Computing $ E[Y^2] $ when $Y$ is a piecewise function of $X sim Pois(2) $












0












$begingroup$


$ X sim Pois(2) $ and let$ Y $ be a random variable : $ Y = begin{cases} 2X , X le 3 \ X , X ge 4 end{cases} $



compute $ E[Y^2] $ ?



I have calculated that $ E[Y] = E[X] + 10e^{-2} $ and I have the identity $ Var(X) = E[X^2] - (E[X])^2 $ and because $ X $ is poisson then $ Var(X) = E[X] $. but I'm not sure if $ Y $ is also a poisson variable?



Any suggestions?






probability poisson-distribution







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$endgroup$

















    0












    $begingroup$


    $ X sim Pois(2) $ and let$ Y $ be a random variable : $ Y = begin{cases} 2X , X le 3 \ X , X ge 4 end{cases} $



    compute $ E[Y^2] $ ?



    I have calculated that $ E[Y] = E[X] + 10e^{-2} $ and I have the identity $ Var(X) = E[X^2] - (E[X])^2 $ and because $ X $ is poisson then $ Var(X) = E[X] $. but I'm not sure if $ Y $ is also a poisson variable?



    Any suggestions?






    probability poisson-distribution







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      $ X sim Pois(2) $ and let$ Y $ be a random variable : $ Y = begin{cases} 2X , X le 3 \ X , X ge 4 end{cases} $



      compute $ E[Y^2] $ ?



      I have calculated that $ E[Y] = E[X] + 10e^{-2} $ and I have the identity $ Var(X) = E[X^2] - (E[X])^2 $ and because $ X $ is poisson then $ Var(X) = E[X] $. but I'm not sure if $ Y $ is also a poisson variable?



      Any suggestions?






      probability poisson-distribution







      share|cite|improve this question











      $endgroup$




      $ X sim Pois(2) $ and let$ Y $ be a random variable : $ Y = begin{cases} 2X , X le 3 \ X , X ge 4 end{cases} $



      compute $ E[Y^2] $ ?



      I have calculated that $ E[Y] = E[X] + 10e^{-2} $ and I have the identity $ Var(X) = E[X^2] - (E[X])^2 $ and because $ X $ is poisson then $ Var(X) = E[X] $. but I'm not sure if $ Y $ is also a poisson variable?



      Any suggestions?






      probability poisson-distribution




      probability poisson-distribution






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 '18 at 16:59









      Nate Eldredge

      63.6k682171




      63.6k682171










      asked Dec 9 '18 at 16:37









      bm1125bm1125

      64916




      64916






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          $Y$ cannot be Poisson, because for example, $$Pr[Y = 3] = Pr[2X = 3] = Pr[X = 3/2] = 0.$$



          If we define $Y$ as in the body of the post, i.e. $$Y = begin{cases} 2X, & X le 3 \ X, & X ge 4, end{cases}$$ then we note that
          $$begin{align*}operatorname{E}[Y^2]
          &= sum_{x=0}^3 (2x)^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
          &= 4 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
          &= 3 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + left(sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} right) \
          &= 3 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=0}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
          &= e^{-lambda} (3lambda + 6 lambda^2 + tfrac{9}{2} lambda^3) + operatorname{E}[X^2] \
          &= 66e^{-2} + operatorname{E}[X]^2 + operatorname{Var}[X] \
          &= 66e^{-2} + lambda^2 + lambda \
          &= 6 + 66e^{-2}.
          end{align*}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Could you explain to me why $ (2x)^2 $ becomes $ 3x^2 $ ?
            $endgroup$
            – bm1125
            Dec 9 '18 at 17:02










          • $begingroup$
            @bm1125 I have expanded the details of that calculation in my edit, but simply put, $4 = 3 + 1.$
            $endgroup$
            – heropup
            Dec 9 '18 at 17:07



















          1












          $begingroup$

          begin{align}
          E[Y^2]
          &= sum_{x=0}^3 (2x)^2 Pr(X=x)+ sum_{x=4}^infty x^2 Pr(X=x)\
          &=sum_{x=0}^3 4x^2 Pr(X=x)+ sum_{x=4}^infty x^2 Pr(X=x)\
          &=sum_{x=0}^3 3x^2 Pr(X=x)+ sum_{x=0}^infty x^2 Pr(X=x)\
          &=3sum_{x=1}^3 x^2 Pr(X=x)+ E[X^2]\
          &= 3left( Pr(X=1) + 4Pr(X=2)+9Pr(X=3) right) + E[X^2]
          end{align}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for your answer! could you explain the second equality? I think you also miss the fact that $ P{X=6} not = P{Y=6} $ as $ P{Y=6} = P{X=3} + P{X=6} $ I believe.
            $endgroup$
            – bm1125
            Dec 9 '18 at 16:55






          • 1




            $begingroup$
            All of the summations should be with respect to $x$.
            $endgroup$
            – heropup
            Dec 9 '18 at 17:00










          • $begingroup$
            oops, i made lots of typos, thanks.
            $endgroup$
            – Siong Thye Goh
            Dec 9 '18 at 17:00










          • $begingroup$
            I use the law of the unconscious statistician $E[g(X)]=sum_x g(x)P(X=x)$. For the line from $4x^$ to $3x^2$, we pass a copy of $x^2$ to the second term, hence there is change of the index.
            $endgroup$
            – Siong Thye Goh
            Dec 9 '18 at 17:02













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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          $Y$ cannot be Poisson, because for example, $$Pr[Y = 3] = Pr[2X = 3] = Pr[X = 3/2] = 0.$$



          If we define $Y$ as in the body of the post, i.e. $$Y = begin{cases} 2X, & X le 3 \ X, & X ge 4, end{cases}$$ then we note that
          $$begin{align*}operatorname{E}[Y^2]
          &= sum_{x=0}^3 (2x)^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
          &= 4 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
          &= 3 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + left(sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} right) \
          &= 3 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=0}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
          &= e^{-lambda} (3lambda + 6 lambda^2 + tfrac{9}{2} lambda^3) + operatorname{E}[X^2] \
          &= 66e^{-2} + operatorname{E}[X]^2 + operatorname{Var}[X] \
          &= 66e^{-2} + lambda^2 + lambda \
          &= 6 + 66e^{-2}.
          end{align*}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Could you explain to me why $ (2x)^2 $ becomes $ 3x^2 $ ?
            $endgroup$
            – bm1125
            Dec 9 '18 at 17:02










          • $begingroup$
            @bm1125 I have expanded the details of that calculation in my edit, but simply put, $4 = 3 + 1.$
            $endgroup$
            – heropup
            Dec 9 '18 at 17:07
















          1












          $begingroup$

          $Y$ cannot be Poisson, because for example, $$Pr[Y = 3] = Pr[2X = 3] = Pr[X = 3/2] = 0.$$



          If we define $Y$ as in the body of the post, i.e. $$Y = begin{cases} 2X, & X le 3 \ X, & X ge 4, end{cases}$$ then we note that
          $$begin{align*}operatorname{E}[Y^2]
          &= sum_{x=0}^3 (2x)^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
          &= 4 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
          &= 3 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + left(sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} right) \
          &= 3 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=0}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
          &= e^{-lambda} (3lambda + 6 lambda^2 + tfrac{9}{2} lambda^3) + operatorname{E}[X^2] \
          &= 66e^{-2} + operatorname{E}[X]^2 + operatorname{Var}[X] \
          &= 66e^{-2} + lambda^2 + lambda \
          &= 6 + 66e^{-2}.
          end{align*}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Could you explain to me why $ (2x)^2 $ becomes $ 3x^2 $ ?
            $endgroup$
            – bm1125
            Dec 9 '18 at 17:02










          • $begingroup$
            @bm1125 I have expanded the details of that calculation in my edit, but simply put, $4 = 3 + 1.$
            $endgroup$
            – heropup
            Dec 9 '18 at 17:07














          1












          1








          1





          $begingroup$

          $Y$ cannot be Poisson, because for example, $$Pr[Y = 3] = Pr[2X = 3] = Pr[X = 3/2] = 0.$$



          If we define $Y$ as in the body of the post, i.e. $$Y = begin{cases} 2X, & X le 3 \ X, & X ge 4, end{cases}$$ then we note that
          $$begin{align*}operatorname{E}[Y^2]
          &= sum_{x=0}^3 (2x)^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
          &= 4 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
          &= 3 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + left(sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} right) \
          &= 3 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=0}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
          &= e^{-lambda} (3lambda + 6 lambda^2 + tfrac{9}{2} lambda^3) + operatorname{E}[X^2] \
          &= 66e^{-2} + operatorname{E}[X]^2 + operatorname{Var}[X] \
          &= 66e^{-2} + lambda^2 + lambda \
          &= 6 + 66e^{-2}.
          end{align*}$$






          share|cite|improve this answer











          $endgroup$



          $Y$ cannot be Poisson, because for example, $$Pr[Y = 3] = Pr[2X = 3] = Pr[X = 3/2] = 0.$$



          If we define $Y$ as in the body of the post, i.e. $$Y = begin{cases} 2X, & X le 3 \ X, & X ge 4, end{cases}$$ then we note that
          $$begin{align*}operatorname{E}[Y^2]
          &= sum_{x=0}^3 (2x)^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
          &= 4 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
          &= 3 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + left(sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} right) \
          &= 3 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=0}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
          &= e^{-lambda} (3lambda + 6 lambda^2 + tfrac{9}{2} lambda^3) + operatorname{E}[X^2] \
          &= 66e^{-2} + operatorname{E}[X]^2 + operatorname{Var}[X] \
          &= 66e^{-2} + lambda^2 + lambda \
          &= 6 + 66e^{-2}.
          end{align*}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 9 '18 at 17:06

























          answered Dec 9 '18 at 16:54









          heropupheropup

          63.8k762102




          63.8k762102












          • $begingroup$
            Could you explain to me why $ (2x)^2 $ becomes $ 3x^2 $ ?
            $endgroup$
            – bm1125
            Dec 9 '18 at 17:02










          • $begingroup$
            @bm1125 I have expanded the details of that calculation in my edit, but simply put, $4 = 3 + 1.$
            $endgroup$
            – heropup
            Dec 9 '18 at 17:07


















          • $begingroup$
            Could you explain to me why $ (2x)^2 $ becomes $ 3x^2 $ ?
            $endgroup$
            – bm1125
            Dec 9 '18 at 17:02










          • $begingroup$
            @bm1125 I have expanded the details of that calculation in my edit, but simply put, $4 = 3 + 1.$
            $endgroup$
            – heropup
            Dec 9 '18 at 17:07
















          $begingroup$
          Could you explain to me why $ (2x)^2 $ becomes $ 3x^2 $ ?
          $endgroup$
          – bm1125
          Dec 9 '18 at 17:02




          $begingroup$
          Could you explain to me why $ (2x)^2 $ becomes $ 3x^2 $ ?
          $endgroup$
          – bm1125
          Dec 9 '18 at 17:02












          $begingroup$
          @bm1125 I have expanded the details of that calculation in my edit, but simply put, $4 = 3 + 1.$
          $endgroup$
          – heropup
          Dec 9 '18 at 17:07




          $begingroup$
          @bm1125 I have expanded the details of that calculation in my edit, but simply put, $4 = 3 + 1.$
          $endgroup$
          – heropup
          Dec 9 '18 at 17:07











          1












          $begingroup$

          begin{align}
          E[Y^2]
          &= sum_{x=0}^3 (2x)^2 Pr(X=x)+ sum_{x=4}^infty x^2 Pr(X=x)\
          &=sum_{x=0}^3 4x^2 Pr(X=x)+ sum_{x=4}^infty x^2 Pr(X=x)\
          &=sum_{x=0}^3 3x^2 Pr(X=x)+ sum_{x=0}^infty x^2 Pr(X=x)\
          &=3sum_{x=1}^3 x^2 Pr(X=x)+ E[X^2]\
          &= 3left( Pr(X=1) + 4Pr(X=2)+9Pr(X=3) right) + E[X^2]
          end{align}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for your answer! could you explain the second equality? I think you also miss the fact that $ P{X=6} not = P{Y=6} $ as $ P{Y=6} = P{X=3} + P{X=6} $ I believe.
            $endgroup$
            – bm1125
            Dec 9 '18 at 16:55






          • 1




            $begingroup$
            All of the summations should be with respect to $x$.
            $endgroup$
            – heropup
            Dec 9 '18 at 17:00










          • $begingroup$
            oops, i made lots of typos, thanks.
            $endgroup$
            – Siong Thye Goh
            Dec 9 '18 at 17:00










          • $begingroup$
            I use the law of the unconscious statistician $E[g(X)]=sum_x g(x)P(X=x)$. For the line from $4x^$ to $3x^2$, we pass a copy of $x^2$ to the second term, hence there is change of the index.
            $endgroup$
            – Siong Thye Goh
            Dec 9 '18 at 17:02


















          1












          $begingroup$

          begin{align}
          E[Y^2]
          &= sum_{x=0}^3 (2x)^2 Pr(X=x)+ sum_{x=4}^infty x^2 Pr(X=x)\
          &=sum_{x=0}^3 4x^2 Pr(X=x)+ sum_{x=4}^infty x^2 Pr(X=x)\
          &=sum_{x=0}^3 3x^2 Pr(X=x)+ sum_{x=0}^infty x^2 Pr(X=x)\
          &=3sum_{x=1}^3 x^2 Pr(X=x)+ E[X^2]\
          &= 3left( Pr(X=1) + 4Pr(X=2)+9Pr(X=3) right) + E[X^2]
          end{align}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for your answer! could you explain the second equality? I think you also miss the fact that $ P{X=6} not = P{Y=6} $ as $ P{Y=6} = P{X=3} + P{X=6} $ I believe.
            $endgroup$
            – bm1125
            Dec 9 '18 at 16:55






          • 1




            $begingroup$
            All of the summations should be with respect to $x$.
            $endgroup$
            – heropup
            Dec 9 '18 at 17:00










          • $begingroup$
            oops, i made lots of typos, thanks.
            $endgroup$
            – Siong Thye Goh
            Dec 9 '18 at 17:00










          • $begingroup$
            I use the law of the unconscious statistician $E[g(X)]=sum_x g(x)P(X=x)$. For the line from $4x^$ to $3x^2$, we pass a copy of $x^2$ to the second term, hence there is change of the index.
            $endgroup$
            – Siong Thye Goh
            Dec 9 '18 at 17:02
















          1












          1








          1





          $begingroup$

          begin{align}
          E[Y^2]
          &= sum_{x=0}^3 (2x)^2 Pr(X=x)+ sum_{x=4}^infty x^2 Pr(X=x)\
          &=sum_{x=0}^3 4x^2 Pr(X=x)+ sum_{x=4}^infty x^2 Pr(X=x)\
          &=sum_{x=0}^3 3x^2 Pr(X=x)+ sum_{x=0}^infty x^2 Pr(X=x)\
          &=3sum_{x=1}^3 x^2 Pr(X=x)+ E[X^2]\
          &= 3left( Pr(X=1) + 4Pr(X=2)+9Pr(X=3) right) + E[X^2]
          end{align}






          share|cite|improve this answer











          $endgroup$



          begin{align}
          E[Y^2]
          &= sum_{x=0}^3 (2x)^2 Pr(X=x)+ sum_{x=4}^infty x^2 Pr(X=x)\
          &=sum_{x=0}^3 4x^2 Pr(X=x)+ sum_{x=4}^infty x^2 Pr(X=x)\
          &=sum_{x=0}^3 3x^2 Pr(X=x)+ sum_{x=0}^infty x^2 Pr(X=x)\
          &=3sum_{x=1}^3 x^2 Pr(X=x)+ E[X^2]\
          &= 3left( Pr(X=1) + 4Pr(X=2)+9Pr(X=3) right) + E[X^2]
          end{align}







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 9 '18 at 17:00

























          answered Dec 9 '18 at 16:46









          Siong Thye GohSiong Thye Goh

          101k1466118




          101k1466118












          • $begingroup$
            Thanks for your answer! could you explain the second equality? I think you also miss the fact that $ P{X=6} not = P{Y=6} $ as $ P{Y=6} = P{X=3} + P{X=6} $ I believe.
            $endgroup$
            – bm1125
            Dec 9 '18 at 16:55






          • 1




            $begingroup$
            All of the summations should be with respect to $x$.
            $endgroup$
            – heropup
            Dec 9 '18 at 17:00










          • $begingroup$
            oops, i made lots of typos, thanks.
            $endgroup$
            – Siong Thye Goh
            Dec 9 '18 at 17:00










          • $begingroup$
            I use the law of the unconscious statistician $E[g(X)]=sum_x g(x)P(X=x)$. For the line from $4x^$ to $3x^2$, we pass a copy of $x^2$ to the second term, hence there is change of the index.
            $endgroup$
            – Siong Thye Goh
            Dec 9 '18 at 17:02




















          • $begingroup$
            Thanks for your answer! could you explain the second equality? I think you also miss the fact that $ P{X=6} not = P{Y=6} $ as $ P{Y=6} = P{X=3} + P{X=6} $ I believe.
            $endgroup$
            – bm1125
            Dec 9 '18 at 16:55






          • 1




            $begingroup$
            All of the summations should be with respect to $x$.
            $endgroup$
            – heropup
            Dec 9 '18 at 17:00










          • $begingroup$
            oops, i made lots of typos, thanks.
            $endgroup$
            – Siong Thye Goh
            Dec 9 '18 at 17:00










          • $begingroup$
            I use the law of the unconscious statistician $E[g(X)]=sum_x g(x)P(X=x)$. For the line from $4x^$ to $3x^2$, we pass a copy of $x^2$ to the second term, hence there is change of the index.
            $endgroup$
            – Siong Thye Goh
            Dec 9 '18 at 17:02


















          $begingroup$
          Thanks for your answer! could you explain the second equality? I think you also miss the fact that $ P{X=6} not = P{Y=6} $ as $ P{Y=6} = P{X=3} + P{X=6} $ I believe.
          $endgroup$
          – bm1125
          Dec 9 '18 at 16:55




          $begingroup$
          Thanks for your answer! could you explain the second equality? I think you also miss the fact that $ P{X=6} not = P{Y=6} $ as $ P{Y=6} = P{X=3} + P{X=6} $ I believe.
          $endgroup$
          – bm1125
          Dec 9 '18 at 16:55




          1




          1




          $begingroup$
          All of the summations should be with respect to $x$.
          $endgroup$
          – heropup
          Dec 9 '18 at 17:00




          $begingroup$
          All of the summations should be with respect to $x$.
          $endgroup$
          – heropup
          Dec 9 '18 at 17:00












          $begingroup$
          oops, i made lots of typos, thanks.
          $endgroup$
          – Siong Thye Goh
          Dec 9 '18 at 17:00




          $begingroup$
          oops, i made lots of typos, thanks.
          $endgroup$
          – Siong Thye Goh
          Dec 9 '18 at 17:00












          $begingroup$
          I use the law of the unconscious statistician $E[g(X)]=sum_x g(x)P(X=x)$. For the line from $4x^$ to $3x^2$, we pass a copy of $x^2$ to the second term, hence there is change of the index.
          $endgroup$
          – Siong Thye Goh
          Dec 9 '18 at 17:02






          $begingroup$
          I use the law of the unconscious statistician $E[g(X)]=sum_x g(x)P(X=x)$. For the line from $4x^$ to $3x^2$, we pass a copy of $x^2$ to the second term, hence there is change of the index.
          $endgroup$
          – Siong Thye Goh
          Dec 9 '18 at 17:02




















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