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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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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. – Carlo Beenakker Jun 3 '13 at 16:12
But $P(\max(X_1,X_2) \le t)= P(\{X_1 \le t\} \cup \{X_2 \le t\}).$ – Mark Jun 4 '13 at 10:05
Mark, I think Uwe is right, just think that $max(X_1,X_2) \leq t$ means $X_1 \leq t$ and $X_2 \leq t$. – parfois Jun 5 '13 at 0:59

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:

  1. This gives an explicit formula in terms of the common distribution function.
  2. If $\mu(X=x)=0$ for all $x$ (for example when $X$ has a density), then the formula is simpler.
  3. This can be extended to $\max(X_1,\dots,X_d)$.
  4. We get an analogous formula for $\min$ instead of $\max$.
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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. – Did Jul 29 '13 at 17:12
@Did I made the correction you suggested. Thank you. – Davide Giraudo Jul 29 '13 at 17:50

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