Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $X=\max(X_1,X_2)$?

share|improve this question
2  
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

1 Answer 1

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 $\max$.
share|improve this answer
2  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.