Problem on Expected Value (Probability)












0












$begingroup$


We are given a set $X = {x_1,...,x_n}$, where $x_i = 2^i$. A sample $S subseteq X$ is drawn by selecting each $x_i$ independently with probability $p_i=frac{1}{2}$. The expected value of the smallest number in sample S is:
(A) $frac{1}{n}$
(B) $2$
(C) $sqrt{n}$
(D) $n$



My approach:



$X = {2^1,2^2,...,2^n}$



Smallest number in sample space is $2$ and probability of $2$ being selected is $frac{1}{2}$. So, I think answer should be $frac{1}{2}$ but it's not in the option.






probability







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    We are given a set $X = {x_1,...,x_n}$, where $x_i = 2^i$. A sample $S subseteq X$ is drawn by selecting each $x_i$ independently with probability $p_i=frac{1}{2}$. The expected value of the smallest number in sample S is:
    (A) $frac{1}{n}$
    (B) $2$
    (C) $sqrt{n}$
    (D) $n$



    My approach:



    $X = {2^1,2^2,...,2^n}$



    Smallest number in sample space is $2$ and probability of $2$ being selected is $frac{1}{2}$. So, I think answer should be $frac{1}{2}$ but it's not in the option.






    probability







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      We are given a set $X = {x_1,...,x_n}$, where $x_i = 2^i$. A sample $S subseteq X$ is drawn by selecting each $x_i$ independently with probability $p_i=frac{1}{2}$. The expected value of the smallest number in sample S is:
      (A) $frac{1}{n}$
      (B) $2$
      (C) $sqrt{n}$
      (D) $n$



      My approach:



      $X = {2^1,2^2,...,2^n}$



      Smallest number in sample space is $2$ and probability of $2$ being selected is $frac{1}{2}$. So, I think answer should be $frac{1}{2}$ but it's not in the option.






      probability







      share|cite|improve this question











      $endgroup$




      We are given a set $X = {x_1,...,x_n}$, where $x_i = 2^i$. A sample $S subseteq X$ is drawn by selecting each $x_i$ independently with probability $p_i=frac{1}{2}$. The expected value of the smallest number in sample S is:
      (A) $frac{1}{n}$
      (B) $2$
      (C) $sqrt{n}$
      (D) $n$



      My approach:



      $X = {2^1,2^2,...,2^n}$



      Smallest number in sample space is $2$ and probability of $2$ being selected is $frac{1}{2}$. So, I think answer should be $frac{1}{2}$ but it's not in the option.






      probability




      probability






      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 10:59









      Jneven

      953322




      953322










      asked Feb 5 '15 at 21:28









      AtineshAtinesh

      4594822




      4594822






















          1 Answer
          1






          active

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          0












          $begingroup$

          The probability that the smallest number in the sample is $2^n$ is $frac{1}{2^n}$ (The $n-1$ numbers
          before $2^n$ must be non-selected and $2^n$ must be selected).



          The expected value is $sum_{i=1}^n x_ip_i$, where $p_i$ is the probability for the value $x_i$.



          Since every product is $1$ and we have $n$ of them, the expected value is $n$.



          So, answer $D$ is correct.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'm little bit confused. In question it is asked "Expected value of the smallest number in sample S" and you are calculating the probability of the smallest sample.
            $endgroup$
            – Atinesh
            Feb 6 '15 at 10:05










          • $begingroup$
            The probability that the smallest number is $2^{n-1}$ is still $frac{1}{2^{n}}$ the $n-2$ numbers before$ 2^{n-1}$ must be non selected and that $2^{n-1}$ and $2^{n}$ could be selected to create the sample giving you the probability $frac{1}{2^{n}}$. In which case the product is not going to be 1 always to get the final answer n. Am I mistaken?
            $endgroup$
            – Satish Ramanathan
            Dec 25 '18 at 13:40












          • $begingroup$
            If $2^{n-1}$ is selected, $2^n$ is not the smallest number in the sample. I edited the answer.
            $endgroup$
            – Peter
            Dec 25 '18 at 13:46














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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          The probability that the smallest number in the sample is $2^n$ is $frac{1}{2^n}$ (The $n-1$ numbers
          before $2^n$ must be non-selected and $2^n$ must be selected).



          The expected value is $sum_{i=1}^n x_ip_i$, where $p_i$ is the probability for the value $x_i$.



          Since every product is $1$ and we have $n$ of them, the expected value is $n$.



          So, answer $D$ is correct.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'm little bit confused. In question it is asked "Expected value of the smallest number in sample S" and you are calculating the probability of the smallest sample.
            $endgroup$
            – Atinesh
            Feb 6 '15 at 10:05










          • $begingroup$
            The probability that the smallest number is $2^{n-1}$ is still $frac{1}{2^{n}}$ the $n-2$ numbers before$ 2^{n-1}$ must be non selected and that $2^{n-1}$ and $2^{n}$ could be selected to create the sample giving you the probability $frac{1}{2^{n}}$. In which case the product is not going to be 1 always to get the final answer n. Am I mistaken?
            $endgroup$
            – Satish Ramanathan
            Dec 25 '18 at 13:40












          • $begingroup$
            If $2^{n-1}$ is selected, $2^n$ is not the smallest number in the sample. I edited the answer.
            $endgroup$
            – Peter
            Dec 25 '18 at 13:46


















          0












          $begingroup$

          The probability that the smallest number in the sample is $2^n$ is $frac{1}{2^n}$ (The $n-1$ numbers
          before $2^n$ must be non-selected and $2^n$ must be selected).



          The expected value is $sum_{i=1}^n x_ip_i$, where $p_i$ is the probability for the value $x_i$.



          Since every product is $1$ and we have $n$ of them, the expected value is $n$.



          So, answer $D$ is correct.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'm little bit confused. In question it is asked "Expected value of the smallest number in sample S" and you are calculating the probability of the smallest sample.
            $endgroup$
            – Atinesh
            Feb 6 '15 at 10:05










          • $begingroup$
            The probability that the smallest number is $2^{n-1}$ is still $frac{1}{2^{n}}$ the $n-2$ numbers before$ 2^{n-1}$ must be non selected and that $2^{n-1}$ and $2^{n}$ could be selected to create the sample giving you the probability $frac{1}{2^{n}}$. In which case the product is not going to be 1 always to get the final answer n. Am I mistaken?
            $endgroup$
            – Satish Ramanathan
            Dec 25 '18 at 13:40












          • $begingroup$
            If $2^{n-1}$ is selected, $2^n$ is not the smallest number in the sample. I edited the answer.
            $endgroup$
            – Peter
            Dec 25 '18 at 13:46
















          0












          0








          0





          $begingroup$

          The probability that the smallest number in the sample is $2^n$ is $frac{1}{2^n}$ (The $n-1$ numbers
          before $2^n$ must be non-selected and $2^n$ must be selected).



          The expected value is $sum_{i=1}^n x_ip_i$, where $p_i$ is the probability for the value $x_i$.



          Since every product is $1$ and we have $n$ of them, the expected value is $n$.



          So, answer $D$ is correct.






          share|cite|improve this answer











          $endgroup$



          The probability that the smallest number in the sample is $2^n$ is $frac{1}{2^n}$ (The $n-1$ numbers
          before $2^n$ must be non-selected and $2^n$ must be selected).



          The expected value is $sum_{i=1}^n x_ip_i$, where $p_i$ is the probability for the value $x_i$.



          Since every product is $1$ and we have $n$ of them, the expected value is $n$.



          So, answer $D$ is correct.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 25 '18 at 13:47

























          answered Feb 5 '15 at 21:36









          PeterPeter

          49.1k1240138




          49.1k1240138












          • $begingroup$
            I'm little bit confused. In question it is asked "Expected value of the smallest number in sample S" and you are calculating the probability of the smallest sample.
            $endgroup$
            – Atinesh
            Feb 6 '15 at 10:05










          • $begingroup$
            The probability that the smallest number is $2^{n-1}$ is still $frac{1}{2^{n}}$ the $n-2$ numbers before$ 2^{n-1}$ must be non selected and that $2^{n-1}$ and $2^{n}$ could be selected to create the sample giving you the probability $frac{1}{2^{n}}$. In which case the product is not going to be 1 always to get the final answer n. Am I mistaken?
            $endgroup$
            – Satish Ramanathan
            Dec 25 '18 at 13:40












          • $begingroup$
            If $2^{n-1}$ is selected, $2^n$ is not the smallest number in the sample. I edited the answer.
            $endgroup$
            – Peter
            Dec 25 '18 at 13:46




















          • $begingroup$
            I'm little bit confused. In question it is asked "Expected value of the smallest number in sample S" and you are calculating the probability of the smallest sample.
            $endgroup$
            – Atinesh
            Feb 6 '15 at 10:05










          • $begingroup$
            The probability that the smallest number is $2^{n-1}$ is still $frac{1}{2^{n}}$ the $n-2$ numbers before$ 2^{n-1}$ must be non selected and that $2^{n-1}$ and $2^{n}$ could be selected to create the sample giving you the probability $frac{1}{2^{n}}$. In which case the product is not going to be 1 always to get the final answer n. Am I mistaken?
            $endgroup$
            – Satish Ramanathan
            Dec 25 '18 at 13:40












          • $begingroup$
            If $2^{n-1}$ is selected, $2^n$ is not the smallest number in the sample. I edited the answer.
            $endgroup$
            – Peter
            Dec 25 '18 at 13:46


















          $begingroup$
          I'm little bit confused. In question it is asked "Expected value of the smallest number in sample S" and you are calculating the probability of the smallest sample.
          $endgroup$
          – Atinesh
          Feb 6 '15 at 10:05




          $begingroup$
          I'm little bit confused. In question it is asked "Expected value of the smallest number in sample S" and you are calculating the probability of the smallest sample.
          $endgroup$
          – Atinesh
          Feb 6 '15 at 10:05












          $begingroup$
          The probability that the smallest number is $2^{n-1}$ is still $frac{1}{2^{n}}$ the $n-2$ numbers before$ 2^{n-1}$ must be non selected and that $2^{n-1}$ and $2^{n}$ could be selected to create the sample giving you the probability $frac{1}{2^{n}}$. In which case the product is not going to be 1 always to get the final answer n. Am I mistaken?
          $endgroup$
          – Satish Ramanathan
          Dec 25 '18 at 13:40






          $begingroup$
          The probability that the smallest number is $2^{n-1}$ is still $frac{1}{2^{n}}$ the $n-2$ numbers before$ 2^{n-1}$ must be non selected and that $2^{n-1}$ and $2^{n}$ could be selected to create the sample giving you the probability $frac{1}{2^{n}}$. In which case the product is not going to be 1 always to get the final answer n. Am I mistaken?
          $endgroup$
          – Satish Ramanathan
          Dec 25 '18 at 13:40














          $begingroup$
          If $2^{n-1}$ is selected, $2^n$ is not the smallest number in the sample. I edited the answer.
          $endgroup$
          – Peter
          Dec 25 '18 at 13:46






          $begingroup$
          If $2^{n-1}$ is selected, $2^n$ is not the smallest number in the sample. I edited the answer.
          $endgroup$
          – Peter
          Dec 25 '18 at 13:46




















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