Expected Value for uniform random permutations












0












$begingroup$



Question:



Let $ngeq 2$ be an integer and let $a_1, a_2, ..., a_n$ be a uniformly random permutation of the set ${1,2,...n}$. Let $X$ be the random variable with value:



$X$ = the number of indices $i$ with $1 leq i leq n-1$ and $a_i < a_{i+1}$



For example, if $n = 6$ and the permutation is $3, 5, 4, 1, 6, 2$ then $X = 2$.
What is the expected value $E(X)$ of $X$? Use indicator variables.




Answer: $frac{n-1}{2}$



I define my indicator variable:



$$
X = left{begin{array}{rc} 1,&text{the number of indices i with 1 $<=$ i $<=$ $n-1$ and $a_i < a_{i+1}$ }{} \ 0,&text{other cases}{}end{array}right.
$$



For $n=2$: {1,2} , {2,1} so $Pr = $ $frac{1}{2}$



$E(X) =$ $P(X_2) =$ $frac{1}{2}$



For $n=3$: {1,2, 3} , {1,3,2} , {2,3,1} so $Pr = $ $frac{3}{6}$



Is this the correct method to solve this? I did this for $n$ values but when I computed there result with the answers, I never got it to be the same. How do I solve this?






probability probability-theory permutations random-variables expected-value







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    0












    $begingroup$



    Question:



    Let $ngeq 2$ be an integer and let $a_1, a_2, ..., a_n$ be a uniformly random permutation of the set ${1,2,...n}$. Let $X$ be the random variable with value:



    $X$ = the number of indices $i$ with $1 leq i leq n-1$ and $a_i < a_{i+1}$



    For example, if $n = 6$ and the permutation is $3, 5, 4, 1, 6, 2$ then $X = 2$.
    What is the expected value $E(X)$ of $X$? Use indicator variables.




    Answer: $frac{n-1}{2}$



    I define my indicator variable:



    $$
    X = left{begin{array}{rc} 1,&text{the number of indices i with 1 $<=$ i $<=$ $n-1$ and $a_i < a_{i+1}$ }{} \ 0,&text{other cases}{}end{array}right.
    $$



    For $n=2$: {1,2} , {2,1} so $Pr = $ $frac{1}{2}$



    $E(X) =$ $P(X_2) =$ $frac{1}{2}$



    For $n=3$: {1,2, 3} , {1,3,2} , {2,3,1} so $Pr = $ $frac{3}{6}$



    Is this the correct method to solve this? I did this for $n$ values but when I computed there result with the answers, I never got it to be the same. How do I solve this?






    probability probability-theory permutations random-variables expected-value







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      1



      $begingroup$



      Question:



      Let $ngeq 2$ be an integer and let $a_1, a_2, ..., a_n$ be a uniformly random permutation of the set ${1,2,...n}$. Let $X$ be the random variable with value:



      $X$ = the number of indices $i$ with $1 leq i leq n-1$ and $a_i < a_{i+1}$



      For example, if $n = 6$ and the permutation is $3, 5, 4, 1, 6, 2$ then $X = 2$.
      What is the expected value $E(X)$ of $X$? Use indicator variables.




      Answer: $frac{n-1}{2}$



      I define my indicator variable:



      $$
      X = left{begin{array}{rc} 1,&text{the number of indices i with 1 $<=$ i $<=$ $n-1$ and $a_i < a_{i+1}$ }{} \ 0,&text{other cases}{}end{array}right.
      $$



      For $n=2$: {1,2} , {2,1} so $Pr = $ $frac{1}{2}$



      $E(X) =$ $P(X_2) =$ $frac{1}{2}$



      For $n=3$: {1,2, 3} , {1,3,2} , {2,3,1} so $Pr = $ $frac{3}{6}$



      Is this the correct method to solve this? I did this for $n$ values but when I computed there result with the answers, I never got it to be the same. How do I solve this?






      probability probability-theory permutations random-variables expected-value







      share|cite|improve this question











      $endgroup$





      Question:



      Let $ngeq 2$ be an integer and let $a_1, a_2, ..., a_n$ be a uniformly random permutation of the set ${1,2,...n}$. Let $X$ be the random variable with value:



      $X$ = the number of indices $i$ with $1 leq i leq n-1$ and $a_i < a_{i+1}$



      For example, if $n = 6$ and the permutation is $3, 5, 4, 1, 6, 2$ then $X = 2$.
      What is the expected value $E(X)$ of $X$? Use indicator variables.




      Answer: $frac{n-1}{2}$



      I define my indicator variable:



      $$
      X = left{begin{array}{rc} 1,&text{the number of indices i with 1 $<=$ i $<=$ $n-1$ and $a_i < a_{i+1}$ }{} \ 0,&text{other cases}{}end{array}right.
      $$



      For $n=2$: {1,2} , {2,1} so $Pr = $ $frac{1}{2}$



      $E(X) =$ $P(X_2) =$ $frac{1}{2}$



      For $n=3$: {1,2, 3} , {1,3,2} , {2,3,1} so $Pr = $ $frac{3}{6}$



      Is this the correct method to solve this? I did this for $n$ values but when I computed there result with the answers, I never got it to be the same. How do I solve this?






      probability probability-theory permutations random-variables expected-value




      probability probability-theory permutations random-variables expected-value






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      share|cite|improve this question













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      edited Dec 2 '18 at 19:27









      greedoid

      39k114797




      39k114797










      asked Dec 2 '18 at 18:38









      TobyToby

      1577




      1577






















          1 Answer
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          $begingroup$

          If you define
          $$I_i = begin{cases} 1 & a_i < a_{i+1} \ 0 & a_i ge a_{i+1}end{cases}$$
          for $1 le i le n-1$,
          then $X = I_1 + I_2 + cdots + I_{n-1}$ so
          $$E[X] = E[I_1 + cdots + I_{n-1}] = E[I_1] + cdots + E[I_{n-1}] = P(a_1 < a_2) + cdots + P(a_{n-1} < a_n).$$
          Can you take it from here?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'm kind of lost on what you are doing. Can you please elaborate.
            $endgroup$
            – Toby
            Dec 2 '18 at 19:00










          • $begingroup$
            I've defined an indicator variable $I_1$ for the event ${a_1 < a_2}$, and so on for the other events ${a_i < a_{i+1}}$. This allows me to write $X$ as the sum of indicator variables. From there, calculating the expectation of $X$ reduces to computing the expectation of each indicator variable.
            $endgroup$
            – angryavian
            Dec 2 '18 at 19:08











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          $begingroup$

          If you define
          $$I_i = begin{cases} 1 & a_i < a_{i+1} \ 0 & a_i ge a_{i+1}end{cases}$$
          for $1 le i le n-1$,
          then $X = I_1 + I_2 + cdots + I_{n-1}$ so
          $$E[X] = E[I_1 + cdots + I_{n-1}] = E[I_1] + cdots + E[I_{n-1}] = P(a_1 < a_2) + cdots + P(a_{n-1} < a_n).$$
          Can you take it from here?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'm kind of lost on what you are doing. Can you please elaborate.
            $endgroup$
            – Toby
            Dec 2 '18 at 19:00










          • $begingroup$
            I've defined an indicator variable $I_1$ for the event ${a_1 < a_2}$, and so on for the other events ${a_i < a_{i+1}}$. This allows me to write $X$ as the sum of indicator variables. From there, calculating the expectation of $X$ reduces to computing the expectation of each indicator variable.
            $endgroup$
            – angryavian
            Dec 2 '18 at 19:08
















          1












          $begingroup$

          If you define
          $$I_i = begin{cases} 1 & a_i < a_{i+1} \ 0 & a_i ge a_{i+1}end{cases}$$
          for $1 le i le n-1$,
          then $X = I_1 + I_2 + cdots + I_{n-1}$ so
          $$E[X] = E[I_1 + cdots + I_{n-1}] = E[I_1] + cdots + E[I_{n-1}] = P(a_1 < a_2) + cdots + P(a_{n-1} < a_n).$$
          Can you take it from here?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'm kind of lost on what you are doing. Can you please elaborate.
            $endgroup$
            – Toby
            Dec 2 '18 at 19:00










          • $begingroup$
            I've defined an indicator variable $I_1$ for the event ${a_1 < a_2}$, and so on for the other events ${a_i < a_{i+1}}$. This allows me to write $X$ as the sum of indicator variables. From there, calculating the expectation of $X$ reduces to computing the expectation of each indicator variable.
            $endgroup$
            – angryavian
            Dec 2 '18 at 19:08














          1












          1








          1





          $begingroup$

          If you define
          $$I_i = begin{cases} 1 & a_i < a_{i+1} \ 0 & a_i ge a_{i+1}end{cases}$$
          for $1 le i le n-1$,
          then $X = I_1 + I_2 + cdots + I_{n-1}$ so
          $$E[X] = E[I_1 + cdots + I_{n-1}] = E[I_1] + cdots + E[I_{n-1}] = P(a_1 < a_2) + cdots + P(a_{n-1} < a_n).$$
          Can you take it from here?






          share|cite|improve this answer











          $endgroup$



          If you define
          $$I_i = begin{cases} 1 & a_i < a_{i+1} \ 0 & a_i ge a_{i+1}end{cases}$$
          for $1 le i le n-1$,
          then $X = I_1 + I_2 + cdots + I_{n-1}$ so
          $$E[X] = E[I_1 + cdots + I_{n-1}] = E[I_1] + cdots + E[I_{n-1}] = P(a_1 < a_2) + cdots + P(a_{n-1} < a_n).$$
          Can you take it from here?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 2 '18 at 19:08

























          answered Dec 2 '18 at 18:45









          angryavianangryavian

          40.2k23280




          40.2k23280












          • $begingroup$
            I'm kind of lost on what you are doing. Can you please elaborate.
            $endgroup$
            – Toby
            Dec 2 '18 at 19:00










          • $begingroup$
            I've defined an indicator variable $I_1$ for the event ${a_1 < a_2}$, and so on for the other events ${a_i < a_{i+1}}$. This allows me to write $X$ as the sum of indicator variables. From there, calculating the expectation of $X$ reduces to computing the expectation of each indicator variable.
            $endgroup$
            – angryavian
            Dec 2 '18 at 19:08


















          • $begingroup$
            I'm kind of lost on what you are doing. Can you please elaborate.
            $endgroup$
            – Toby
            Dec 2 '18 at 19:00










          • $begingroup$
            I've defined an indicator variable $I_1$ for the event ${a_1 < a_2}$, and so on for the other events ${a_i < a_{i+1}}$. This allows me to write $X$ as the sum of indicator variables. From there, calculating the expectation of $X$ reduces to computing the expectation of each indicator variable.
            $endgroup$
            – angryavian
            Dec 2 '18 at 19:08
















          $begingroup$
          I'm kind of lost on what you are doing. Can you please elaborate.
          $endgroup$
          – Toby
          Dec 2 '18 at 19:00




          $begingroup$
          I'm kind of lost on what you are doing. Can you please elaborate.
          $endgroup$
          – Toby
          Dec 2 '18 at 19:00












          $begingroup$
          I've defined an indicator variable $I_1$ for the event ${a_1 < a_2}$, and so on for the other events ${a_i < a_{i+1}}$. This allows me to write $X$ as the sum of indicator variables. From there, calculating the expectation of $X$ reduces to computing the expectation of each indicator variable.
          $endgroup$
          – angryavian
          Dec 2 '18 at 19:08




          $begingroup$
          I've defined an indicator variable $I_1$ for the event ${a_1 < a_2}$, and so on for the other events ${a_i < a_{i+1}}$. This allows me to write $X$ as the sum of indicator variables. From there, calculating the expectation of $X$ reduces to computing the expectation of each indicator variable.
          $endgroup$
          – angryavian
          Dec 2 '18 at 19:08


















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