Calculate the variance of this random variable












0












$begingroup$


Let $p_1$ have $p_x(x) = 0.3*0.7^X$. Let $p_2$ have the exact same $p_x(x) = 0.3*0.7^X$. Assume this is valid for all $xgeq 0$.



Assume that $p_1,p_2$ are independent random variables.



Let $Y=(p_1)(p_2)$. Calculate $var(Y)$.




Here is my attempt:



$var(Y)=[E(Y^2)]-[E(Y)]^2=[E(p_1p_2)^2]-[E(p_1p_2)]^2=[E(p_1^2p_2^2)]-[E(p_1p_2)]^2$



Usually my process for this is to calculate $E[Y], E[Y^2]$, and then apply the formula (squaring $E[Y]$ as a whole, of course]



So, $E[Y]=E[p_1p_2]=E[p_1]E[p_2]$. Calculating one will give me the other, as they are the exact same.



I notice that $p_1$ is $Geo(theta=0.3)=theta(1-theta)^X$. The expectation is given by $(1-theta)/theta=0.7/0.3=2.33$



This helps me with $E[Y]=(2.33)(2.33)$, but how do I calculate $E[Y^2]$?






probability sequences-and-series random-variables power-series variance







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$endgroup$

















    0












    $begingroup$


    Let $p_1$ have $p_x(x) = 0.3*0.7^X$. Let $p_2$ have the exact same $p_x(x) = 0.3*0.7^X$. Assume this is valid for all $xgeq 0$.



    Assume that $p_1,p_2$ are independent random variables.



    Let $Y=(p_1)(p_2)$. Calculate $var(Y)$.




    Here is my attempt:



    $var(Y)=[E(Y^2)]-[E(Y)]^2=[E(p_1p_2)^2]-[E(p_1p_2)]^2=[E(p_1^2p_2^2)]-[E(p_1p_2)]^2$



    Usually my process for this is to calculate $E[Y], E[Y^2]$, and then apply the formula (squaring $E[Y]$ as a whole, of course]



    So, $E[Y]=E[p_1p_2]=E[p_1]E[p_2]$. Calculating one will give me the other, as they are the exact same.



    I notice that $p_1$ is $Geo(theta=0.3)=theta(1-theta)^X$. The expectation is given by $(1-theta)/theta=0.7/0.3=2.33$



    This helps me with $E[Y]=(2.33)(2.33)$, but how do I calculate $E[Y^2]$?






    probability sequences-and-series random-variables power-series variance







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $p_1$ have $p_x(x) = 0.3*0.7^X$. Let $p_2$ have the exact same $p_x(x) = 0.3*0.7^X$. Assume this is valid for all $xgeq 0$.



      Assume that $p_1,p_2$ are independent random variables.



      Let $Y=(p_1)(p_2)$. Calculate $var(Y)$.




      Here is my attempt:



      $var(Y)=[E(Y^2)]-[E(Y)]^2=[E(p_1p_2)^2]-[E(p_1p_2)]^2=[E(p_1^2p_2^2)]-[E(p_1p_2)]^2$



      Usually my process for this is to calculate $E[Y], E[Y^2]$, and then apply the formula (squaring $E[Y]$ as a whole, of course]



      So, $E[Y]=E[p_1p_2]=E[p_1]E[p_2]$. Calculating one will give me the other, as they are the exact same.



      I notice that $p_1$ is $Geo(theta=0.3)=theta(1-theta)^X$. The expectation is given by $(1-theta)/theta=0.7/0.3=2.33$



      This helps me with $E[Y]=(2.33)(2.33)$, but how do I calculate $E[Y^2]$?






      probability sequences-and-series random-variables power-series variance







      share|cite|improve this question











      $endgroup$




      Let $p_1$ have $p_x(x) = 0.3*0.7^X$. Let $p_2$ have the exact same $p_x(x) = 0.3*0.7^X$. Assume this is valid for all $xgeq 0$.



      Assume that $p_1,p_2$ are independent random variables.



      Let $Y=(p_1)(p_2)$. Calculate $var(Y)$.




      Here is my attempt:



      $var(Y)=[E(Y^2)]-[E(Y)]^2=[E(p_1p_2)^2]-[E(p_1p_2)]^2=[E(p_1^2p_2^2)]-[E(p_1p_2)]^2$



      Usually my process for this is to calculate $E[Y], E[Y^2]$, and then apply the formula (squaring $E[Y]$ as a whole, of course]



      So, $E[Y]=E[p_1p_2]=E[p_1]E[p_2]$. Calculating one will give me the other, as they are the exact same.



      I notice that $p_1$ is $Geo(theta=0.3)=theta(1-theta)^X$. The expectation is given by $(1-theta)/theta=0.7/0.3=2.33$



      This helps me with $E[Y]=(2.33)(2.33)$, but how do I calculate $E[Y^2]$?






      probability sequences-and-series random-variables power-series variance




      probability sequences-and-series random-variables power-series variance






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Nov 24 '18 at 1:36







      K Split X

















      asked Nov 24 '18 at 1:23









      K Split XK Split X

      4,21111031




      4,21111031






















          3 Answers
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          1












          $begingroup$

          begin{align}
          E[Y^2] &= E[p_1^2p_2^2]\
          &=E[p_1^2]E[p_2^2]\
          &= (Var(p_1)+E[p_1]^2)^2
          end{align}






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Notice that if $p_1$ and $p_2$ are independent then so are $p_1 ^2$ and $p_2 ^2$. Therfore $$
            E (p_1 ^2 p_2 ^2) = E (p_1 ^2)E( p_2 ^2)
            $$






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              $$E(Y^2){=E(p_1^2p_2^2)\=E(p_1^2)E(p_2^2)}$$according to Geometric distribution(https://en.wikipedia.org/wiki/Geometric_distribution) we obtain $$E(p_1^2)=E(p_2^2)={0.7times 1.7over 0.3^2}={119over 9}$$therefore $$E(Y^2)={14161over 81}approx 174.83$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                How did you calculate $E(p_1^2)$? The formula on wikipedia is for $E(X)$, not $E(X^2)$?
                $endgroup$
                – K Split X
                Nov 24 '18 at 16:19










              • $begingroup$
                I calculated $E(X^2)$ from the variance and the mean using $$E(X^2)=mu^2+sigma^2$$. Also i spotted a mistake in my answer that i will fix it soon........
                $endgroup$
                – Mostafa Ayaz
                Nov 24 '18 at 16:27











              Your Answer





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              3 Answers
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              3 Answers
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              1












              $begingroup$

              begin{align}
              E[Y^2] &= E[p_1^2p_2^2]\
              &=E[p_1^2]E[p_2^2]\
              &= (Var(p_1)+E[p_1]^2)^2
              end{align}






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                begin{align}
                E[Y^2] &= E[p_1^2p_2^2]\
                &=E[p_1^2]E[p_2^2]\
                &= (Var(p_1)+E[p_1]^2)^2
                end{align}






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  begin{align}
                  E[Y^2] &= E[p_1^2p_2^2]\
                  &=E[p_1^2]E[p_2^2]\
                  &= (Var(p_1)+E[p_1]^2)^2
                  end{align}






                  share|cite|improve this answer









                  $endgroup$



                  begin{align}
                  E[Y^2] &= E[p_1^2p_2^2]\
                  &=E[p_1^2]E[p_2^2]\
                  &= (Var(p_1)+E[p_1]^2)^2
                  end{align}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 24 '18 at 1:48









                  Siong Thye GohSiong Thye Goh

                  100k1465117




                  100k1465117























                      0












                      $begingroup$

                      Notice that if $p_1$ and $p_2$ are independent then so are $p_1 ^2$ and $p_2 ^2$. Therfore $$
                      E (p_1 ^2 p_2 ^2) = E (p_1 ^2)E( p_2 ^2)
                      $$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Notice that if $p_1$ and $p_2$ are independent then so are $p_1 ^2$ and $p_2 ^2$. Therfore $$
                        E (p_1 ^2 p_2 ^2) = E (p_1 ^2)E( p_2 ^2)
                        $$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Notice that if $p_1$ and $p_2$ are independent then so are $p_1 ^2$ and $p_2 ^2$. Therfore $$
                          E (p_1 ^2 p_2 ^2) = E (p_1 ^2)E( p_2 ^2)
                          $$






                          share|cite|improve this answer









                          $endgroup$



                          Notice that if $p_1$ and $p_2$ are independent then so are $p_1 ^2$ and $p_2 ^2$. Therfore $$
                          E (p_1 ^2 p_2 ^2) = E (p_1 ^2)E( p_2 ^2)
                          $$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 24 '18 at 1:48









                          user609189user609189

                          314




                          314























                              0












                              $begingroup$

                              $$E(Y^2){=E(p_1^2p_2^2)\=E(p_1^2)E(p_2^2)}$$according to Geometric distribution(https://en.wikipedia.org/wiki/Geometric_distribution) we obtain $$E(p_1^2)=E(p_2^2)={0.7times 1.7over 0.3^2}={119over 9}$$therefore $$E(Y^2)={14161over 81}approx 174.83$$






                              share|cite|improve this answer











                              $endgroup$













                              • $begingroup$
                                How did you calculate $E(p_1^2)$? The formula on wikipedia is for $E(X)$, not $E(X^2)$?
                                $endgroup$
                                – K Split X
                                Nov 24 '18 at 16:19










                              • $begingroup$
                                I calculated $E(X^2)$ from the variance and the mean using $$E(X^2)=mu^2+sigma^2$$. Also i spotted a mistake in my answer that i will fix it soon........
                                $endgroup$
                                – Mostafa Ayaz
                                Nov 24 '18 at 16:27
















                              0












                              $begingroup$

                              $$E(Y^2){=E(p_1^2p_2^2)\=E(p_1^2)E(p_2^2)}$$according to Geometric distribution(https://en.wikipedia.org/wiki/Geometric_distribution) we obtain $$E(p_1^2)=E(p_2^2)={0.7times 1.7over 0.3^2}={119over 9}$$therefore $$E(Y^2)={14161over 81}approx 174.83$$






                              share|cite|improve this answer











                              $endgroup$













                              • $begingroup$
                                How did you calculate $E(p_1^2)$? The formula on wikipedia is for $E(X)$, not $E(X^2)$?
                                $endgroup$
                                – K Split X
                                Nov 24 '18 at 16:19










                              • $begingroup$
                                I calculated $E(X^2)$ from the variance and the mean using $$E(X^2)=mu^2+sigma^2$$. Also i spotted a mistake in my answer that i will fix it soon........
                                $endgroup$
                                – Mostafa Ayaz
                                Nov 24 '18 at 16:27














                              0












                              0








                              0





                              $begingroup$

                              $$E(Y^2){=E(p_1^2p_2^2)\=E(p_1^2)E(p_2^2)}$$according to Geometric distribution(https://en.wikipedia.org/wiki/Geometric_distribution) we obtain $$E(p_1^2)=E(p_2^2)={0.7times 1.7over 0.3^2}={119over 9}$$therefore $$E(Y^2)={14161over 81}approx 174.83$$






                              share|cite|improve this answer











                              $endgroup$



                              $$E(Y^2){=E(p_1^2p_2^2)\=E(p_1^2)E(p_2^2)}$$according to Geometric distribution(https://en.wikipedia.org/wiki/Geometric_distribution) we obtain $$E(p_1^2)=E(p_2^2)={0.7times 1.7over 0.3^2}={119over 9}$$therefore $$E(Y^2)={14161over 81}approx 174.83$$







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Nov 24 '18 at 16:31

























                              answered Nov 24 '18 at 9:41









                              Mostafa AyazMostafa Ayaz

                              14.8k3938




                              14.8k3938












                              • $begingroup$
                                How did you calculate $E(p_1^2)$? The formula on wikipedia is for $E(X)$, not $E(X^2)$?
                                $endgroup$
                                – K Split X
                                Nov 24 '18 at 16:19










                              • $begingroup$
                                I calculated $E(X^2)$ from the variance and the mean using $$E(X^2)=mu^2+sigma^2$$. Also i spotted a mistake in my answer that i will fix it soon........
                                $endgroup$
                                – Mostafa Ayaz
                                Nov 24 '18 at 16:27


















                              • $begingroup$
                                How did you calculate $E(p_1^2)$? The formula on wikipedia is for $E(X)$, not $E(X^2)$?
                                $endgroup$
                                – K Split X
                                Nov 24 '18 at 16:19










                              • $begingroup$
                                I calculated $E(X^2)$ from the variance and the mean using $$E(X^2)=mu^2+sigma^2$$. Also i spotted a mistake in my answer that i will fix it soon........
                                $endgroup$
                                – Mostafa Ayaz
                                Nov 24 '18 at 16:27
















                              $begingroup$
                              How did you calculate $E(p_1^2)$? The formula on wikipedia is for $E(X)$, not $E(X^2)$?
                              $endgroup$
                              – K Split X
                              Nov 24 '18 at 16:19




                              $begingroup$
                              How did you calculate $E(p_1^2)$? The formula on wikipedia is for $E(X)$, not $E(X^2)$?
                              $endgroup$
                              – K Split X
                              Nov 24 '18 at 16:19












                              $begingroup$
                              I calculated $E(X^2)$ from the variance and the mean using $$E(X^2)=mu^2+sigma^2$$. Also i spotted a mistake in my answer that i will fix it soon........
                              $endgroup$
                              – Mostafa Ayaz
                              Nov 24 '18 at 16:27




                              $begingroup$
                              I calculated $E(X^2)$ from the variance and the mean using $$E(X^2)=mu^2+sigma^2$$. Also i spotted a mistake in my answer that i will fix it soon........
                              $endgroup$
                              – Mostafa Ayaz
                              Nov 24 '18 at 16:27


















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