Continuous Random Variables question












1












$begingroup$



The continuous random variable $X$ has p.d.f. $$f(x) =begin{cases}
frac{2(7-x)}{25},&text{for $2leq xleq 7$}\ 0,& text{otherwise}
end{cases} $$



The function $g(X)$ is defined by $g(x) = x^2 + x$



Find the value of $E[g(X)]$.




Is the correct answer to this question $37/2$ or $397/6$? I disagree with the answer the textbook has given.



I believe the answer to be $37/2$, this is for 2 main reasons: Firstly, if you compute the integral of $(x^2 + x) (14/25 - 2x/25)$ with limits $2$ and $7$ then you get $37/2$. Secondly, an easier way. $E[X]$ in this case is $11/3$, and $E[X^2]$ is $89/6$ add them together: $89/6 + 11/3 = 37/2$. However, the textbook says the answer is $397/6$, and then follows up the question using this value.






probability random-variables expected-value







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can you explain why do you disagree?
    $endgroup$
    – Thomas Shelby
    Dec 2 '18 at 2:06






  • 1




    $begingroup$
    I believe the answer to be 37/2, this is for 2 main reasons: Firstly, if you compute the integral of (x^2 + x) (14/25 - 2x/25) with limits 2 and 7 then you get 37/2 Secondly, an easier way. E(X) in this case is 11/3, and E(X^2) is 89/6 add them together: 89/6 + 11/3 = 37/2 This is why I think the answer is 37/2. However, the textbook says the answer is 397/6, and then follows up the question using this.
    $endgroup$
    – user546944
    Dec 2 '18 at 2:11










  • $begingroup$
    My calculation also yielded $18.5$. The textbook answer may be wrong.
    $endgroup$
    – Thomas Shelby
    Dec 2 '18 at 2:32










  • $begingroup$
    Your definition of $f(x)$ uses $gt$ not =. Why?
    $endgroup$
    – herb steinberg
    Dec 9 '18 at 3:42
















1












$begingroup$



The continuous random variable $X$ has p.d.f. $$f(x) =begin{cases}
frac{2(7-x)}{25},&text{for $2leq xleq 7$}\ 0,& text{otherwise}
end{cases} $$



The function $g(X)$ is defined by $g(x) = x^2 + x$



Find the value of $E[g(X)]$.




Is the correct answer to this question $37/2$ or $397/6$? I disagree with the answer the textbook has given.



I believe the answer to be $37/2$, this is for 2 main reasons: Firstly, if you compute the integral of $(x^2 + x) (14/25 - 2x/25)$ with limits $2$ and $7$ then you get $37/2$. Secondly, an easier way. $E[X]$ in this case is $11/3$, and $E[X^2]$ is $89/6$ add them together: $89/6 + 11/3 = 37/2$. However, the textbook says the answer is $397/6$, and then follows up the question using this value.






probability random-variables expected-value







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can you explain why do you disagree?
    $endgroup$
    – Thomas Shelby
    Dec 2 '18 at 2:06






  • 1




    $begingroup$
    I believe the answer to be 37/2, this is for 2 main reasons: Firstly, if you compute the integral of (x^2 + x) (14/25 - 2x/25) with limits 2 and 7 then you get 37/2 Secondly, an easier way. E(X) in this case is 11/3, and E(X^2) is 89/6 add them together: 89/6 + 11/3 = 37/2 This is why I think the answer is 37/2. However, the textbook says the answer is 397/6, and then follows up the question using this.
    $endgroup$
    – user546944
    Dec 2 '18 at 2:11










  • $begingroup$
    My calculation also yielded $18.5$. The textbook answer may be wrong.
    $endgroup$
    – Thomas Shelby
    Dec 2 '18 at 2:32










  • $begingroup$
    Your definition of $f(x)$ uses $gt$ not =. Why?
    $endgroup$
    – herb steinberg
    Dec 9 '18 at 3:42














1












1








1





$begingroup$



The continuous random variable $X$ has p.d.f. $$f(x) =begin{cases}
frac{2(7-x)}{25},&text{for $2leq xleq 7$}\ 0,& text{otherwise}
end{cases} $$



The function $g(X)$ is defined by $g(x) = x^2 + x$



Find the value of $E[g(X)]$.




Is the correct answer to this question $37/2$ or $397/6$? I disagree with the answer the textbook has given.



I believe the answer to be $37/2$, this is for 2 main reasons: Firstly, if you compute the integral of $(x^2 + x) (14/25 - 2x/25)$ with limits $2$ and $7$ then you get $37/2$. Secondly, an easier way. $E[X]$ in this case is $11/3$, and $E[X^2]$ is $89/6$ add them together: $89/6 + 11/3 = 37/2$. However, the textbook says the answer is $397/6$, and then follows up the question using this value.






probability random-variables expected-value







share|cite|improve this question











$endgroup$





The continuous random variable $X$ has p.d.f. $$f(x) =begin{cases}
frac{2(7-x)}{25},&text{for $2leq xleq 7$}\ 0,& text{otherwise}
end{cases} $$



The function $g(X)$ is defined by $g(x) = x^2 + x$



Find the value of $E[g(X)]$.




Is the correct answer to this question $37/2$ or $397/6$? I disagree with the answer the textbook has given.



I believe the answer to be $37/2$, this is for 2 main reasons: Firstly, if you compute the integral of $(x^2 + x) (14/25 - 2x/25)$ with limits $2$ and $7$ then you get $37/2$. Secondly, an easier way. $E[X]$ in this case is $11/3$, and $E[X^2]$ is $89/6$ add them together: $89/6 + 11/3 = 37/2$. However, the textbook says the answer is $397/6$, and then follows up the question using this value.






probability random-variables expected-value




probability random-variables expected-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 4:10









André 3000

12.6k22243




12.6k22243










asked Dec 2 '18 at 2:05







user546944



















  • $begingroup$
    Can you explain why do you disagree?
    $endgroup$
    – Thomas Shelby
    Dec 2 '18 at 2:06






  • 1




    $begingroup$
    I believe the answer to be 37/2, this is for 2 main reasons: Firstly, if you compute the integral of (x^2 + x) (14/25 - 2x/25) with limits 2 and 7 then you get 37/2 Secondly, an easier way. E(X) in this case is 11/3, and E(X^2) is 89/6 add them together: 89/6 + 11/3 = 37/2 This is why I think the answer is 37/2. However, the textbook says the answer is 397/6, and then follows up the question using this.
    $endgroup$
    – user546944
    Dec 2 '18 at 2:11










  • $begingroup$
    My calculation also yielded $18.5$. The textbook answer may be wrong.
    $endgroup$
    – Thomas Shelby
    Dec 2 '18 at 2:32










  • $begingroup$
    Your definition of $f(x)$ uses $gt$ not =. Why?
    $endgroup$
    – herb steinberg
    Dec 9 '18 at 3:42


















  • $begingroup$
    Can you explain why do you disagree?
    $endgroup$
    – Thomas Shelby
    Dec 2 '18 at 2:06






  • 1




    $begingroup$
    I believe the answer to be 37/2, this is for 2 main reasons: Firstly, if you compute the integral of (x^2 + x) (14/25 - 2x/25) with limits 2 and 7 then you get 37/2 Secondly, an easier way. E(X) in this case is 11/3, and E(X^2) is 89/6 add them together: 89/6 + 11/3 = 37/2 This is why I think the answer is 37/2. However, the textbook says the answer is 397/6, and then follows up the question using this.
    $endgroup$
    – user546944
    Dec 2 '18 at 2:11










  • $begingroup$
    My calculation also yielded $18.5$. The textbook answer may be wrong.
    $endgroup$
    – Thomas Shelby
    Dec 2 '18 at 2:32










  • $begingroup$
    Your definition of $f(x)$ uses $gt$ not =. Why?
    $endgroup$
    – herb steinberg
    Dec 9 '18 at 3:42
















$begingroup$
Can you explain why do you disagree?
$endgroup$
– Thomas Shelby
Dec 2 '18 at 2:06




$begingroup$
Can you explain why do you disagree?
$endgroup$
– Thomas Shelby
Dec 2 '18 at 2:06




1




1




$begingroup$
I believe the answer to be 37/2, this is for 2 main reasons: Firstly, if you compute the integral of (x^2 + x) (14/25 - 2x/25) with limits 2 and 7 then you get 37/2 Secondly, an easier way. E(X) in this case is 11/3, and E(X^2) is 89/6 add them together: 89/6 + 11/3 = 37/2 This is why I think the answer is 37/2. However, the textbook says the answer is 397/6, and then follows up the question using this.
$endgroup$
– user546944
Dec 2 '18 at 2:11




$begingroup$
I believe the answer to be 37/2, this is for 2 main reasons: Firstly, if you compute the integral of (x^2 + x) (14/25 - 2x/25) with limits 2 and 7 then you get 37/2 Secondly, an easier way. E(X) in this case is 11/3, and E(X^2) is 89/6 add them together: 89/6 + 11/3 = 37/2 This is why I think the answer is 37/2. However, the textbook says the answer is 397/6, and then follows up the question using this.
$endgroup$
– user546944
Dec 2 '18 at 2:11












$begingroup$
My calculation also yielded $18.5$. The textbook answer may be wrong.
$endgroup$
– Thomas Shelby
Dec 2 '18 at 2:32




$begingroup$
My calculation also yielded $18.5$. The textbook answer may be wrong.
$endgroup$
– Thomas Shelby
Dec 2 '18 at 2:32












$begingroup$
Your definition of $f(x)$ uses $gt$ not =. Why?
$endgroup$
– herb steinberg
Dec 9 '18 at 3:42




$begingroup$
Your definition of $f(x)$ uses $gt$ not =. Why?
$endgroup$
– herb steinberg
Dec 9 '18 at 3:42










1 Answer
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$begingroup$

In addition to your calculations, which appear to be correct:



With probability $1$ we have that $2 < x leq 7.$
Within the interval $2 < x leq 7,$
the function $g(x) = x^2 + x$ has its maximum at $x = 7.$
With probability $1,$ therefore, $g(x) leq g(7) = 56.$
This implies that $E[g(x)] leq 56.$



Observing that $frac{397}{6} > 66,$ we know that $E[g(x)] neq frac{397}{6}.$






share|cite|improve this answer









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    1 Answer
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    active

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    1 Answer
    1






    active

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    active

    oldest

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    2












    $begingroup$

    In addition to your calculations, which appear to be correct:



    With probability $1$ we have that $2 < x leq 7.$
    Within the interval $2 < x leq 7,$
    the function $g(x) = x^2 + x$ has its maximum at $x = 7.$
    With probability $1,$ therefore, $g(x) leq g(7) = 56.$
    This implies that $E[g(x)] leq 56.$



    Observing that $frac{397}{6} > 66,$ we know that $E[g(x)] neq frac{397}{6}.$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      In addition to your calculations, which appear to be correct:



      With probability $1$ we have that $2 < x leq 7.$
      Within the interval $2 < x leq 7,$
      the function $g(x) = x^2 + x$ has its maximum at $x = 7.$
      With probability $1,$ therefore, $g(x) leq g(7) = 56.$
      This implies that $E[g(x)] leq 56.$



      Observing that $frac{397}{6} > 66,$ we know that $E[g(x)] neq frac{397}{6}.$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        In addition to your calculations, which appear to be correct:



        With probability $1$ we have that $2 < x leq 7.$
        Within the interval $2 < x leq 7,$
        the function $g(x) = x^2 + x$ has its maximum at $x = 7.$
        With probability $1,$ therefore, $g(x) leq g(7) = 56.$
        This implies that $E[g(x)] leq 56.$



        Observing that $frac{397}{6} > 66,$ we know that $E[g(x)] neq frac{397}{6}.$






        share|cite|improve this answer









        $endgroup$



        In addition to your calculations, which appear to be correct:



        With probability $1$ we have that $2 < x leq 7.$
        Within the interval $2 < x leq 7,$
        the function $g(x) = x^2 + x$ has its maximum at $x = 7.$
        With probability $1,$ therefore, $g(x) leq g(7) = 56.$
        This implies that $E[g(x)] leq 56.$



        Observing that $frac{397}{6} > 66,$ we know that $E[g(x)] neq frac{397}{6}.$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 2:47









        David KDavid K

        54.4k343118




        54.4k343118






























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