Find the standard deviation of $|X−Y|$












2












$begingroup$



Let $X$ and $Y$ be independent random variables with a Bernoulli Distribution $Ber(1/3)$. Find the standard deviation of $|X−Y|$.




The Standard Deviation in the square root of the variance. For a $Ber(1/3)$ the $Var(X)=Var(Y)=1/3(1-1/3)=2/9$, now how can I calculate $sigma=sqrt{Var(|X−Y|)}$? Because my idea was to subtract the two $Var$ but then the result will be $0$, but it should be $frac{2sqrt{5}}{9}$, how can I solve it?






probability statistics







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    $Ber$ = bernoulli? There are only $4$ outcomes to consider, so it's easy to do this from the definition.
    $endgroup$
    – Robert Israel
    Jan 25 at 16:18










  • $begingroup$
    @RobertIsrael yes it's bernoulli, which four cases?
    $endgroup$
    – FTAC
    Jan 25 at 16:20
















2












$begingroup$



Let $X$ and $Y$ be independent random variables with a Bernoulli Distribution $Ber(1/3)$. Find the standard deviation of $|X−Y|$.




The Standard Deviation in the square root of the variance. For a $Ber(1/3)$ the $Var(X)=Var(Y)=1/3(1-1/3)=2/9$, now how can I calculate $sigma=sqrt{Var(|X−Y|)}$? Because my idea was to subtract the two $Var$ but then the result will be $0$, but it should be $frac{2sqrt{5}}{9}$, how can I solve it?






probability statistics







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    $Ber$ = bernoulli? There are only $4$ outcomes to consider, so it's easy to do this from the definition.
    $endgroup$
    – Robert Israel
    Jan 25 at 16:18










  • $begingroup$
    @RobertIsrael yes it's bernoulli, which four cases?
    $endgroup$
    – FTAC
    Jan 25 at 16:20














2












2








2


1



$begingroup$



Let $X$ and $Y$ be independent random variables with a Bernoulli Distribution $Ber(1/3)$. Find the standard deviation of $|X−Y|$.




The Standard Deviation in the square root of the variance. For a $Ber(1/3)$ the $Var(X)=Var(Y)=1/3(1-1/3)=2/9$, now how can I calculate $sigma=sqrt{Var(|X−Y|)}$? Because my idea was to subtract the two $Var$ but then the result will be $0$, but it should be $frac{2sqrt{5}}{9}$, how can I solve it?






probability statistics







share|cite|improve this question











$endgroup$





Let $X$ and $Y$ be independent random variables with a Bernoulli Distribution $Ber(1/3)$. Find the standard deviation of $|X−Y|$.




The Standard Deviation in the square root of the variance. For a $Ber(1/3)$ the $Var(X)=Var(Y)=1/3(1-1/3)=2/9$, now how can I calculate $sigma=sqrt{Var(|X−Y|)}$? Because my idea was to subtract the two $Var$ but then the result will be $0$, but it should be $frac{2sqrt{5}}{9}$, how can I solve it?






probability statistics




probability statistics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 25 at 16:19







FTAC

















asked Jan 25 at 16:06









FTACFTAC

2649




2649








  • 3




    $begingroup$
    $Ber$ = bernoulli? There are only $4$ outcomes to consider, so it's easy to do this from the definition.
    $endgroup$
    – Robert Israel
    Jan 25 at 16:18










  • $begingroup$
    @RobertIsrael yes it's bernoulli, which four cases?
    $endgroup$
    – FTAC
    Jan 25 at 16:20














  • 3




    $begingroup$
    $Ber$ = bernoulli? There are only $4$ outcomes to consider, so it's easy to do this from the definition.
    $endgroup$
    – Robert Israel
    Jan 25 at 16:18










  • $begingroup$
    @RobertIsrael yes it's bernoulli, which four cases?
    $endgroup$
    – FTAC
    Jan 25 at 16:20








3




3




$begingroup$
$Ber$ = bernoulli? There are only $4$ outcomes to consider, so it's easy to do this from the definition.
$endgroup$
– Robert Israel
Jan 25 at 16:18




$begingroup$
$Ber$ = bernoulli? There are only $4$ outcomes to consider, so it's easy to do this from the definition.
$endgroup$
– Robert Israel
Jan 25 at 16:18












$begingroup$
@RobertIsrael yes it's bernoulli, which four cases?
$endgroup$
– FTAC
Jan 25 at 16:20




$begingroup$
@RobertIsrael yes it's bernoulli, which four cases?
$endgroup$
– FTAC
Jan 25 at 16:20










4 Answers
4






active

oldest

votes


















2












$begingroup$

begin{equation}
|X-Y|= begin{cases}
1 & text{iff} quad (X=1 quad wedgequad Y=0)quad vee quad (X=0 quad wedgequad Y=1) \
0 & text{iff} quad (X=1 quad wedgequad Y=1)quad vee quad (X=0 quad wedgequad Y=0)
end{cases}
end{equation}
So $|X-Y|$ is another Bernoulli random variable and



$Pr(|X-Y|=1)=Pr(X=1)Pr(Y=0)+Pr(X=0)Pr(Y=1)=(1/3)(2/3)+(2/3)(1/3)=4/9=p$



and then



$Var(|X-Y|)=(4/9)(5/9)=20/81 implies SD=sqrt{20/81}=(2sqrt{5})/9$,
as required.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thanks for the answer, what does Pr mean?
    $endgroup$
    – FTAC
    Jan 25 at 17:00






  • 2




    $begingroup$
    @FabioTaccaliti The same as $P()$ (in statistical context). $P(X=0)$ or $Pr(X=0)$ is the probability that the random variable $X$ takes the value $0$.
    $endgroup$
    – callculus
    Jan 25 at 17:25





















2












$begingroup$

Hints:




  • $X$ could be $1$ and $Y$ could be $0$, or the other way round, or both $1$, or both $0$.


  • In the first two cases their difference would be $1$ while in the last two it would be $0$.


  • So $|X-Y|$ is another Bernoulli random variable, and you need to find the probability it is $1$ and then calculate the variance and take its square root







share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    In general, let $X,Ysim B(p)$. Refer to the table:
    $$begin{array}{c|c|c|c|c|c} X & P(X)&Y&P(Y)&|X-Y|&P(|X-Y|)&|X-Y|^2&P(|X-Y|^2)\
    hline
    0&q&0&q&0&q^2&0&q^2\
    0&q&1&p&1&qp&1&qp\
    1&p&0&q&1&pq&1&pq\
    1&p&1&p&0&p^2&0&p^2\
    end{array}\
    text{1-method}: |X-Y|sim B(2pq); \
    Var(|X-Y|)=2pqcdot (1-2pq).\
    text{2-method}: Var(|X-Y|)=mathbb E(|X-Y|^2)-[mathbb E(|X-Y|)]^2=\
    2pq-[2pq]^2 =2pqcdot (1-2pq).$$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      The easiest thing is to use a table to obtain the distribution.



      The pdf of $|X-Y|$



      $$begin{array}{|c|c|c|c|} hline X/Y & 0left(p=frac13right)&1left(p=frac23right) \ hline 0left(p=frac13right) & 0 &1 \ hline 1left(p=frac23right) & 1 &0 \ hline end{array}$$



      Now you can use the well known formulas to obtain $Var(|X-Y|)$




      • $E(|X-Y|)=sumlimits_{x=0}^{1}sumlimits_{y=0}^{1} |x-y|cdot p(x,y)$


      • $E(|X-Y|^2)=sumlimits_{x=0}^{1}sumlimits_{y=0}^{1} |x-y|^2cdot
        p(x,y)$


      • $Var(Z)=E(Z^2)-E^2(Z)$







      share|cite|improve this answer









      $endgroup$













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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        begin{equation}
        |X-Y|= begin{cases}
        1 & text{iff} quad (X=1 quad wedgequad Y=0)quad vee quad (X=0 quad wedgequad Y=1) \
        0 & text{iff} quad (X=1 quad wedgequad Y=1)quad vee quad (X=0 quad wedgequad Y=0)
        end{cases}
        end{equation}
        So $|X-Y|$ is another Bernoulli random variable and



        $Pr(|X-Y|=1)=Pr(X=1)Pr(Y=0)+Pr(X=0)Pr(Y=1)=(1/3)(2/3)+(2/3)(1/3)=4/9=p$



        and then



        $Var(|X-Y|)=(4/9)(5/9)=20/81 implies SD=sqrt{20/81}=(2sqrt{5})/9$,
        as required.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          thanks for the answer, what does Pr mean?
          $endgroup$
          – FTAC
          Jan 25 at 17:00






        • 2




          $begingroup$
          @FabioTaccaliti The same as $P()$ (in statistical context). $P(X=0)$ or $Pr(X=0)$ is the probability that the random variable $X$ takes the value $0$.
          $endgroup$
          – callculus
          Jan 25 at 17:25


















        2












        $begingroup$

        begin{equation}
        |X-Y|= begin{cases}
        1 & text{iff} quad (X=1 quad wedgequad Y=0)quad vee quad (X=0 quad wedgequad Y=1) \
        0 & text{iff} quad (X=1 quad wedgequad Y=1)quad vee quad (X=0 quad wedgequad Y=0)
        end{cases}
        end{equation}
        So $|X-Y|$ is another Bernoulli random variable and



        $Pr(|X-Y|=1)=Pr(X=1)Pr(Y=0)+Pr(X=0)Pr(Y=1)=(1/3)(2/3)+(2/3)(1/3)=4/9=p$



        and then



        $Var(|X-Y|)=(4/9)(5/9)=20/81 implies SD=sqrt{20/81}=(2sqrt{5})/9$,
        as required.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          thanks for the answer, what does Pr mean?
          $endgroup$
          – FTAC
          Jan 25 at 17:00






        • 2




          $begingroup$
          @FabioTaccaliti The same as $P()$ (in statistical context). $P(X=0)$ or $Pr(X=0)$ is the probability that the random variable $X$ takes the value $0$.
          $endgroup$
          – callculus
          Jan 25 at 17:25
















        2












        2








        2





        $begingroup$

        begin{equation}
        |X-Y|= begin{cases}
        1 & text{iff} quad (X=1 quad wedgequad Y=0)quad vee quad (X=0 quad wedgequad Y=1) \
        0 & text{iff} quad (X=1 quad wedgequad Y=1)quad vee quad (X=0 quad wedgequad Y=0)
        end{cases}
        end{equation}
        So $|X-Y|$ is another Bernoulli random variable and



        $Pr(|X-Y|=1)=Pr(X=1)Pr(Y=0)+Pr(X=0)Pr(Y=1)=(1/3)(2/3)+(2/3)(1/3)=4/9=p$



        and then



        $Var(|X-Y|)=(4/9)(5/9)=20/81 implies SD=sqrt{20/81}=(2sqrt{5})/9$,
        as required.






        share|cite|improve this answer









        $endgroup$



        begin{equation}
        |X-Y|= begin{cases}
        1 & text{iff} quad (X=1 quad wedgequad Y=0)quad vee quad (X=0 quad wedgequad Y=1) \
        0 & text{iff} quad (X=1 quad wedgequad Y=1)quad vee quad (X=0 quad wedgequad Y=0)
        end{cases}
        end{equation}
        So $|X-Y|$ is another Bernoulli random variable and



        $Pr(|X-Y|=1)=Pr(X=1)Pr(Y=0)+Pr(X=0)Pr(Y=1)=(1/3)(2/3)+(2/3)(1/3)=4/9=p$



        and then



        $Var(|X-Y|)=(4/9)(5/9)=20/81 implies SD=sqrt{20/81}=(2sqrt{5})/9$,
        as required.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 25 at 16:55









        amator2357amator2357

        729




        729












        • $begingroup$
          thanks for the answer, what does Pr mean?
          $endgroup$
          – FTAC
          Jan 25 at 17:00






        • 2




          $begingroup$
          @FabioTaccaliti The same as $P()$ (in statistical context). $P(X=0)$ or $Pr(X=0)$ is the probability that the random variable $X$ takes the value $0$.
          $endgroup$
          – callculus
          Jan 25 at 17:25




















        • $begingroup$
          thanks for the answer, what does Pr mean?
          $endgroup$
          – FTAC
          Jan 25 at 17:00






        • 2




          $begingroup$
          @FabioTaccaliti The same as $P()$ (in statistical context). $P(X=0)$ or $Pr(X=0)$ is the probability that the random variable $X$ takes the value $0$.
          $endgroup$
          – callculus
          Jan 25 at 17:25


















        $begingroup$
        thanks for the answer, what does Pr mean?
        $endgroup$
        – FTAC
        Jan 25 at 17:00




        $begingroup$
        thanks for the answer, what does Pr mean?
        $endgroup$
        – FTAC
        Jan 25 at 17:00




        2




        2




        $begingroup$
        @FabioTaccaliti The same as $P()$ (in statistical context). $P(X=0)$ or $Pr(X=0)$ is the probability that the random variable $X$ takes the value $0$.
        $endgroup$
        – callculus
        Jan 25 at 17:25






        $begingroup$
        @FabioTaccaliti The same as $P()$ (in statistical context). $P(X=0)$ or $Pr(X=0)$ is the probability that the random variable $X$ takes the value $0$.
        $endgroup$
        – callculus
        Jan 25 at 17:25













        2












        $begingroup$

        Hints:




        • $X$ could be $1$ and $Y$ could be $0$, or the other way round, or both $1$, or both $0$.


        • In the first two cases their difference would be $1$ while in the last two it would be $0$.


        • So $|X-Y|$ is another Bernoulli random variable, and you need to find the probability it is $1$ and then calculate the variance and take its square root







        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          Hints:




          • $X$ could be $1$ and $Y$ could be $0$, or the other way round, or both $1$, or both $0$.


          • In the first two cases their difference would be $1$ while in the last two it would be $0$.


          • So $|X-Y|$ is another Bernoulli random variable, and you need to find the probability it is $1$ and then calculate the variance and take its square root







          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            Hints:




            • $X$ could be $1$ and $Y$ could be $0$, or the other way round, or both $1$, or both $0$.


            • In the first two cases their difference would be $1$ while in the last two it would be $0$.


            • So $|X-Y|$ is another Bernoulli random variable, and you need to find the probability it is $1$ and then calculate the variance and take its square root







            share|cite|improve this answer









            $endgroup$



            Hints:




            • $X$ could be $1$ and $Y$ could be $0$, or the other way round, or both $1$, or both $0$.


            • In the first two cases their difference would be $1$ while in the last two it would be $0$.


            • So $|X-Y|$ is another Bernoulli random variable, and you need to find the probability it is $1$ and then calculate the variance and take its square root








            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 25 at 16:37









            HenryHenry

            101k481168




            101k481168























                2












                $begingroup$

                In general, let $X,Ysim B(p)$. Refer to the table:
                $$begin{array}{c|c|c|c|c|c} X & P(X)&Y&P(Y)&|X-Y|&P(|X-Y|)&|X-Y|^2&P(|X-Y|^2)\
                hline
                0&q&0&q&0&q^2&0&q^2\
                0&q&1&p&1&qp&1&qp\
                1&p&0&q&1&pq&1&pq\
                1&p&1&p&0&p^2&0&p^2\
                end{array}\
                text{1-method}: |X-Y|sim B(2pq); \
                Var(|X-Y|)=2pqcdot (1-2pq).\
                text{2-method}: Var(|X-Y|)=mathbb E(|X-Y|^2)-[mathbb E(|X-Y|)]^2=\
                2pq-[2pq]^2 =2pqcdot (1-2pq).$$






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  In general, let $X,Ysim B(p)$. Refer to the table:
                  $$begin{array}{c|c|c|c|c|c} X & P(X)&Y&P(Y)&|X-Y|&P(|X-Y|)&|X-Y|^2&P(|X-Y|^2)\
                  hline
                  0&q&0&q&0&q^2&0&q^2\
                  0&q&1&p&1&qp&1&qp\
                  1&p&0&q&1&pq&1&pq\
                  1&p&1&p&0&p^2&0&p^2\
                  end{array}\
                  text{1-method}: |X-Y|sim B(2pq); \
                  Var(|X-Y|)=2pqcdot (1-2pq).\
                  text{2-method}: Var(|X-Y|)=mathbb E(|X-Y|^2)-[mathbb E(|X-Y|)]^2=\
                  2pq-[2pq]^2 =2pqcdot (1-2pq).$$






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    In general, let $X,Ysim B(p)$. Refer to the table:
                    $$begin{array}{c|c|c|c|c|c} X & P(X)&Y&P(Y)&|X-Y|&P(|X-Y|)&|X-Y|^2&P(|X-Y|^2)\
                    hline
                    0&q&0&q&0&q^2&0&q^2\
                    0&q&1&p&1&qp&1&qp\
                    1&p&0&q&1&pq&1&pq\
                    1&p&1&p&0&p^2&0&p^2\
                    end{array}\
                    text{1-method}: |X-Y|sim B(2pq); \
                    Var(|X-Y|)=2pqcdot (1-2pq).\
                    text{2-method}: Var(|X-Y|)=mathbb E(|X-Y|^2)-[mathbb E(|X-Y|)]^2=\
                    2pq-[2pq]^2 =2pqcdot (1-2pq).$$






                    share|cite|improve this answer









                    $endgroup$



                    In general, let $X,Ysim B(p)$. Refer to the table:
                    $$begin{array}{c|c|c|c|c|c} X & P(X)&Y&P(Y)&|X-Y|&P(|X-Y|)&|X-Y|^2&P(|X-Y|^2)\
                    hline
                    0&q&0&q&0&q^2&0&q^2\
                    0&q&1&p&1&qp&1&qp\
                    1&p&0&q&1&pq&1&pq\
                    1&p&1&p&0&p^2&0&p^2\
                    end{array}\
                    text{1-method}: |X-Y|sim B(2pq); \
                    Var(|X-Y|)=2pqcdot (1-2pq).\
                    text{2-method}: Var(|X-Y|)=mathbb E(|X-Y|^2)-[mathbb E(|X-Y|)]^2=\
                    2pq-[2pq]^2 =2pqcdot (1-2pq).$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 25 at 17:15









                    farruhotafarruhota

                    20.8k2741




                    20.8k2741























                        1












                        $begingroup$

                        The easiest thing is to use a table to obtain the distribution.



                        The pdf of $|X-Y|$



                        $$begin{array}{|c|c|c|c|} hline X/Y & 0left(p=frac13right)&1left(p=frac23right) \ hline 0left(p=frac13right) & 0 &1 \ hline 1left(p=frac23right) & 1 &0 \ hline end{array}$$



                        Now you can use the well known formulas to obtain $Var(|X-Y|)$




                        • $E(|X-Y|)=sumlimits_{x=0}^{1}sumlimits_{y=0}^{1} |x-y|cdot p(x,y)$


                        • $E(|X-Y|^2)=sumlimits_{x=0}^{1}sumlimits_{y=0}^{1} |x-y|^2cdot
                          p(x,y)$


                        • $Var(Z)=E(Z^2)-E^2(Z)$







                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          The easiest thing is to use a table to obtain the distribution.



                          The pdf of $|X-Y|$



                          $$begin{array}{|c|c|c|c|} hline X/Y & 0left(p=frac13right)&1left(p=frac23right) \ hline 0left(p=frac13right) & 0 &1 \ hline 1left(p=frac23right) & 1 &0 \ hline end{array}$$



                          Now you can use the well known formulas to obtain $Var(|X-Y|)$




                          • $E(|X-Y|)=sumlimits_{x=0}^{1}sumlimits_{y=0}^{1} |x-y|cdot p(x,y)$


                          • $E(|X-Y|^2)=sumlimits_{x=0}^{1}sumlimits_{y=0}^{1} |x-y|^2cdot
                            p(x,y)$


                          • $Var(Z)=E(Z^2)-E^2(Z)$







                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            The easiest thing is to use a table to obtain the distribution.



                            The pdf of $|X-Y|$



                            $$begin{array}{|c|c|c|c|} hline X/Y & 0left(p=frac13right)&1left(p=frac23right) \ hline 0left(p=frac13right) & 0 &1 \ hline 1left(p=frac23right) & 1 &0 \ hline end{array}$$



                            Now you can use the well known formulas to obtain $Var(|X-Y|)$




                            • $E(|X-Y|)=sumlimits_{x=0}^{1}sumlimits_{y=0}^{1} |x-y|cdot p(x,y)$


                            • $E(|X-Y|^2)=sumlimits_{x=0}^{1}sumlimits_{y=0}^{1} |x-y|^2cdot
                              p(x,y)$


                            • $Var(Z)=E(Z^2)-E^2(Z)$







                            share|cite|improve this answer









                            $endgroup$



                            The easiest thing is to use a table to obtain the distribution.



                            The pdf of $|X-Y|$



                            $$begin{array}{|c|c|c|c|} hline X/Y & 0left(p=frac13right)&1left(p=frac23right) \ hline 0left(p=frac13right) & 0 &1 \ hline 1left(p=frac23right) & 1 &0 \ hline end{array}$$



                            Now you can use the well known formulas to obtain $Var(|X-Y|)$




                            • $E(|X-Y|)=sumlimits_{x=0}^{1}sumlimits_{y=0}^{1} |x-y|cdot p(x,y)$


                            • $E(|X-Y|^2)=sumlimits_{x=0}^{1}sumlimits_{y=0}^{1} |x-y|^2cdot
                              p(x,y)$


                            • $Var(Z)=E(Z^2)-E^2(Z)$








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                            answered Jan 25 at 17:03









                            callculuscallculus

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