Expected value of maximum of i.i.d. random variables












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$begingroup$


Let $n$ be a fixed positive integer. For a distribution $F$ on positive real numbers, let $a_n(F)$ be the expected value of the maximum of $n$ i.i.d. random variables drawn from $F$, and let $a_{n+1}(F)$ be the expected value of the maximum of $n+1$ i.i.d. random variables drawn from $F$. What is $$max_{F}frac{a_{n+1}(F)}{a_n(F)}?$$



For example, if $F$ is the uniform distribution on $(0,1)$, then $a_n(F)=frac{n}{n+1}$ and $a_{n+1}(F)=frac{n+1}{n+2}$. My guess is that the maximum is $frac{n+1}{n}$. How can we show it, or is there a theorem stating this?






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$endgroup$

















    0












    $begingroup$


    Let $n$ be a fixed positive integer. For a distribution $F$ on positive real numbers, let $a_n(F)$ be the expected value of the maximum of $n$ i.i.d. random variables drawn from $F$, and let $a_{n+1}(F)$ be the expected value of the maximum of $n+1$ i.i.d. random variables drawn from $F$. What is $$max_{F}frac{a_{n+1}(F)}{a_n(F)}?$$



    For example, if $F$ is the uniform distribution on $(0,1)$, then $a_n(F)=frac{n}{n+1}$ and $a_{n+1}(F)=frac{n+1}{n+2}$. My guess is that the maximum is $frac{n+1}{n}$. How can we show it, or is there a theorem stating this?






    probability







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $n$ be a fixed positive integer. For a distribution $F$ on positive real numbers, let $a_n(F)$ be the expected value of the maximum of $n$ i.i.d. random variables drawn from $F$, and let $a_{n+1}(F)$ be the expected value of the maximum of $n+1$ i.i.d. random variables drawn from $F$. What is $$max_{F}frac{a_{n+1}(F)}{a_n(F)}?$$



      For example, if $F$ is the uniform distribution on $(0,1)$, then $a_n(F)=frac{n}{n+1}$ and $a_{n+1}(F)=frac{n+1}{n+2}$. My guess is that the maximum is $frac{n+1}{n}$. How can we show it, or is there a theorem stating this?






      probability







      share|cite|improve this question









      $endgroup$




      Let $n$ be a fixed positive integer. For a distribution $F$ on positive real numbers, let $a_n(F)$ be the expected value of the maximum of $n$ i.i.d. random variables drawn from $F$, and let $a_{n+1}(F)$ be the expected value of the maximum of $n+1$ i.i.d. random variables drawn from $F$. What is $$max_{F}frac{a_{n+1}(F)}{a_n(F)}?$$



      For example, if $F$ is the uniform distribution on $(0,1)$, then $a_n(F)=frac{n}{n+1}$ and $a_{n+1}(F)=frac{n+1}{n+2}$. My guess is that the maximum is $frac{n+1}{n}$. How can we show it, or is there a theorem stating this?






      probability




      probability






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Jan 25 at 16:03









      pi66pi66

      4,0201239




      4,0201239






















          1 Answer
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          $begingroup$

          If $X_1, ldots, X_n$ are iid on the positive reals with cdf $F$, the cdf of $M_n = max(X_1, ldots, X_n)$ is
          $$ mathbb F_n(x) = mathbb P(M_n le x) = mathbb P(text{all } X_n le x) = F(x)^n $$
          and so
          $$ mathbb E[M_n] = int_0^infty (1 - F(x)^n); dx $$



          Now for any $0 le t le 1$ and positive integer $n$, $$1 - t^{n+1} le frac{n+1}{n} (1 - t^n)$$
          since $g(t) = (n+1) (1 - t^n) - n (1-t^{n+1})$ is nonincreasing on $[0,1]$ (its derivative is $n(n+1)(t^{n-1}-t^n) le 0$).
          Thus it is indeed true that $$mathbb E[M_{n+1}] le frac{n+1}{n} mathbb E[M_n]$$



          To see that the bound is sharp, consider a Bernoulli distribution with parameter $p to 1-$. We have
          $$lim_{p to 1-}frac{mathbb E[M_{n+1}]}{mathbb E[M_n]} = lim_{p to 1-} frac{1-p^{n+1}}{1-p^n} = frac{n+1}{n} $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How does your formula for $mathbb E[M_n]$ follow from the definition $mathbb E[X]=int_0^infty xf(x)dx$?
            $endgroup$
            – pi66
            Jan 25 at 17:02










          • $begingroup$
            Integration by parts.
            $endgroup$
            – Robert Israel
            Jan 25 at 17:48










          • $begingroup$
            I think you get $xF(x)^nmid_0^infty - int_0^infty F(x)^n dx$. How do you take care of the first term?
            $endgroup$
            – pi66
            Jan 25 at 18:18










          • $begingroup$
            Use $F(x)^n - 1$ instead of $F(x)^n$. Then $left.x (F(x)^n-1)right|_0^infty = 0$.
            $endgroup$
            – Robert Israel
            Jan 27 at 1:51












          • $begingroup$
            Sorry, I'm still confused. What $f(x)$ do you use then? And in $x(F(x)^n-1)$, if you take $x=infty$, you get $inftycdot 0$ which is undefined, don't you?
            $endgroup$
            – pi66
            Jan 27 at 10:33











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          $begingroup$

          If $X_1, ldots, X_n$ are iid on the positive reals with cdf $F$, the cdf of $M_n = max(X_1, ldots, X_n)$ is
          $$ mathbb F_n(x) = mathbb P(M_n le x) = mathbb P(text{all } X_n le x) = F(x)^n $$
          and so
          $$ mathbb E[M_n] = int_0^infty (1 - F(x)^n); dx $$



          Now for any $0 le t le 1$ and positive integer $n$, $$1 - t^{n+1} le frac{n+1}{n} (1 - t^n)$$
          since $g(t) = (n+1) (1 - t^n) - n (1-t^{n+1})$ is nonincreasing on $[0,1]$ (its derivative is $n(n+1)(t^{n-1}-t^n) le 0$).
          Thus it is indeed true that $$mathbb E[M_{n+1}] le frac{n+1}{n} mathbb E[M_n]$$



          To see that the bound is sharp, consider a Bernoulli distribution with parameter $p to 1-$. We have
          $$lim_{p to 1-}frac{mathbb E[M_{n+1}]}{mathbb E[M_n]} = lim_{p to 1-} frac{1-p^{n+1}}{1-p^n} = frac{n+1}{n} $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How does your formula for $mathbb E[M_n]$ follow from the definition $mathbb E[X]=int_0^infty xf(x)dx$?
            $endgroup$
            – pi66
            Jan 25 at 17:02










          • $begingroup$
            Integration by parts.
            $endgroup$
            – Robert Israel
            Jan 25 at 17:48










          • $begingroup$
            I think you get $xF(x)^nmid_0^infty - int_0^infty F(x)^n dx$. How do you take care of the first term?
            $endgroup$
            – pi66
            Jan 25 at 18:18










          • $begingroup$
            Use $F(x)^n - 1$ instead of $F(x)^n$. Then $left.x (F(x)^n-1)right|_0^infty = 0$.
            $endgroup$
            – Robert Israel
            Jan 27 at 1:51












          • $begingroup$
            Sorry, I'm still confused. What $f(x)$ do you use then? And in $x(F(x)^n-1)$, if you take $x=infty$, you get $inftycdot 0$ which is undefined, don't you?
            $endgroup$
            – pi66
            Jan 27 at 10:33
















          1












          $begingroup$

          If $X_1, ldots, X_n$ are iid on the positive reals with cdf $F$, the cdf of $M_n = max(X_1, ldots, X_n)$ is
          $$ mathbb F_n(x) = mathbb P(M_n le x) = mathbb P(text{all } X_n le x) = F(x)^n $$
          and so
          $$ mathbb E[M_n] = int_0^infty (1 - F(x)^n); dx $$



          Now for any $0 le t le 1$ and positive integer $n$, $$1 - t^{n+1} le frac{n+1}{n} (1 - t^n)$$
          since $g(t) = (n+1) (1 - t^n) - n (1-t^{n+1})$ is nonincreasing on $[0,1]$ (its derivative is $n(n+1)(t^{n-1}-t^n) le 0$).
          Thus it is indeed true that $$mathbb E[M_{n+1}] le frac{n+1}{n} mathbb E[M_n]$$



          To see that the bound is sharp, consider a Bernoulli distribution with parameter $p to 1-$. We have
          $$lim_{p to 1-}frac{mathbb E[M_{n+1}]}{mathbb E[M_n]} = lim_{p to 1-} frac{1-p^{n+1}}{1-p^n} = frac{n+1}{n} $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How does your formula for $mathbb E[M_n]$ follow from the definition $mathbb E[X]=int_0^infty xf(x)dx$?
            $endgroup$
            – pi66
            Jan 25 at 17:02










          • $begingroup$
            Integration by parts.
            $endgroup$
            – Robert Israel
            Jan 25 at 17:48










          • $begingroup$
            I think you get $xF(x)^nmid_0^infty - int_0^infty F(x)^n dx$. How do you take care of the first term?
            $endgroup$
            – pi66
            Jan 25 at 18:18










          • $begingroup$
            Use $F(x)^n - 1$ instead of $F(x)^n$. Then $left.x (F(x)^n-1)right|_0^infty = 0$.
            $endgroup$
            – Robert Israel
            Jan 27 at 1:51












          • $begingroup$
            Sorry, I'm still confused. What $f(x)$ do you use then? And in $x(F(x)^n-1)$, if you take $x=infty$, you get $inftycdot 0$ which is undefined, don't you?
            $endgroup$
            – pi66
            Jan 27 at 10:33














          1












          1








          1





          $begingroup$

          If $X_1, ldots, X_n$ are iid on the positive reals with cdf $F$, the cdf of $M_n = max(X_1, ldots, X_n)$ is
          $$ mathbb F_n(x) = mathbb P(M_n le x) = mathbb P(text{all } X_n le x) = F(x)^n $$
          and so
          $$ mathbb E[M_n] = int_0^infty (1 - F(x)^n); dx $$



          Now for any $0 le t le 1$ and positive integer $n$, $$1 - t^{n+1} le frac{n+1}{n} (1 - t^n)$$
          since $g(t) = (n+1) (1 - t^n) - n (1-t^{n+1})$ is nonincreasing on $[0,1]$ (its derivative is $n(n+1)(t^{n-1}-t^n) le 0$).
          Thus it is indeed true that $$mathbb E[M_{n+1}] le frac{n+1}{n} mathbb E[M_n]$$



          To see that the bound is sharp, consider a Bernoulli distribution with parameter $p to 1-$. We have
          $$lim_{p to 1-}frac{mathbb E[M_{n+1}]}{mathbb E[M_n]} = lim_{p to 1-} frac{1-p^{n+1}}{1-p^n} = frac{n+1}{n} $$






          share|cite|improve this answer









          $endgroup$



          If $X_1, ldots, X_n$ are iid on the positive reals with cdf $F$, the cdf of $M_n = max(X_1, ldots, X_n)$ is
          $$ mathbb F_n(x) = mathbb P(M_n le x) = mathbb P(text{all } X_n le x) = F(x)^n $$
          and so
          $$ mathbb E[M_n] = int_0^infty (1 - F(x)^n); dx $$



          Now for any $0 le t le 1$ and positive integer $n$, $$1 - t^{n+1} le frac{n+1}{n} (1 - t^n)$$
          since $g(t) = (n+1) (1 - t^n) - n (1-t^{n+1})$ is nonincreasing on $[0,1]$ (its derivative is $n(n+1)(t^{n-1}-t^n) le 0$).
          Thus it is indeed true that $$mathbb E[M_{n+1}] le frac{n+1}{n} mathbb E[M_n]$$



          To see that the bound is sharp, consider a Bernoulli distribution with parameter $p to 1-$. We have
          $$lim_{p to 1-}frac{mathbb E[M_{n+1}]}{mathbb E[M_n]} = lim_{p to 1-} frac{1-p^{n+1}}{1-p^n} = frac{n+1}{n} $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 25 at 16:48









          Robert IsraelRobert Israel

          327k23216470




          327k23216470












          • $begingroup$
            How does your formula for $mathbb E[M_n]$ follow from the definition $mathbb E[X]=int_0^infty xf(x)dx$?
            $endgroup$
            – pi66
            Jan 25 at 17:02










          • $begingroup$
            Integration by parts.
            $endgroup$
            – Robert Israel
            Jan 25 at 17:48










          • $begingroup$
            I think you get $xF(x)^nmid_0^infty - int_0^infty F(x)^n dx$. How do you take care of the first term?
            $endgroup$
            – pi66
            Jan 25 at 18:18










          • $begingroup$
            Use $F(x)^n - 1$ instead of $F(x)^n$. Then $left.x (F(x)^n-1)right|_0^infty = 0$.
            $endgroup$
            – Robert Israel
            Jan 27 at 1:51












          • $begingroup$
            Sorry, I'm still confused. What $f(x)$ do you use then? And in $x(F(x)^n-1)$, if you take $x=infty$, you get $inftycdot 0$ which is undefined, don't you?
            $endgroup$
            – pi66
            Jan 27 at 10:33


















          • $begingroup$
            How does your formula for $mathbb E[M_n]$ follow from the definition $mathbb E[X]=int_0^infty xf(x)dx$?
            $endgroup$
            – pi66
            Jan 25 at 17:02










          • $begingroup$
            Integration by parts.
            $endgroup$
            – Robert Israel
            Jan 25 at 17:48










          • $begingroup$
            I think you get $xF(x)^nmid_0^infty - int_0^infty F(x)^n dx$. How do you take care of the first term?
            $endgroup$
            – pi66
            Jan 25 at 18:18










          • $begingroup$
            Use $F(x)^n - 1$ instead of $F(x)^n$. Then $left.x (F(x)^n-1)right|_0^infty = 0$.
            $endgroup$
            – Robert Israel
            Jan 27 at 1:51












          • $begingroup$
            Sorry, I'm still confused. What $f(x)$ do you use then? And in $x(F(x)^n-1)$, if you take $x=infty$, you get $inftycdot 0$ which is undefined, don't you?
            $endgroup$
            – pi66
            Jan 27 at 10:33
















          $begingroup$
          How does your formula for $mathbb E[M_n]$ follow from the definition $mathbb E[X]=int_0^infty xf(x)dx$?
          $endgroup$
          – pi66
          Jan 25 at 17:02




          $begingroup$
          How does your formula for $mathbb E[M_n]$ follow from the definition $mathbb E[X]=int_0^infty xf(x)dx$?
          $endgroup$
          – pi66
          Jan 25 at 17:02












          $begingroup$
          Integration by parts.
          $endgroup$
          – Robert Israel
          Jan 25 at 17:48




          $begingroup$
          Integration by parts.
          $endgroup$
          – Robert Israel
          Jan 25 at 17:48












          $begingroup$
          I think you get $xF(x)^nmid_0^infty - int_0^infty F(x)^n dx$. How do you take care of the first term?
          $endgroup$
          – pi66
          Jan 25 at 18:18




          $begingroup$
          I think you get $xF(x)^nmid_0^infty - int_0^infty F(x)^n dx$. How do you take care of the first term?
          $endgroup$
          – pi66
          Jan 25 at 18:18












          $begingroup$
          Use $F(x)^n - 1$ instead of $F(x)^n$. Then $left.x (F(x)^n-1)right|_0^infty = 0$.
          $endgroup$
          – Robert Israel
          Jan 27 at 1:51






          $begingroup$
          Use $F(x)^n - 1$ instead of $F(x)^n$. Then $left.x (F(x)^n-1)right|_0^infty = 0$.
          $endgroup$
          – Robert Israel
          Jan 27 at 1:51














          $begingroup$
          Sorry, I'm still confused. What $f(x)$ do you use then? And in $x(F(x)^n-1)$, if you take $x=infty$, you get $inftycdot 0$ which is undefined, don't you?
          $endgroup$
          – pi66
          Jan 27 at 10:33




          $begingroup$
          Sorry, I'm still confused. What $f(x)$ do you use then? And in $x(F(x)^n-1)$, if you take $x=infty$, you get $inftycdot 0$ which is undefined, don't you?
          $endgroup$
          – pi66
          Jan 27 at 10:33


















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