Compute $ E[|X^2-16|] $ where $ X sim U $












0












$begingroup$


$ X sim U (-4,7) $ with $U$ a continuous uniform distribution.



X is continuous variable. Compute $ E[|X^2 - 16|] $



So: $$ E[|X^2 - 16|] = begin{cases} E[X^2 - 16] , 4< X \ E[-X^2+16], X < 4 end{cases} \ $$ My attempt is to calculate each case and then sum them. $$E[X^2 - 16] = sum_{i=4}^7 (X_i^2-16) cdot frac{1}{11} = frac{1}{11} cdot sum_{j=1}^4 (j+3)^2-16 = frac{1}{11} cdot sum_{j=1}^4 j^2 +6j - 7\ = -frac{28}{11}cdot (sum_{j=1}^4 j^2 + 6cdotsum_{j=1}^4 j) = - frac{28}{11} ( frac{4 cdot 5 cdot 9}{6} + 6 cdot frac{4 cdot 5}{2} )$$



But I'm not sure if that's the correct way to compute?






probability-theory uniform-distribution







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    0












    $begingroup$


    $ X sim U (-4,7) $ with $U$ a continuous uniform distribution.



    X is continuous variable. Compute $ E[|X^2 - 16|] $



    So: $$ E[|X^2 - 16|] = begin{cases} E[X^2 - 16] , 4< X \ E[-X^2+16], X < 4 end{cases} \ $$ My attempt is to calculate each case and then sum them. $$E[X^2 - 16] = sum_{i=4}^7 (X_i^2-16) cdot frac{1}{11} = frac{1}{11} cdot sum_{j=1}^4 (j+3)^2-16 = frac{1}{11} cdot sum_{j=1}^4 j^2 +6j - 7\ = -frac{28}{11}cdot (sum_{j=1}^4 j^2 + 6cdotsum_{j=1}^4 j) = - frac{28}{11} ( frac{4 cdot 5 cdot 9}{6} + 6 cdot frac{4 cdot 5}{2} )$$



    But I'm not sure if that's the correct way to compute?






    probability-theory uniform-distribution







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      $ X sim U (-4,7) $ with $U$ a continuous uniform distribution.



      X is continuous variable. Compute $ E[|X^2 - 16|] $



      So: $$ E[|X^2 - 16|] = begin{cases} E[X^2 - 16] , 4< X \ E[-X^2+16], X < 4 end{cases} \ $$ My attempt is to calculate each case and then sum them. $$E[X^2 - 16] = sum_{i=4}^7 (X_i^2-16) cdot frac{1}{11} = frac{1}{11} cdot sum_{j=1}^4 (j+3)^2-16 = frac{1}{11} cdot sum_{j=1}^4 j^2 +6j - 7\ = -frac{28}{11}cdot (sum_{j=1}^4 j^2 + 6cdotsum_{j=1}^4 j) = - frac{28}{11} ( frac{4 cdot 5 cdot 9}{6} + 6 cdot frac{4 cdot 5}{2} )$$



      But I'm not sure if that's the correct way to compute?






      probability-theory uniform-distribution







      share|cite|improve this question











      $endgroup$




      $ X sim U (-4,7) $ with $U$ a continuous uniform distribution.



      X is continuous variable. Compute $ E[|X^2 - 16|] $



      So: $$ E[|X^2 - 16|] = begin{cases} E[X^2 - 16] , 4< X \ E[-X^2+16], X < 4 end{cases} \ $$ My attempt is to calculate each case and then sum them. $$E[X^2 - 16] = sum_{i=4}^7 (X_i^2-16) cdot frac{1}{11} = frac{1}{11} cdot sum_{j=1}^4 (j+3)^2-16 = frac{1}{11} cdot sum_{j=1}^4 j^2 +6j - 7\ = -frac{28}{11}cdot (sum_{j=1}^4 j^2 + 6cdotsum_{j=1}^4 j) = - frac{28}{11} ( frac{4 cdot 5 cdot 9}{6} + 6 cdot frac{4 cdot 5}{2} )$$



      But I'm not sure if that's the correct way to compute?






      probability-theory uniform-distribution




      probability-theory uniform-distribution






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      share|cite|improve this question













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      edited Dec 18 '18 at 17:13









      quid♦

      37.1k95093




      37.1k95093










      asked Dec 18 '18 at 8:23









      bm1125bm1125

      64916




      64916






















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          $begingroup$

          $X$ is a random variable and not a constant. You cannot split the computation into the cases $4<X$ and $X<4$. (The probabilities of these events will have an effect on the answer). Also, you treating a continuous random variable as a discrete one. The finite sums you are considering are not related to uniform distribution at all. The correct formula for $E|X^{2}-16|$ is $frac 1 {11} int_{-4}^{7} |x^{2}-16|, dx$. Now split the integral into integral form $-4$ to $4$ and $4$ to $7$.






          share|cite|improve this answer











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          • $begingroup$
            Thanks a lot for pointing that out for me!
            $endgroup$
            – bm1125
            Dec 18 '18 at 8:59











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          $begingroup$

          $X$ is a random variable and not a constant. You cannot split the computation into the cases $4<X$ and $X<4$. (The probabilities of these events will have an effect on the answer). Also, you treating a continuous random variable as a discrete one. The finite sums you are considering are not related to uniform distribution at all. The correct formula for $E|X^{2}-16|$ is $frac 1 {11} int_{-4}^{7} |x^{2}-16|, dx$. Now split the integral into integral form $-4$ to $4$ and $4$ to $7$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks a lot for pointing that out for me!
            $endgroup$
            – bm1125
            Dec 18 '18 at 8:59
















          3












          $begingroup$

          $X$ is a random variable and not a constant. You cannot split the computation into the cases $4<X$ and $X<4$. (The probabilities of these events will have an effect on the answer). Also, you treating a continuous random variable as a discrete one. The finite sums you are considering are not related to uniform distribution at all. The correct formula for $E|X^{2}-16|$ is $frac 1 {11} int_{-4}^{7} |x^{2}-16|, dx$. Now split the integral into integral form $-4$ to $4$ and $4$ to $7$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks a lot for pointing that out for me!
            $endgroup$
            – bm1125
            Dec 18 '18 at 8:59














          3












          3








          3





          $begingroup$

          $X$ is a random variable and not a constant. You cannot split the computation into the cases $4<X$ and $X<4$. (The probabilities of these events will have an effect on the answer). Also, you treating a continuous random variable as a discrete one. The finite sums you are considering are not related to uniform distribution at all. The correct formula for $E|X^{2}-16|$ is $frac 1 {11} int_{-4}^{7} |x^{2}-16|, dx$. Now split the integral into integral form $-4$ to $4$ and $4$ to $7$.






          share|cite|improve this answer











          $endgroup$



          $X$ is a random variable and not a constant. You cannot split the computation into the cases $4<X$ and $X<4$. (The probabilities of these events will have an effect on the answer). Also, you treating a continuous random variable as a discrete one. The finite sums you are considering are not related to uniform distribution at all. The correct formula for $E|X^{2}-16|$ is $frac 1 {11} int_{-4}^{7} |x^{2}-16|, dx$. Now split the integral into integral form $-4$ to $4$ and $4$ to $7$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 18 '18 at 8:40

























          answered Dec 18 '18 at 8:28









          Kavi Rama MurthyKavi Rama Murthy

          60k42161




          60k42161












          • $begingroup$
            Thanks a lot for pointing that out for me!
            $endgroup$
            – bm1125
            Dec 18 '18 at 8:59


















          • $begingroup$
            Thanks a lot for pointing that out for me!
            $endgroup$
            – bm1125
            Dec 18 '18 at 8:59
















          $begingroup$
          Thanks a lot for pointing that out for me!
          $endgroup$
          – bm1125
          Dec 18 '18 at 8:59




          $begingroup$
          Thanks a lot for pointing that out for me!
          $endgroup$
          – bm1125
          Dec 18 '18 at 8:59


















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