Expected value of maximum of i.i.d. random variables
$begingroup$
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
asked Jan 25 at 16:03
pi66pi66
4,0201239
$endgroup$
add a comment |
$begingroup$
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
asked Jan 25 at 16:03
pi66pi66
4,0201239
$endgroup$
add a comment |
$begingroup$
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
asked Jan 25 at 16:03
pi66pi66
4,0201239
$endgroup$
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
asked Jan 25 at 16:03
pi66pi66
4,0201239
asked Jan 25 at 16:03
pi66pi66
4,0201239
asked Jan 25 at 16:03
pi66pi66
4,0201239
asked Jan 25 at 16:03
pi66pi66
4,0201239
asked Jan 25 at 16:03
pi66pi66
4,0201239
4,0201239
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
$$lim_{p to 1-}frac{mathbb E[M_{n+1}]}{mathbb E[M_n]} = lim_{p to 1-} frac{1-p^{n+1}}{1-p^n} = frac{n+1}{n} $$
answered Jan 25 at 16:48
Robert IsraelRobert Israel
327k23216470
$endgroup$
$begingroup$
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
|
show 1 more comment
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1 Answer
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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
$$lim_{p to 1-}frac{mathbb E[M_{n+1}]}{mathbb E[M_n]} = lim_{p to 1-} frac{1-p^{n+1}}{1-p^n} = frac{n+1}{n} $$
answered Jan 25 at 16:48
Robert IsraelRobert Israel
327k23216470
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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
$$lim_{p to 1-}frac{mathbb E[M_{n+1}]}{mathbb E[M_n]} = lim_{p to 1-} frac{1-p^{n+1}}{1-p^n} = frac{n+1}{n} $$
answered Jan 25 at 16:48
Robert IsraelRobert Israel
327k23216470
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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
$$lim_{p to 1-}frac{mathbb E[M_{n+1}]}{mathbb E[M_n]} = lim_{p to 1-} frac{1-p^{n+1}}{1-p^n} = frac{n+1}{n} $$
answered Jan 25 at 16:48
Robert IsraelRobert Israel
327k23216470
$endgroup$
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
$$lim_{p to 1-}frac{mathbb E[M_{n+1}]}{mathbb E[M_n]} = lim_{p to 1-} frac{1-p^{n+1}}{1-p^n} = frac{n+1}{n} $$
answered Jan 25 at 16:48
Robert IsraelRobert Israel
327k23216470
answered Jan 25 at 16:48
Robert IsraelRobert Israel
327k23216470
answered Jan 25 at 16:48
Robert IsraelRobert Israel
327k23216470
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
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