Statistics: Point estimation












-1












$begingroup$


If we have a sample of $x=2$ from a $Po(6 cdot lambda)$ distribution.



How do we calculate $lambda*$ and $d(lambda*)$?



I think that $lambda* = frac{2}{6} = frac{1}{3} $ but I am not sure about the other one.






statistics parameter-estimation







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is $lambda*$?
    $endgroup$
    – Sean Roberson
    Dec 4 '18 at 18:04










  • $begingroup$
    could you define $lambda*$ and $d(lambda*)$ more specifically?
    $endgroup$
    – MoonKnight
    Dec 4 '18 at 18:05










  • $begingroup$
    $lambda*$ is the parameter that we estimate and $d(lambda*)$ is the standard deviation.
    $endgroup$
    – Clone
    Dec 4 '18 at 18:20
















-1












$begingroup$


If we have a sample of $x=2$ from a $Po(6 cdot lambda)$ distribution.



How do we calculate $lambda*$ and $d(lambda*)$?



I think that $lambda* = frac{2}{6} = frac{1}{3} $ but I am not sure about the other one.






statistics parameter-estimation







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is $lambda*$?
    $endgroup$
    – Sean Roberson
    Dec 4 '18 at 18:04










  • $begingroup$
    could you define $lambda*$ and $d(lambda*)$ more specifically?
    $endgroup$
    – MoonKnight
    Dec 4 '18 at 18:05










  • $begingroup$
    $lambda*$ is the parameter that we estimate and $d(lambda*)$ is the standard deviation.
    $endgroup$
    – Clone
    Dec 4 '18 at 18:20














-1












-1








-1


1



$begingroup$


If we have a sample of $x=2$ from a $Po(6 cdot lambda)$ distribution.



How do we calculate $lambda*$ and $d(lambda*)$?



I think that $lambda* = frac{2}{6} = frac{1}{3} $ but I am not sure about the other one.






statistics parameter-estimation







share|cite|improve this question









$endgroup$




If we have a sample of $x=2$ from a $Po(6 cdot lambda)$ distribution.



How do we calculate $lambda*$ and $d(lambda*)$?



I think that $lambda* = frac{2}{6} = frac{1}{3} $ but I am not sure about the other one.






statistics parameter-estimation




statistics parameter-estimation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 4 '18 at 17:57









CloneClone

99




99












  • $begingroup$
    What is $lambda*$?
    $endgroup$
    – Sean Roberson
    Dec 4 '18 at 18:04










  • $begingroup$
    could you define $lambda*$ and $d(lambda*)$ more specifically?
    $endgroup$
    – MoonKnight
    Dec 4 '18 at 18:05










  • $begingroup$
    $lambda*$ is the parameter that we estimate and $d(lambda*)$ is the standard deviation.
    $endgroup$
    – Clone
    Dec 4 '18 at 18:20


















  • $begingroup$
    What is $lambda*$?
    $endgroup$
    – Sean Roberson
    Dec 4 '18 at 18:04










  • $begingroup$
    could you define $lambda*$ and $d(lambda*)$ more specifically?
    $endgroup$
    – MoonKnight
    Dec 4 '18 at 18:05










  • $begingroup$
    $lambda*$ is the parameter that we estimate and $d(lambda*)$ is the standard deviation.
    $endgroup$
    – Clone
    Dec 4 '18 at 18:20
















$begingroup$
What is $lambda*$?
$endgroup$
– Sean Roberson
Dec 4 '18 at 18:04




$begingroup$
What is $lambda*$?
$endgroup$
– Sean Roberson
Dec 4 '18 at 18:04












$begingroup$
could you define $lambda*$ and $d(lambda*)$ more specifically?
$endgroup$
– MoonKnight
Dec 4 '18 at 18:05




$begingroup$
could you define $lambda*$ and $d(lambda*)$ more specifically?
$endgroup$
– MoonKnight
Dec 4 '18 at 18:05












$begingroup$
$lambda*$ is the parameter that we estimate and $d(lambda*)$ is the standard deviation.
$endgroup$
– Clone
Dec 4 '18 at 18:20




$begingroup$
$lambda*$ is the parameter that we estimate and $d(lambda*)$ is the standard deviation.
$endgroup$
– Clone
Dec 4 '18 at 18:20










1 Answer
1






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oldest

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0












$begingroup$

Whatever you may mean by $lambda *$ and $d(lambda *),$ which I suppose are defined in your text or notes, the following
information may be helpful.



If $X_1$ and $X_2$ are independently $mathsf{Pois}(6lambda),$ Then $$S = X_1 + X_2 sim mathsf{Pois}(12lambda).$$ Then $E(S) = 12lambda$ and $Var(S) = 12lambda.$



So $S$ is a point estimate of $12 lambda$ and
$S/12$ is a point estimate of $lambda.$






share|cite|improve this answer









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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    0












    $begingroup$

    Whatever you may mean by $lambda *$ and $d(lambda *),$ which I suppose are defined in your text or notes, the following
    information may be helpful.



    If $X_1$ and $X_2$ are independently $mathsf{Pois}(6lambda),$ Then $$S = X_1 + X_2 sim mathsf{Pois}(12lambda).$$ Then $E(S) = 12lambda$ and $Var(S) = 12lambda.$



    So $S$ is a point estimate of $12 lambda$ and
    $S/12$ is a point estimate of $lambda.$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Whatever you may mean by $lambda *$ and $d(lambda *),$ which I suppose are defined in your text or notes, the following
      information may be helpful.



      If $X_1$ and $X_2$ are independently $mathsf{Pois}(6lambda),$ Then $$S = X_1 + X_2 sim mathsf{Pois}(12lambda).$$ Then $E(S) = 12lambda$ and $Var(S) = 12lambda.$



      So $S$ is a point estimate of $12 lambda$ and
      $S/12$ is a point estimate of $lambda.$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Whatever you may mean by $lambda *$ and $d(lambda *),$ which I suppose are defined in your text or notes, the following
        information may be helpful.



        If $X_1$ and $X_2$ are independently $mathsf{Pois}(6lambda),$ Then $$S = X_1 + X_2 sim mathsf{Pois}(12lambda).$$ Then $E(S) = 12lambda$ and $Var(S) = 12lambda.$



        So $S$ is a point estimate of $12 lambda$ and
        $S/12$ is a point estimate of $lambda.$






        share|cite|improve this answer









        $endgroup$



        Whatever you may mean by $lambda *$ and $d(lambda *),$ which I suppose are defined in your text or notes, the following
        information may be helpful.



        If $X_1$ and $X_2$ are independently $mathsf{Pois}(6lambda),$ Then $$S = X_1 + X_2 sim mathsf{Pois}(12lambda).$$ Then $E(S) = 12lambda$ and $Var(S) = 12lambda.$



        So $S$ is a point estimate of $12 lambda$ and
        $S/12$ is a point estimate of $lambda.$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 20:44









        BruceETBruceET

        35.3k71440




        35.3k71440






























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