Proving mean of sample minimum of U[0,1] is 1/(n+1) without calculus












0












$begingroup$


Let $U : mathbb{R} times mathbb{R} nrightarrow mathrm{dist}[mathbb{R}]$ denote the parametrized family of uniform distributions where $U(a, b)$ is the uniform distribution with minimum $a$ and maximum $b$ . $U$ is a partial function defined whenever $a lt b$ .



Let's define $Y_1, Y_2, dots Y_n$ as an $n$-element sample drawn from $U(0,1)$ . Let $Y_{(1)}$ denote the sample minimum. Let $f$ denote the pdf of the sample minimum of $n$ elements drawn from $U(0, 1)$ and let $F$ denote the corresponding cdf.



I want to demonstrate that its expected value is $frac{1}{n+1}$ in a way that's as simple as possible, ideally without using calculus.



$$ mathrm{E}[Y_{(1)}] = frac{1}{n+1} $$



Here's one way to do it, which does use calculus:



$$ mathrm{E}[Y_{(1)}] tag{1}$$



use known formula for expectation in terms of cdf



$$ int_{s=0}^infty 1-F(s) mathrm{d}s tag{2}$$



$F(s) = 1$ when $s ge 1$ .



$$ int_{s=0}^{1} 1 - F(s) mathrm{d}s tag{3} $$



$F(t)$ is the probability that the sample minimum is less than $t$.



$$ int_{s=0}^{1} 1 - mathbb{P}[Y_1 le s lor cdots lor Y_n le s] mathrm{d}s tag{4} $$



$1-mathbb{P}[psi]$ is $mathbb{P}[lnot psi]$ .



$$ int_{s=0}^{1} mathbb{P}[Y_1 ge s land cdots land Y_n ge s] mathrm{d}s tag{5} $$



replace integrand with product.



$$ int_{s=0}^{1} (1-s)^n mathrm{d}s tag{6} $$



change variable $ t = 1-s $ and swap bounds of integral.



$$ int_{t=0}^{1} t^n mathrm{d}t tag{7} $$



simplify



$$ frac{1}{n+1} tag{8} $$



ΟΕΔ






alternative-proof order-statistics







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Let $U : mathbb{R} times mathbb{R} nrightarrow mathrm{dist}[mathbb{R}]$ denote the parametrized family of uniform distributions where $U(a, b)$ is the uniform distribution with minimum $a$ and maximum $b$ . $U$ is a partial function defined whenever $a lt b$ .



    Let's define $Y_1, Y_2, dots Y_n$ as an $n$-element sample drawn from $U(0,1)$ . Let $Y_{(1)}$ denote the sample minimum. Let $f$ denote the pdf of the sample minimum of $n$ elements drawn from $U(0, 1)$ and let $F$ denote the corresponding cdf.



    I want to demonstrate that its expected value is $frac{1}{n+1}$ in a way that's as simple as possible, ideally without using calculus.



    $$ mathrm{E}[Y_{(1)}] = frac{1}{n+1} $$



    Here's one way to do it, which does use calculus:



    $$ mathrm{E}[Y_{(1)}] tag{1}$$



    use known formula for expectation in terms of cdf



    $$ int_{s=0}^infty 1-F(s) mathrm{d}s tag{2}$$



    $F(s) = 1$ when $s ge 1$ .



    $$ int_{s=0}^{1} 1 - F(s) mathrm{d}s tag{3} $$



    $F(t)$ is the probability that the sample minimum is less than $t$.



    $$ int_{s=0}^{1} 1 - mathbb{P}[Y_1 le s lor cdots lor Y_n le s] mathrm{d}s tag{4} $$



    $1-mathbb{P}[psi]$ is $mathbb{P}[lnot psi]$ .



    $$ int_{s=0}^{1} mathbb{P}[Y_1 ge s land cdots land Y_n ge s] mathrm{d}s tag{5} $$



    replace integrand with product.



    $$ int_{s=0}^{1} (1-s)^n mathrm{d}s tag{6} $$



    change variable $ t = 1-s $ and swap bounds of integral.



    $$ int_{t=0}^{1} t^n mathrm{d}t tag{7} $$



    simplify



    $$ frac{1}{n+1} tag{8} $$



    ΟΕΔ






    alternative-proof order-statistics







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $U : mathbb{R} times mathbb{R} nrightarrow mathrm{dist}[mathbb{R}]$ denote the parametrized family of uniform distributions where $U(a, b)$ is the uniform distribution with minimum $a$ and maximum $b$ . $U$ is a partial function defined whenever $a lt b$ .



      Let's define $Y_1, Y_2, dots Y_n$ as an $n$-element sample drawn from $U(0,1)$ . Let $Y_{(1)}$ denote the sample minimum. Let $f$ denote the pdf of the sample minimum of $n$ elements drawn from $U(0, 1)$ and let $F$ denote the corresponding cdf.



      I want to demonstrate that its expected value is $frac{1}{n+1}$ in a way that's as simple as possible, ideally without using calculus.



      $$ mathrm{E}[Y_{(1)}] = frac{1}{n+1} $$



      Here's one way to do it, which does use calculus:



      $$ mathrm{E}[Y_{(1)}] tag{1}$$



      use known formula for expectation in terms of cdf



      $$ int_{s=0}^infty 1-F(s) mathrm{d}s tag{2}$$



      $F(s) = 1$ when $s ge 1$ .



      $$ int_{s=0}^{1} 1 - F(s) mathrm{d}s tag{3} $$



      $F(t)$ is the probability that the sample minimum is less than $t$.



      $$ int_{s=0}^{1} 1 - mathbb{P}[Y_1 le s lor cdots lor Y_n le s] mathrm{d}s tag{4} $$



      $1-mathbb{P}[psi]$ is $mathbb{P}[lnot psi]$ .



      $$ int_{s=0}^{1} mathbb{P}[Y_1 ge s land cdots land Y_n ge s] mathrm{d}s tag{5} $$



      replace integrand with product.



      $$ int_{s=0}^{1} (1-s)^n mathrm{d}s tag{6} $$



      change variable $ t = 1-s $ and swap bounds of integral.



      $$ int_{t=0}^{1} t^n mathrm{d}t tag{7} $$



      simplify



      $$ frac{1}{n+1} tag{8} $$



      ΟΕΔ






      alternative-proof order-statistics







      share|cite|improve this question











      $endgroup$




      Let $U : mathbb{R} times mathbb{R} nrightarrow mathrm{dist}[mathbb{R}]$ denote the parametrized family of uniform distributions where $U(a, b)$ is the uniform distribution with minimum $a$ and maximum $b$ . $U$ is a partial function defined whenever $a lt b$ .



      Let's define $Y_1, Y_2, dots Y_n$ as an $n$-element sample drawn from $U(0,1)$ . Let $Y_{(1)}$ denote the sample minimum. Let $f$ denote the pdf of the sample minimum of $n$ elements drawn from $U(0, 1)$ and let $F$ denote the corresponding cdf.



      I want to demonstrate that its expected value is $frac{1}{n+1}$ in a way that's as simple as possible, ideally without using calculus.



      $$ mathrm{E}[Y_{(1)}] = frac{1}{n+1} $$



      Here's one way to do it, which does use calculus:



      $$ mathrm{E}[Y_{(1)}] tag{1}$$



      use known formula for expectation in terms of cdf



      $$ int_{s=0}^infty 1-F(s) mathrm{d}s tag{2}$$



      $F(s) = 1$ when $s ge 1$ .



      $$ int_{s=0}^{1} 1 - F(s) mathrm{d}s tag{3} $$



      $F(t)$ is the probability that the sample minimum is less than $t$.



      $$ int_{s=0}^{1} 1 - mathbb{P}[Y_1 le s lor cdots lor Y_n le s] mathrm{d}s tag{4} $$



      $1-mathbb{P}[psi]$ is $mathbb{P}[lnot psi]$ .



      $$ int_{s=0}^{1} mathbb{P}[Y_1 ge s land cdots land Y_n ge s] mathrm{d}s tag{5} $$



      replace integrand with product.



      $$ int_{s=0}^{1} (1-s)^n mathrm{d}s tag{6} $$



      change variable $ t = 1-s $ and swap bounds of integral.



      $$ int_{t=0}^{1} t^n mathrm{d}t tag{7} $$



      simplify



      $$ frac{1}{n+1} tag{8} $$



      ΟΕΔ






      alternative-proof order-statistics




      alternative-proof order-statistics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 25 '18 at 1:15







      Gregory Nisbet

















      asked Dec 24 '18 at 1:49









      Gregory NisbetGregory Nisbet

      709512




      709512






















          1 Answer
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          $begingroup$

          Take a circle of length $1$, and choose $n+1$ points independently and uniformly on it. Cut the circle at the first point, and unwrap to the interval $[0,1]$; the remaining $n$ points are distributed uniformly and independently on that interval. Now, those points cut $[0,1]$ into $n+1$ subintervals, in order. What's the expected length of each of them? Well, pull back to the circle - there was nothing special about us cutting at the first point. Cut at one of the other points, and we cycle the subintervals around, while keeping the same picture of $n$ uniform points after the cut. Thus, the lengths of all $n+1$ subintervals are identically distributed. In particular, they have the same mean. As their sum is $1$, the mean of each must be $frac1{n+1}$ by linearity of expectation. Done.



          This also works to calculate the expected value of all of the other order statistics; the $k$th smallest has expected value $frac{k}{n+1}$.






          share|cite|improve this answer









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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Take a circle of length $1$, and choose $n+1$ points independently and uniformly on it. Cut the circle at the first point, and unwrap to the interval $[0,1]$; the remaining $n$ points are distributed uniformly and independently on that interval. Now, those points cut $[0,1]$ into $n+1$ subintervals, in order. What's the expected length of each of them? Well, pull back to the circle - there was nothing special about us cutting at the first point. Cut at one of the other points, and we cycle the subintervals around, while keeping the same picture of $n$ uniform points after the cut. Thus, the lengths of all $n+1$ subintervals are identically distributed. In particular, they have the same mean. As their sum is $1$, the mean of each must be $frac1{n+1}$ by linearity of expectation. Done.



            This also works to calculate the expected value of all of the other order statistics; the $k$th smallest has expected value $frac{k}{n+1}$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Take a circle of length $1$, and choose $n+1$ points independently and uniformly on it. Cut the circle at the first point, and unwrap to the interval $[0,1]$; the remaining $n$ points are distributed uniformly and independently on that interval. Now, those points cut $[0,1]$ into $n+1$ subintervals, in order. What's the expected length of each of them? Well, pull back to the circle - there was nothing special about us cutting at the first point. Cut at one of the other points, and we cycle the subintervals around, while keeping the same picture of $n$ uniform points after the cut. Thus, the lengths of all $n+1$ subintervals are identically distributed. In particular, they have the same mean. As their sum is $1$, the mean of each must be $frac1{n+1}$ by linearity of expectation. Done.



              This also works to calculate the expected value of all of the other order statistics; the $k$th smallest has expected value $frac{k}{n+1}$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Take a circle of length $1$, and choose $n+1$ points independently and uniformly on it. Cut the circle at the first point, and unwrap to the interval $[0,1]$; the remaining $n$ points are distributed uniformly and independently on that interval. Now, those points cut $[0,1]$ into $n+1$ subintervals, in order. What's the expected length of each of them? Well, pull back to the circle - there was nothing special about us cutting at the first point. Cut at one of the other points, and we cycle the subintervals around, while keeping the same picture of $n$ uniform points after the cut. Thus, the lengths of all $n+1$ subintervals are identically distributed. In particular, they have the same mean. As their sum is $1$, the mean of each must be $frac1{n+1}$ by linearity of expectation. Done.



                This also works to calculate the expected value of all of the other order statistics; the $k$th smallest has expected value $frac{k}{n+1}$.






                share|cite|improve this answer









                $endgroup$



                Take a circle of length $1$, and choose $n+1$ points independently and uniformly on it. Cut the circle at the first point, and unwrap to the interval $[0,1]$; the remaining $n$ points are distributed uniformly and independently on that interval. Now, those points cut $[0,1]$ into $n+1$ subintervals, in order. What's the expected length of each of them? Well, pull back to the circle - there was nothing special about us cutting at the first point. Cut at one of the other points, and we cycle the subintervals around, while keeping the same picture of $n$ uniform points after the cut. Thus, the lengths of all $n+1$ subintervals are identically distributed. In particular, they have the same mean. As their sum is $1$, the mean of each must be $frac1{n+1}$ by linearity of expectation. Done.



                This also works to calculate the expected value of all of the other order statistics; the $k$th smallest has expected value $frac{k}{n+1}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 24 '18 at 2:00









                jmerryjmerry

                9,7681225




                9,7681225






























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