$E(X^2)$ of discrete uniform distributionHow is Efron–Stein inequality usefulFor $X_i$ iid $ Var( sum_i=1^n X_i )= sum_i=1^n Var (X_i)$?Convergence in distribution from PGFCalculating variance of marginal distributionConsistent estimator of the meanHard time factoring Normal Distribution based on transformation problem.Expectation and Variance of a RatioDeriving which estimator is preferred for success probability $p$?Counting pairs of $k$-cycles with $r$ common edges (Variance of number of $k$-cycles)Higher Moments of a Cauchy Distribution

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$E(X^2)$ of discrete uniform distribution


How is Efron–Stein inequality usefulFor $X_i$ iid $ Var( sum_i=1^n X_i )= sum_i=1^n Var (X_i)$?Convergence in distribution from PGFCalculating variance of marginal distributionConsistent estimator of the meanHard time factoring Normal Distribution based on transformation problem.Expectation and Variance of a RatioDeriving which estimator is preferred for success probability $p$?Counting pairs of $k$-cycles with $r$ common edges (Variance of number of $k$-cycles)Higher Moments of a Cauchy Distribution













1












$begingroup$


I have a discrete uniform distribution (from IFoA formulae) with parameters a, b, h, where a and b are the start/end points, and h is the interval between each value. The p.d.f. is given as $frachb-a+h$. I manage to find out $$E[X] = sum_xin X xf(x) = sum_xin X fracxhb-a+h = frachb-a+hsum_xin X x\= frachb-a+htimes fraca+b2times fracb-a+hh = fraca+b2.$$ However, I cannot figure out the second moment ($E[X^2]$) and Variance by using the same method. I wonder if it is possible to find them using the same way, I know they could be found by using the M.G.F., I just want to know if it is possible.



$Var(X) = frac(b-a)(b-a+2h)12$ is given by the formulae.



Here is what I have got so far:
$$sum_xin X x^2f(x) = sum_xin X fracx^2 hb-a+h = frachb-a+hsum_xin X x^2=frachb-a+hsum_k=0^fracb-a+hh(hk+a)^2$$



edit:
beginalign*
E[X^2] &= sum_xin X x^2f(x)\
&=frach^3b-a+hsum_k=0^fracb-a+hhk^2+frac2ah^2b-a+hsum_k=0^fracb-a+hhk+fraca^2hb-a+hsum_k=0^fracb-a+hh1\
&=frach^3b-a+hfrac(fracb-ah)(fracb-ah+1)(2(fracb-ah)+1)6+frac2ah^2b-a+hfrac(fracb-ah+1)(fracb-ah)2+fraca^2hb-a+hbigg(fracb-ah+1bigg)\
&= frach^3b-a+hfrac(fracb-ah)(fracb-a+hh)(frac2(b-a)+hh)6+frac2ah^2b-a+hfrac(fracb-a+hh)(fracb-ah)2+fraca^2hb-a+hbigg(fracb-a+hhbigg)\
&= frac(b-a)(2(b-a)+h)6+a(b-a)+a^2\
Var(X) &= E[X^2] - E[X]^2\
&= frac(b-a)(2(b-a)+h)6+a(b-a)+a^2 - bigg(fraca+b2bigg)^2\
&= frac2(b-a)(2(b-a)+h)+12a(b-a)+12a^212 - frac3(a+b)^212\
&= frac4b^2-4ab+2bh-4ab+4a^2-2ah+12ab-12a^2+12a^2-3a^2-6ab-3b^212\
&= frac(a-b)^2+2h(b-a)12 \
&= frac(b-a)^2+2h(b-a)12\
&= frac(b-a)(b-a+2h)12
endalign*






probability uniform-distribution







share|cite|improve this question











$endgroup$











  • $begingroup$
    Would it help if you knew that $sum_x=0^n x^2 = fracn(n+1)(2n+1)6$?
    $endgroup$
    – Paul
    Oct 2 '18 at 14:33










  • $begingroup$
    It would be helpful for $h = 1$, but I cannot imply it for $hneq1$.
    $endgroup$
    – VincentN
    Oct 2 '18 at 14:36











  • $begingroup$
    You can apply it for $x'=x/h$.
    $endgroup$
    – Paul
    Oct 2 '18 at 14:38










  • $begingroup$
    I don't quite get what you mean by that.
    $endgroup$
    – VincentN
    Oct 2 '18 at 14:53










  • $begingroup$
    I'll put it in an answer.
    $endgroup$
    – Paul
    Oct 2 '18 at 14:57















1












$begingroup$


I have a discrete uniform distribution (from IFoA formulae) with parameters a, b, h, where a and b are the start/end points, and h is the interval between each value. The p.d.f. is given as $frachb-a+h$. I manage to find out $$E[X] = sum_xin X xf(x) = sum_xin X fracxhb-a+h = frachb-a+hsum_xin X x\= frachb-a+htimes fraca+b2times fracb-a+hh = fraca+b2.$$ However, I cannot figure out the second moment ($E[X^2]$) and Variance by using the same method. I wonder if it is possible to find them using the same way, I know they could be found by using the M.G.F., I just want to know if it is possible.



$Var(X) = frac(b-a)(b-a+2h)12$ is given by the formulae.



Here is what I have got so far:
$$sum_xin X x^2f(x) = sum_xin X fracx^2 hb-a+h = frachb-a+hsum_xin X x^2=frachb-a+hsum_k=0^fracb-a+hh(hk+a)^2$$



edit:
beginalign*
E[X^2] &= sum_xin X x^2f(x)\
&=frach^3b-a+hsum_k=0^fracb-a+hhk^2+frac2ah^2b-a+hsum_k=0^fracb-a+hhk+fraca^2hb-a+hsum_k=0^fracb-a+hh1\
&=frach^3b-a+hfrac(fracb-ah)(fracb-ah+1)(2(fracb-ah)+1)6+frac2ah^2b-a+hfrac(fracb-ah+1)(fracb-ah)2+fraca^2hb-a+hbigg(fracb-ah+1bigg)\
&= frach^3b-a+hfrac(fracb-ah)(fracb-a+hh)(frac2(b-a)+hh)6+frac2ah^2b-a+hfrac(fracb-a+hh)(fracb-ah)2+fraca^2hb-a+hbigg(fracb-a+hhbigg)\
&= frac(b-a)(2(b-a)+h)6+a(b-a)+a^2\
Var(X) &= E[X^2] - E[X]^2\
&= frac(b-a)(2(b-a)+h)6+a(b-a)+a^2 - bigg(fraca+b2bigg)^2\
&= frac2(b-a)(2(b-a)+h)+12a(b-a)+12a^212 - frac3(a+b)^212\
&= frac4b^2-4ab+2bh-4ab+4a^2-2ah+12ab-12a^2+12a^2-3a^2-6ab-3b^212\
&= frac(a-b)^2+2h(b-a)12 \
&= frac(b-a)^2+2h(b-a)12\
&= frac(b-a)(b-a+2h)12
endalign*






probability uniform-distribution







share|cite|improve this question











$endgroup$











  • $begingroup$
    Would it help if you knew that $sum_x=0^n x^2 = fracn(n+1)(2n+1)6$?
    $endgroup$
    – Paul
    Oct 2 '18 at 14:33










  • $begingroup$
    It would be helpful for $h = 1$, but I cannot imply it for $hneq1$.
    $endgroup$
    – VincentN
    Oct 2 '18 at 14:36











  • $begingroup$
    You can apply it for $x'=x/h$.
    $endgroup$
    – Paul
    Oct 2 '18 at 14:38










  • $begingroup$
    I don't quite get what you mean by that.
    $endgroup$
    – VincentN
    Oct 2 '18 at 14:53










  • $begingroup$
    I'll put it in an answer.
    $endgroup$
    – Paul
    Oct 2 '18 at 14:57













1












1








1





$begingroup$


I have a discrete uniform distribution (from IFoA formulae) with parameters a, b, h, where a and b are the start/end points, and h is the interval between each value. The p.d.f. is given as $frachb-a+h$. I manage to find out $$E[X] = sum_xin X xf(x) = sum_xin X fracxhb-a+h = frachb-a+hsum_xin X x\= frachb-a+htimes fraca+b2times fracb-a+hh = fraca+b2.$$ However, I cannot figure out the second moment ($E[X^2]$) and Variance by using the same method. I wonder if it is possible to find them using the same way, I know they could be found by using the M.G.F., I just want to know if it is possible.



$Var(X) = frac(b-a)(b-a+2h)12$ is given by the formulae.



Here is what I have got so far:
$$sum_xin X x^2f(x) = sum_xin X fracx^2 hb-a+h = frachb-a+hsum_xin X x^2=frachb-a+hsum_k=0^fracb-a+hh(hk+a)^2$$



edit:
beginalign*
E[X^2] &= sum_xin X x^2f(x)\
&=frach^3b-a+hsum_k=0^fracb-a+hhk^2+frac2ah^2b-a+hsum_k=0^fracb-a+hhk+fraca^2hb-a+hsum_k=0^fracb-a+hh1\
&=frach^3b-a+hfrac(fracb-ah)(fracb-ah+1)(2(fracb-ah)+1)6+frac2ah^2b-a+hfrac(fracb-ah+1)(fracb-ah)2+fraca^2hb-a+hbigg(fracb-ah+1bigg)\
&= frach^3b-a+hfrac(fracb-ah)(fracb-a+hh)(frac2(b-a)+hh)6+frac2ah^2b-a+hfrac(fracb-a+hh)(fracb-ah)2+fraca^2hb-a+hbigg(fracb-a+hhbigg)\
&= frac(b-a)(2(b-a)+h)6+a(b-a)+a^2\
Var(X) &= E[X^2] - E[X]^2\
&= frac(b-a)(2(b-a)+h)6+a(b-a)+a^2 - bigg(fraca+b2bigg)^2\
&= frac2(b-a)(2(b-a)+h)+12a(b-a)+12a^212 - frac3(a+b)^212\
&= frac4b^2-4ab+2bh-4ab+4a^2-2ah+12ab-12a^2+12a^2-3a^2-6ab-3b^212\
&= frac(a-b)^2+2h(b-a)12 \
&= frac(b-a)^2+2h(b-a)12\
&= frac(b-a)(b-a+2h)12
endalign*






probability uniform-distribution







share|cite|improve this question











$endgroup$




I have a discrete uniform distribution (from IFoA formulae) with parameters a, b, h, where a and b are the start/end points, and h is the interval between each value. The p.d.f. is given as $frachb-a+h$. I manage to find out $$E[X] = sum_xin X xf(x) = sum_xin X fracxhb-a+h = frachb-a+hsum_xin X x\= frachb-a+htimes fraca+b2times fracb-a+hh = fraca+b2.$$ However, I cannot figure out the second moment ($E[X^2]$) and Variance by using the same method. I wonder if it is possible to find them using the same way, I know they could be found by using the M.G.F., I just want to know if it is possible.



$Var(X) = frac(b-a)(b-a+2h)12$ is given by the formulae.



Here is what I have got so far:
$$sum_xin X x^2f(x) = sum_xin X fracx^2 hb-a+h = frachb-a+hsum_xin X x^2=frachb-a+hsum_k=0^fracb-a+hh(hk+a)^2$$



edit:
beginalign*
E[X^2] &= sum_xin X x^2f(x)\
&=frach^3b-a+hsum_k=0^fracb-a+hhk^2+frac2ah^2b-a+hsum_k=0^fracb-a+hhk+fraca^2hb-a+hsum_k=0^fracb-a+hh1\
&=frach^3b-a+hfrac(fracb-ah)(fracb-ah+1)(2(fracb-ah)+1)6+frac2ah^2b-a+hfrac(fracb-ah+1)(fracb-ah)2+fraca^2hb-a+hbigg(fracb-ah+1bigg)\
&= frach^3b-a+hfrac(fracb-ah)(fracb-a+hh)(frac2(b-a)+hh)6+frac2ah^2b-a+hfrac(fracb-a+hh)(fracb-ah)2+fraca^2hb-a+hbigg(fracb-a+hhbigg)\
&= frac(b-a)(2(b-a)+h)6+a(b-a)+a^2\
Var(X) &= E[X^2] - E[X]^2\
&= frac(b-a)(2(b-a)+h)6+a(b-a)+a^2 - bigg(fraca+b2bigg)^2\
&= frac2(b-a)(2(b-a)+h)+12a(b-a)+12a^212 - frac3(a+b)^212\
&= frac4b^2-4ab+2bh-4ab+4a^2-2ah+12ab-12a^2+12a^2-3a^2-6ab-3b^212\
&= frac(a-b)^2+2h(b-a)12 \
&= frac(b-a)^2+2h(b-a)12\
&= frac(b-a)(b-a+2h)12
endalign*






probability uniform-distribution




probability uniform-distribution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 15 at 19:15







VincentN

















asked Oct 2 '18 at 14:26









VincentNVincentN

155




155











  • $begingroup$
    Would it help if you knew that $sum_x=0^n x^2 = fracn(n+1)(2n+1)6$?
    $endgroup$
    – Paul
    Oct 2 '18 at 14:33










  • $begingroup$
    It would be helpful for $h = 1$, but I cannot imply it for $hneq1$.
    $endgroup$
    – VincentN
    Oct 2 '18 at 14:36











  • $begingroup$
    You can apply it for $x'=x/h$.
    $endgroup$
    – Paul
    Oct 2 '18 at 14:38










  • $begingroup$
    I don't quite get what you mean by that.
    $endgroup$
    – VincentN
    Oct 2 '18 at 14:53










  • $begingroup$
    I'll put it in an answer.
    $endgroup$
    – Paul
    Oct 2 '18 at 14:57
















  • $begingroup$
    Would it help if you knew that $sum_x=0^n x^2 = fracn(n+1)(2n+1)6$?
    $endgroup$
    – Paul
    Oct 2 '18 at 14:33










  • $begingroup$
    It would be helpful for $h = 1$, but I cannot imply it for $hneq1$.
    $endgroup$
    – VincentN
    Oct 2 '18 at 14:36











  • $begingroup$
    You can apply it for $x'=x/h$.
    $endgroup$
    – Paul
    Oct 2 '18 at 14:38










  • $begingroup$
    I don't quite get what you mean by that.
    $endgroup$
    – VincentN
    Oct 2 '18 at 14:53










  • $begingroup$
    I'll put it in an answer.
    $endgroup$
    – Paul
    Oct 2 '18 at 14:57















$begingroup$
Would it help if you knew that $sum_x=0^n x^2 = fracn(n+1)(2n+1)6$?
$endgroup$
– Paul
Oct 2 '18 at 14:33




$begingroup$
Would it help if you knew that $sum_x=0^n x^2 = fracn(n+1)(2n+1)6$?
$endgroup$
– Paul
Oct 2 '18 at 14:33












$begingroup$
It would be helpful for $h = 1$, but I cannot imply it for $hneq1$.
$endgroup$
– VincentN
Oct 2 '18 at 14:36





$begingroup$
It would be helpful for $h = 1$, but I cannot imply it for $hneq1$.
$endgroup$
– VincentN
Oct 2 '18 at 14:36













$begingroup$
You can apply it for $x'=x/h$.
$endgroup$
– Paul
Oct 2 '18 at 14:38




$begingroup$
You can apply it for $x'=x/h$.
$endgroup$
– Paul
Oct 2 '18 at 14:38












$begingroup$
I don't quite get what you mean by that.
$endgroup$
– VincentN
Oct 2 '18 at 14:53




$begingroup$
I don't quite get what you mean by that.
$endgroup$
– VincentN
Oct 2 '18 at 14:53












$begingroup$
I'll put it in an answer.
$endgroup$
– Paul
Oct 2 '18 at 14:57




$begingroup$
I'll put it in an answer.
$endgroup$
– Paul
Oct 2 '18 at 14:57










1 Answer
1






active

oldest

votes


















0












$begingroup$

With your edit, you are almost there. You can simplify
$$frachb-a+hsum_k=0^b-a(hk+a)^2$$
$$=frachb-a+hsum_k=0^b-aleft((hk)^2+2ahk+a^2 right)$$
$$=frachb-a+hsum_k=0^b-a(hk)^2+frachb-a+hsum_k=0^b-a(2ahk)+frachb-a+hsum_k=0^b-aa^2$$
$$=frach^3b-a+hsum_k=0^b-ak^2+frac2ah^2b-a+hsum_k=0^b-ak+fraca^2hb-a+hsum_k=0^b-a1$$
For the first term you can apply
$$sum_x=0^n x^2 = fracn(n+1)(2n+1)6$$
and apply the similar formulas that you did previously for the other 2 terms.






share|cite|improve this answer









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    0












    $begingroup$

    With your edit, you are almost there. You can simplify
    $$frachb-a+hsum_k=0^b-a(hk+a)^2$$
    $$=frachb-a+hsum_k=0^b-aleft((hk)^2+2ahk+a^2 right)$$
    $$=frachb-a+hsum_k=0^b-a(hk)^2+frachb-a+hsum_k=0^b-a(2ahk)+frachb-a+hsum_k=0^b-aa^2$$
    $$=frach^3b-a+hsum_k=0^b-ak^2+frac2ah^2b-a+hsum_k=0^b-ak+fraca^2hb-a+hsum_k=0^b-a1$$
    For the first term you can apply
    $$sum_x=0^n x^2 = fracn(n+1)(2n+1)6$$
    and apply the similar formulas that you did previously for the other 2 terms.






    share|cite|improve this answer









    $endgroup$

















      0












      $begingroup$

      With your edit, you are almost there. You can simplify
      $$frachb-a+hsum_k=0^b-a(hk+a)^2$$
      $$=frachb-a+hsum_k=0^b-aleft((hk)^2+2ahk+a^2 right)$$
      $$=frachb-a+hsum_k=0^b-a(hk)^2+frachb-a+hsum_k=0^b-a(2ahk)+frachb-a+hsum_k=0^b-aa^2$$
      $$=frach^3b-a+hsum_k=0^b-ak^2+frac2ah^2b-a+hsum_k=0^b-ak+fraca^2hb-a+hsum_k=0^b-a1$$
      For the first term you can apply
      $$sum_x=0^n x^2 = fracn(n+1)(2n+1)6$$
      and apply the similar formulas that you did previously for the other 2 terms.






      share|cite|improve this answer









      $endgroup$















        0












        0








        0





        $begingroup$

        With your edit, you are almost there. You can simplify
        $$frachb-a+hsum_k=0^b-a(hk+a)^2$$
        $$=frachb-a+hsum_k=0^b-aleft((hk)^2+2ahk+a^2 right)$$
        $$=frachb-a+hsum_k=0^b-a(hk)^2+frachb-a+hsum_k=0^b-a(2ahk)+frachb-a+hsum_k=0^b-aa^2$$
        $$=frach^3b-a+hsum_k=0^b-ak^2+frac2ah^2b-a+hsum_k=0^b-ak+fraca^2hb-a+hsum_k=0^b-a1$$
        For the first term you can apply
        $$sum_x=0^n x^2 = fracn(n+1)(2n+1)6$$
        and apply the similar formulas that you did previously for the other 2 terms.






        share|cite|improve this answer









        $endgroup$



        With your edit, you are almost there. You can simplify
        $$frachb-a+hsum_k=0^b-a(hk+a)^2$$
        $$=frachb-a+hsum_k=0^b-aleft((hk)^2+2ahk+a^2 right)$$
        $$=frachb-a+hsum_k=0^b-a(hk)^2+frachb-a+hsum_k=0^b-a(2ahk)+frachb-a+hsum_k=0^b-aa^2$$
        $$=frach^3b-a+hsum_k=0^b-ak^2+frac2ah^2b-a+hsum_k=0^b-ak+fraca^2hb-a+hsum_k=0^b-a1$$
        For the first term you can apply
        $$sum_x=0^n x^2 = fracn(n+1)(2n+1)6$$
        and apply the similar formulas that you did previously for the other 2 terms.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Oct 2 '18 at 15:08









        PaulPaul

        6,23721023




        6,23721023



























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