Expected number of offers until house is sold

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I am selling my house, and have decided to accept the first offering
exceeding $K$ dollars. Assuming that offers are independent rv with
common distribution $F$, find the expected number of offers received
before I sell the house.




Try



I call $X$ to be number of offers receiver before house is sold. Suppose we have $n$ such offers and call them $X_1,X_2,...,X_n$. Therefore, $X = sum X_i$. We have that



$$ E(X) = E(X_1) + .. + E(X_n) $$



Since all $X_i$ have common distribution $F$, then



$$ E(X_i) = intlimits_0^K x f(x) $$



where $f = F' $. So,



$$ E(X) = n intlimits_0^K x f(x) $$



is this correct?






probability







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    up vote
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    I am selling my house, and have decided to accept the first offering
    exceeding $K$ dollars. Assuming that offers are independent rv with
    common distribution $F$, find the expected number of offers received
    before I sell the house.




    Try



    I call $X$ to be number of offers receiver before house is sold. Suppose we have $n$ such offers and call them $X_1,X_2,...,X_n$. Therefore, $X = sum X_i$. We have that



    $$ E(X) = E(X_1) + .. + E(X_n) $$



    Since all $X_i$ have common distribution $F$, then



    $$ E(X_i) = intlimits_0^K x f(x) $$



    where $f = F' $. So,



    $$ E(X) = n intlimits_0^K x f(x) $$



    is this correct?






    probability







    share|cite|improve this question























      up vote
      4
      down vote

      favorite









      up vote
      4
      down vote

      favorite












      I am selling my house, and have decided to accept the first offering
      exceeding $K$ dollars. Assuming that offers are independent rv with
      common distribution $F$, find the expected number of offers received
      before I sell the house.




      Try



      I call $X$ to be number of offers receiver before house is sold. Suppose we have $n$ such offers and call them $X_1,X_2,...,X_n$. Therefore, $X = sum X_i$. We have that



      $$ E(X) = E(X_1) + .. + E(X_n) $$



      Since all $X_i$ have common distribution $F$, then



      $$ E(X_i) = intlimits_0^K x f(x) $$



      where $f = F' $. So,



      $$ E(X) = n intlimits_0^K x f(x) $$



      is this correct?






      probability







      share|cite|improve this question














      I am selling my house, and have decided to accept the first offering
      exceeding $K$ dollars. Assuming that offers are independent rv with
      common distribution $F$, find the expected number of offers received
      before I sell the house.




      Try



      I call $X$ to be number of offers receiver before house is sold. Suppose we have $n$ such offers and call them $X_1,X_2,...,X_n$. Therefore, $X = sum X_i$. We have that



      $$ E(X) = E(X_1) + .. + E(X_n) $$



      Since all $X_i$ have common distribution $F$, then



      $$ E(X_i) = intlimits_0^K x f(x) $$



      where $f = F' $. So,



      $$ E(X) = n intlimits_0^K x f(x) $$



      is this correct?






      probability




      probability






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      asked 2 hours ago









      Jimmy Sabater

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          This is completely the wrong way to do it. $F(K)$ gives the probability that one offer is less than $K$, i.e. the offer fails, so $1-F(K)$ is the probability of an offer succeeding. Now the number of offers received follows a geometric distribution with success rate $1-F(K)$, so the expected number of offers is its reciprocal, $frac11-F(K)$.






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            up vote
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            down vote



            accepted










            This is completely the wrong way to do it. $F(K)$ gives the probability that one offer is less than $K$, i.e. the offer fails, so $1-F(K)$ is the probability of an offer succeeding. Now the number of offers received follows a geometric distribution with success rate $1-F(K)$, so the expected number of offers is its reciprocal, $frac11-F(K)$.






            share|cite|improve this answer
























              up vote
              3
              down vote



              accepted










              This is completely the wrong way to do it. $F(K)$ gives the probability that one offer is less than $K$, i.e. the offer fails, so $1-F(K)$ is the probability of an offer succeeding. Now the number of offers received follows a geometric distribution with success rate $1-F(K)$, so the expected number of offers is its reciprocal, $frac11-F(K)$.






              share|cite|improve this answer






















                up vote
                3
                down vote



                accepted







                up vote
                3
                down vote



                accepted






                This is completely the wrong way to do it. $F(K)$ gives the probability that one offer is less than $K$, i.e. the offer fails, so $1-F(K)$ is the probability of an offer succeeding. Now the number of offers received follows a geometric distribution with success rate $1-F(K)$, so the expected number of offers is its reciprocal, $frac11-F(K)$.






                share|cite|improve this answer












                This is completely the wrong way to do it. $F(K)$ gives the probability that one offer is less than $K$, i.e. the offer fails, so $1-F(K)$ is the probability of an offer succeeding. Now the number of offers received follows a geometric distribution with success rate $1-F(K)$, so the expected number of offers is its reciprocal, $frac11-F(K)$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                Parcly Taxel

                38k137097




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