented Jun 4, 2013 at 10:05

$\begingroup$ Mark, I think Uwe is right, just think that $max(X_1,X_2) \leq t$ means $X_1 \leq t$ and $X_2 \leq t$. $\endgroup$– parfoisCommented Jun 5, 2013 at 0:59
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1 Answer
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Assume that the random variables $X$ and $Y$ are defined on the probability space $(\Omega,\mathcal F,\mu)$. Let $\Delta:=\{(x,y)\in\Bbb R^2,x\lt y\}$. We have by independence $$ E\left[e^{it\max(X,Y)}\right]=\int_{\Bbb R^2}e^{it\max(x,y)}\mathrm d\mu_X\otimes\mu_Y(x,y). $$ Splitting over $\Delta$ and its complement, and denoting $F$ the common cumulative distribution function of $X$ and $Y$, we thus get $$E\left[e^{it\max(X,Y)}\right]=2E\left[F(X)e^{itX}\right]\int_{\Bbb R}\mu(X=x)e^{itx}\mathrm d\mu_X(x).$$
Some remarks:
 This gives an explicit formula in terms of the common distribution function.
 If $\mu(X=x)=0$ for all $x$ (for example when $X$ has a density), then the formula is simpler.
 This can be extended to $\max(X_1,\dots,X_d)$.
 We get an analogous formula for $\min$ instead of $\max$.

3$\begingroup$ The last $\mathrm d\mu$ should be $\mathrm d\mu_X$. And if $F$ is defined by $F(x)=P(X\leqslant x)$ (the càdlàg choice, if you wish), then the $+$ sign before this integral should be a $$ sign. $\endgroup$– DidCommented Jul 29, 2013 at 17:12

$\begingroup$ @Did I made the correction you suggested. Thank you. $\endgroup$ Commented Jul 29, 2013 at 17:50

$\begingroup$ I suppose there is no hope of any simplification when $X,Y$ are neither independent, not identically distributed? $\endgroup$ Commented Dec 15, 2019 at 21:36