Expected minimum absolute difference to a given point correctly computed?











up vote
0
down vote

favorite












I am trying to compute the expected absolute difference between $n$ iid uniformly between $[- epsilon, epsilon]$ sampled points to a given point $- epsilon< c < epsilon$:



$$
mathrm{E}(mathrm{min}(|x_1-c|,|x_2-c|,...,|x_n-c|))
$$



I know, that the expected minimum of n iid uniformly between $[0, epsilon]$ sampled points is $frac{epsilon}{n+1}$.



Can I simply put my coordinate systems origin onto c and say I have an expected minimum value of $frac{epsilon - c}{n+1}$ on the right, and $frac{epsilon + c}{n+1}$ on the left, and for the overall distribution I have an weighted average (wheigted with the width, so $epsilon-c$ and $epsilon+c$ respectively) of $frac{1}{2 epsilon} cdotfrac{(epsilon-c)^2 + (epsilon+c)^2}{n+1}$?



If not, how would I go about this problem?










share|cite|improve this question


























    up vote
    0
    down vote

    favorite












    I am trying to compute the expected absolute difference between $n$ iid uniformly between $[- epsilon, epsilon]$ sampled points to a given point $- epsilon< c < epsilon$:



    $$
    mathrm{E}(mathrm{min}(|x_1-c|,|x_2-c|,...,|x_n-c|))
    $$



    I know, that the expected minimum of n iid uniformly between $[0, epsilon]$ sampled points is $frac{epsilon}{n+1}$.



    Can I simply put my coordinate systems origin onto c and say I have an expected minimum value of $frac{epsilon - c}{n+1}$ on the right, and $frac{epsilon + c}{n+1}$ on the left, and for the overall distribution I have an weighted average (wheigted with the width, so $epsilon-c$ and $epsilon+c$ respectively) of $frac{1}{2 epsilon} cdotfrac{(epsilon-c)^2 + (epsilon+c)^2}{n+1}$?



    If not, how would I go about this problem?










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I am trying to compute the expected absolute difference between $n$ iid uniformly between $[- epsilon, epsilon]$ sampled points to a given point $- epsilon< c < epsilon$:



      $$
      mathrm{E}(mathrm{min}(|x_1-c|,|x_2-c|,...,|x_n-c|))
      $$



      I know, that the expected minimum of n iid uniformly between $[0, epsilon]$ sampled points is $frac{epsilon}{n+1}$.



      Can I simply put my coordinate systems origin onto c and say I have an expected minimum value of $frac{epsilon - c}{n+1}$ on the right, and $frac{epsilon + c}{n+1}$ on the left, and for the overall distribution I have an weighted average (wheigted with the width, so $epsilon-c$ and $epsilon+c$ respectively) of $frac{1}{2 epsilon} cdotfrac{(epsilon-c)^2 + (epsilon+c)^2}{n+1}$?



      If not, how would I go about this problem?










      share|cite|improve this question













      I am trying to compute the expected absolute difference between $n$ iid uniformly between $[- epsilon, epsilon]$ sampled points to a given point $- epsilon< c < epsilon$:



      $$
      mathrm{E}(mathrm{min}(|x_1-c|,|x_2-c|,...,|x_n-c|))
      $$



      I know, that the expected minimum of n iid uniformly between $[0, epsilon]$ sampled points is $frac{epsilon}{n+1}$.



      Can I simply put my coordinate systems origin onto c and say I have an expected minimum value of $frac{epsilon - c}{n+1}$ on the right, and $frac{epsilon + c}{n+1}$ on the left, and for the overall distribution I have an weighted average (wheigted with the width, so $epsilon-c$ and $epsilon+c$ respectively) of $frac{1}{2 epsilon} cdotfrac{(epsilon-c)^2 + (epsilon+c)^2}{n+1}$?



      If not, how would I go about this problem?







      probability probability-theory probability-distributions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 16 at 0:04









      Luca Thiede

      1486




      1486






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          Since $min|X_i - c|$ is non-negative and you want to compute its expectation, we can compute its survival function first. Consider the survival function of individual $|X_i - c|$:



          $$ begin{align}
          S(x)&=Pr{|X_i - c| > x} \
          &= Pr{X_i > c + x} + Pr{X_i < c - x} \
          &= begin{cases}
          1 & text {if} & x leq 0 \
          displaystyle 1 - frac {x} {epsilon} & text{if} & 0 < x leq epsilon - |c|\
          displaystyle frac {epsilon + |c| - x} {2epsilon} & text {if} & epsilon - |c| < x < epsilon + |c| \
          0 & text{if} & x geq epsilon + |c|
          end{cases}
          end{align}$$



          Then the survival function of $min |X_i - c|$ will be $S(x)^n$ and thus
          $$ begin{align}
          E[min|X_i - c|]
          &= int_0^{+infty}S(x)^ndx \
          &= int_0^{epsilon - |c|} left(1 - frac {x} {epsilon} right)^n dx
          + int_{epsilon - |c|}^{epsilon + |c|} left(frac {epsilon + |c| - x} {2epsilon} right)^n dx \
          &= left. frac {-epsilon} {n+1}left(1 - frac {x} {epsilon} right)^{n+1} right|_0^{epsilon - |c|} + left.frac {-2epsilon} {n+1} left(frac {epsilon + |c| - x} {2epsilon} right)^{n+1} right|_{epsilon - |c|}^{epsilon + |c|} \
          &= frac {epsilon} {n+1}left(1 - frac {|c|^{n+1}} {epsilon^{n+1}} + 2frac {|c|^{n+1}} {epsilon^{n+1}}right) \
          &= frac {epsilon} {n+1}left(1 + frac {|c|^{n+1}} {epsilon^{n+1}} right)
          end{align}$$



          As you see, the answer is minimized when $c = 0$ and it reduced back to the previous solution.






          share|cite|improve this answer





















            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000520%2fexpected-minimum-absolute-difference-to-a-given-point-correctly-computed%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            1
            down vote



            accepted










            Since $min|X_i - c|$ is non-negative and you want to compute its expectation, we can compute its survival function first. Consider the survival function of individual $|X_i - c|$:



            $$ begin{align}
            S(x)&=Pr{|X_i - c| > x} \
            &= Pr{X_i > c + x} + Pr{X_i < c - x} \
            &= begin{cases}
            1 & text {if} & x leq 0 \
            displaystyle 1 - frac {x} {epsilon} & text{if} & 0 < x leq epsilon - |c|\
            displaystyle frac {epsilon + |c| - x} {2epsilon} & text {if} & epsilon - |c| < x < epsilon + |c| \
            0 & text{if} & x geq epsilon + |c|
            end{cases}
            end{align}$$



            Then the survival function of $min |X_i - c|$ will be $S(x)^n$ and thus
            $$ begin{align}
            E[min|X_i - c|]
            &= int_0^{+infty}S(x)^ndx \
            &= int_0^{epsilon - |c|} left(1 - frac {x} {epsilon} right)^n dx
            + int_{epsilon - |c|}^{epsilon + |c|} left(frac {epsilon + |c| - x} {2epsilon} right)^n dx \
            &= left. frac {-epsilon} {n+1}left(1 - frac {x} {epsilon} right)^{n+1} right|_0^{epsilon - |c|} + left.frac {-2epsilon} {n+1} left(frac {epsilon + |c| - x} {2epsilon} right)^{n+1} right|_{epsilon - |c|}^{epsilon + |c|} \
            &= frac {epsilon} {n+1}left(1 - frac {|c|^{n+1}} {epsilon^{n+1}} + 2frac {|c|^{n+1}} {epsilon^{n+1}}right) \
            &= frac {epsilon} {n+1}left(1 + frac {|c|^{n+1}} {epsilon^{n+1}} right)
            end{align}$$



            As you see, the answer is minimized when $c = 0$ and it reduced back to the previous solution.






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              Since $min|X_i - c|$ is non-negative and you want to compute its expectation, we can compute its survival function first. Consider the survival function of individual $|X_i - c|$:



              $$ begin{align}
              S(x)&=Pr{|X_i - c| > x} \
              &= Pr{X_i > c + x} + Pr{X_i < c - x} \
              &= begin{cases}
              1 & text {if} & x leq 0 \
              displaystyle 1 - frac {x} {epsilon} & text{if} & 0 < x leq epsilon - |c|\
              displaystyle frac {epsilon + |c| - x} {2epsilon} & text {if} & epsilon - |c| < x < epsilon + |c| \
              0 & text{if} & x geq epsilon + |c|
              end{cases}
              end{align}$$



              Then the survival function of $min |X_i - c|$ will be $S(x)^n$ and thus
              $$ begin{align}
              E[min|X_i - c|]
              &= int_0^{+infty}S(x)^ndx \
              &= int_0^{epsilon - |c|} left(1 - frac {x} {epsilon} right)^n dx
              + int_{epsilon - |c|}^{epsilon + |c|} left(frac {epsilon + |c| - x} {2epsilon} right)^n dx \
              &= left. frac {-epsilon} {n+1}left(1 - frac {x} {epsilon} right)^{n+1} right|_0^{epsilon - |c|} + left.frac {-2epsilon} {n+1} left(frac {epsilon + |c| - x} {2epsilon} right)^{n+1} right|_{epsilon - |c|}^{epsilon + |c|} \
              &= frac {epsilon} {n+1}left(1 - frac {|c|^{n+1}} {epsilon^{n+1}} + 2frac {|c|^{n+1}} {epsilon^{n+1}}right) \
              &= frac {epsilon} {n+1}left(1 + frac {|c|^{n+1}} {epsilon^{n+1}} right)
              end{align}$$



              As you see, the answer is minimized when $c = 0$ and it reduced back to the previous solution.






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Since $min|X_i - c|$ is non-negative and you want to compute its expectation, we can compute its survival function first. Consider the survival function of individual $|X_i - c|$:



                $$ begin{align}
                S(x)&=Pr{|X_i - c| > x} \
                &= Pr{X_i > c + x} + Pr{X_i < c - x} \
                &= begin{cases}
                1 & text {if} & x leq 0 \
                displaystyle 1 - frac {x} {epsilon} & text{if} & 0 < x leq epsilon - |c|\
                displaystyle frac {epsilon + |c| - x} {2epsilon} & text {if} & epsilon - |c| < x < epsilon + |c| \
                0 & text{if} & x geq epsilon + |c|
                end{cases}
                end{align}$$



                Then the survival function of $min |X_i - c|$ will be $S(x)^n$ and thus
                $$ begin{align}
                E[min|X_i - c|]
                &= int_0^{+infty}S(x)^ndx \
                &= int_0^{epsilon - |c|} left(1 - frac {x} {epsilon} right)^n dx
                + int_{epsilon - |c|}^{epsilon + |c|} left(frac {epsilon + |c| - x} {2epsilon} right)^n dx \
                &= left. frac {-epsilon} {n+1}left(1 - frac {x} {epsilon} right)^{n+1} right|_0^{epsilon - |c|} + left.frac {-2epsilon} {n+1} left(frac {epsilon + |c| - x} {2epsilon} right)^{n+1} right|_{epsilon - |c|}^{epsilon + |c|} \
                &= frac {epsilon} {n+1}left(1 - frac {|c|^{n+1}} {epsilon^{n+1}} + 2frac {|c|^{n+1}} {epsilon^{n+1}}right) \
                &= frac {epsilon} {n+1}left(1 + frac {|c|^{n+1}} {epsilon^{n+1}} right)
                end{align}$$



                As you see, the answer is minimized when $c = 0$ and it reduced back to the previous solution.






                share|cite|improve this answer












                Since $min|X_i - c|$ is non-negative and you want to compute its expectation, we can compute its survival function first. Consider the survival function of individual $|X_i - c|$:



                $$ begin{align}
                S(x)&=Pr{|X_i - c| > x} \
                &= Pr{X_i > c + x} + Pr{X_i < c - x} \
                &= begin{cases}
                1 & text {if} & x leq 0 \
                displaystyle 1 - frac {x} {epsilon} & text{if} & 0 < x leq epsilon - |c|\
                displaystyle frac {epsilon + |c| - x} {2epsilon} & text {if} & epsilon - |c| < x < epsilon + |c| \
                0 & text{if} & x geq epsilon + |c|
                end{cases}
                end{align}$$



                Then the survival function of $min |X_i - c|$ will be $S(x)^n$ and thus
                $$ begin{align}
                E[min|X_i - c|]
                &= int_0^{+infty}S(x)^ndx \
                &= int_0^{epsilon - |c|} left(1 - frac {x} {epsilon} right)^n dx
                + int_{epsilon - |c|}^{epsilon + |c|} left(frac {epsilon + |c| - x} {2epsilon} right)^n dx \
                &= left. frac {-epsilon} {n+1}left(1 - frac {x} {epsilon} right)^{n+1} right|_0^{epsilon - |c|} + left.frac {-2epsilon} {n+1} left(frac {epsilon + |c| - x} {2epsilon} right)^{n+1} right|_{epsilon - |c|}^{epsilon + |c|} \
                &= frac {epsilon} {n+1}left(1 - frac {|c|^{n+1}} {epsilon^{n+1}} + 2frac {|c|^{n+1}} {epsilon^{n+1}}right) \
                &= frac {epsilon} {n+1}left(1 + frac {|c|^{n+1}} {epsilon^{n+1}} right)
                end{align}$$



                As you see, the answer is minimized when $c = 0$ and it reduced back to the previous solution.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 16 at 9:22









                BGM

                3,715148




                3,715148






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000520%2fexpected-minimum-absolute-difference-to-a-given-point-correctly-computed%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    How to change which sound is reproduced for terminal bell?

                    Can I use Tabulator js library in my java Spring + Thymeleaf project?

                    Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents