Compute $ P(X_i | sum X_i = 45) $ where $ X_i sim mathrm{Pois}$Independence of Poisson random variables...

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Compute $ P(X_i | sum X_i = 45) $ where $ X_i sim mathrm{Pois}$


Independence of Poisson random variables coming from Poisson samplingContradiction when applying CLT to Poisson random variables?A sum of a random number of Poisson random variablesConditional Distribution $X_1sim {rm Poi}(mu_1)mid sum X_i$Binomial and Poisson DistributionsConverse statements for convolutions of probability distributionsShow that $(X_1 mid X_1 + X_2 + dots + X_k)$ has binomial distribution, where $X_i$ has Poisson distribution $lambda_i$Compute $ P{X_1 = 3 | X = 10 } $, $ X sim Pois(10) $$ X_i $ is discrete random variable, Compute $ sum X_i = 97 $Marginal distribution of $X$ when $X|m sim Pois(m)$ and $M sim Gamma (2,1) $













2












$begingroup$


Let $ X_i sim mathrm{Pois}(10) $ be independent random variables $ i = 1, 2,3,4,5 $ .



What is the probability of $ X_1 = 9 $ given $ sum X_i = 45 $



now $ sum X_i sim mathrm{Pois}(50) $ and so I thought



$$ P(X_1 = 9 | sum_{i=1}^5 X_i = 45 ) = P(X_1 =9 | sum_{i=2}^5X_i = 36) = frac{P(X_1 =9 , sum_{i=2}^5 X_i = 36)}{sum_{i=1}^5 X_i = 45 } = frac{e^{-5}cdot frac{5^9}{9!}cdot e^{-40}cdotfrac{40^{36}}{36!}}{e^{-50}cdotfrac{50^{45}}{45!}} = frac{0.03626 cdot 0.053939}{0.0458} = 0.04269 $$



But apparently that is wrong and the solution is using binomial distribution with parameters $ n = 45, p = 1/5 $ and final answer is $ 0.14724 $



Is there anyway I can solve it with the Poisson distribution?






probability poisson-distribution







share|cite|improve this question











$endgroup$












  • $begingroup$
    You can't change the condition to $sumlimits_{i=2}^{5}X_i = 36$. However, it is true that $$Pleft(X_1 = 9, Bigg{lvert},sumlimits_{i=1}^{5}X_i = 45right) =color{blue}{frac{Pleft(X_1 = 9, sumlimits_{i=2}^{5}X_i = 36 right)}{Pleft(sumlimits_{i=1}^{5}X_i = 45right)}}$$ (this is different to $Pleft(X_1 = 9,Bigg{lvert} ,sumlimits_{i=2}^{5}X_i = 36right)$). What did you get when you tried to calculate that ratio of probabilities?
    $endgroup$
    – Minus One-Twelfth
    Mar 19 at 23:58












  • $begingroup$
    Thanks for the answer! I got $ 0.0427 $
    $endgroup$
    – bm1125
    Mar 20 at 0:05






  • 1




    $begingroup$
    Maybe show us your working out that led to that answer? That way we can see if there was anything wrong with it.
    $endgroup$
    – Minus One-Twelfth
    Mar 20 at 0:06






  • 1




    $begingroup$
    Note that $5times 5 =25$, not $50$.
    $endgroup$
    – Minus One-Twelfth
    Mar 20 at 0:14








  • 1




    $begingroup$
    OK, well, that changes your calculations, doesn't it?
    $endgroup$
    – Gerry Myerson
    Mar 20 at 0:16
















2












$begingroup$


Let $ X_i sim mathrm{Pois}(10) $ be independent random variables $ i = 1, 2,3,4,5 $ .



What is the probability of $ X_1 = 9 $ given $ sum X_i = 45 $



now $ sum X_i sim mathrm{Pois}(50) $ and so I thought



$$ P(X_1 = 9 | sum_{i=1}^5 X_i = 45 ) = P(X_1 =9 | sum_{i=2}^5X_i = 36) = frac{P(X_1 =9 , sum_{i=2}^5 X_i = 36)}{sum_{i=1}^5 X_i = 45 } = frac{e^{-5}cdot frac{5^9}{9!}cdot e^{-40}cdotfrac{40^{36}}{36!}}{e^{-50}cdotfrac{50^{45}}{45!}} = frac{0.03626 cdot 0.053939}{0.0458} = 0.04269 $$



But apparently that is wrong and the solution is using binomial distribution with parameters $ n = 45, p = 1/5 $ and final answer is $ 0.14724 $



Is there anyway I can solve it with the Poisson distribution?






probability poisson-distribution







share|cite|improve this question











$endgroup$












  • $begingroup$
    You can't change the condition to $sumlimits_{i=2}^{5}X_i = 36$. However, it is true that $$Pleft(X_1 = 9, Bigg{lvert},sumlimits_{i=1}^{5}X_i = 45right) =color{blue}{frac{Pleft(X_1 = 9, sumlimits_{i=2}^{5}X_i = 36 right)}{Pleft(sumlimits_{i=1}^{5}X_i = 45right)}}$$ (this is different to $Pleft(X_1 = 9,Bigg{lvert} ,sumlimits_{i=2}^{5}X_i = 36right)$). What did you get when you tried to calculate that ratio of probabilities?
    $endgroup$
    – Minus One-Twelfth
    Mar 19 at 23:58












  • $begingroup$
    Thanks for the answer! I got $ 0.0427 $
    $endgroup$
    – bm1125
    Mar 20 at 0:05






  • 1




    $begingroup$
    Maybe show us your working out that led to that answer? That way we can see if there was anything wrong with it.
    $endgroup$
    – Minus One-Twelfth
    Mar 20 at 0:06






  • 1




    $begingroup$
    Note that $5times 5 =25$, not $50$.
    $endgroup$
    – Minus One-Twelfth
    Mar 20 at 0:14








  • 1




    $begingroup$
    OK, well, that changes your calculations, doesn't it?
    $endgroup$
    – Gerry Myerson
    Mar 20 at 0:16














2












2








2





$begingroup$


Let $ X_i sim mathrm{Pois}(10) $ be independent random variables $ i = 1, 2,3,4,5 $ .



What is the probability of $ X_1 = 9 $ given $ sum X_i = 45 $



now $ sum X_i sim mathrm{Pois}(50) $ and so I thought



$$ P(X_1 = 9 | sum_{i=1}^5 X_i = 45 ) = P(X_1 =9 | sum_{i=2}^5X_i = 36) = frac{P(X_1 =9 , sum_{i=2}^5 X_i = 36)}{sum_{i=1}^5 X_i = 45 } = frac{e^{-5}cdot frac{5^9}{9!}cdot e^{-40}cdotfrac{40^{36}}{36!}}{e^{-50}cdotfrac{50^{45}}{45!}} = frac{0.03626 cdot 0.053939}{0.0458} = 0.04269 $$



But apparently that is wrong and the solution is using binomial distribution with parameters $ n = 45, p = 1/5 $ and final answer is $ 0.14724 $



Is there anyway I can solve it with the Poisson distribution?






probability poisson-distribution







share|cite|improve this question











$endgroup$




Let $ X_i sim mathrm{Pois}(10) $ be independent random variables $ i = 1, 2,3,4,5 $ .



What is the probability of $ X_1 = 9 $ given $ sum X_i = 45 $



now $ sum X_i sim mathrm{Pois}(50) $ and so I thought



$$ P(X_1 = 9 | sum_{i=1}^5 X_i = 45 ) = P(X_1 =9 | sum_{i=2}^5X_i = 36) = frac{P(X_1 =9 , sum_{i=2}^5 X_i = 36)}{sum_{i=1}^5 X_i = 45 } = frac{e^{-5}cdot frac{5^9}{9!}cdot e^{-40}cdotfrac{40^{36}}{36!}}{e^{-50}cdotfrac{50^{45}}{45!}} = frac{0.03626 cdot 0.053939}{0.0458} = 0.04269 $$



But apparently that is wrong and the solution is using binomial distribution with parameters $ n = 45, p = 1/5 $ and final answer is $ 0.14724 $



Is there anyway I can solve it with the Poisson distribution?






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 Mar 20 at 0:14







bm1125

















asked Mar 19 at 23:51









bm1125bm1125

68516




68516












  • $begingroup$
    You can't change the condition to $sumlimits_{i=2}^{5}X_i = 36$. However, it is true that $$Pleft(X_1 = 9, Bigg{lvert},sumlimits_{i=1}^{5}X_i = 45right) =color{blue}{frac{Pleft(X_1 = 9, sumlimits_{i=2}^{5}X_i = 36 right)}{Pleft(sumlimits_{i=1}^{5}X_i = 45right)}}$$ (this is different to $Pleft(X_1 = 9,Bigg{lvert} ,sumlimits_{i=2}^{5}X_i = 36right)$). What did you get when you tried to calculate that ratio of probabilities?
    $endgroup$
    – Minus One-Twelfth
    Mar 19 at 23:58












  • $begingroup$
    Thanks for the answer! I got $ 0.0427 $
    $endgroup$
    – bm1125
    Mar 20 at 0:05






  • 1




    $begingroup$
    Maybe show us your working out that led to that answer? That way we can see if there was anything wrong with it.
    $endgroup$
    – Minus One-Twelfth
    Mar 20 at 0:06






  • 1




    $begingroup$
    Note that $5times 5 =25$, not $50$.
    $endgroup$
    – Minus One-Twelfth
    Mar 20 at 0:14








  • 1




    $begingroup$
    OK, well, that changes your calculations, doesn't it?
    $endgroup$
    – Gerry Myerson
    Mar 20 at 0:16


















  • $begingroup$
    You can't change the condition to $sumlimits_{i=2}^{5}X_i = 36$. However, it is true that $$Pleft(X_1 = 9, Bigg{lvert},sumlimits_{i=1}^{5}X_i = 45right) =color{blue}{frac{Pleft(X_1 = 9, sumlimits_{i=2}^{5}X_i = 36 right)}{Pleft(sumlimits_{i=1}^{5}X_i = 45right)}}$$ (this is different to $Pleft(X_1 = 9,Bigg{lvert} ,sumlimits_{i=2}^{5}X_i = 36right)$). What did you get when you tried to calculate that ratio of probabilities?
    $endgroup$
    – Minus One-Twelfth
    Mar 19 at 23:58












  • $begingroup$
    Thanks for the answer! I got $ 0.0427 $
    $endgroup$
    – bm1125
    Mar 20 at 0:05






  • 1




    $begingroup$
    Maybe show us your working out that led to that answer? That way we can see if there was anything wrong with it.
    $endgroup$
    – Minus One-Twelfth
    Mar 20 at 0:06






  • 1




    $begingroup$
    Note that $5times 5 =25$, not $50$.
    $endgroup$
    – Minus One-Twelfth
    Mar 20 at 0:14








  • 1




    $begingroup$
    OK, well, that changes your calculations, doesn't it?
    $endgroup$
    – Gerry Myerson
    Mar 20 at 0:16
















$begingroup$
You can't change the condition to $sumlimits_{i=2}^{5}X_i = 36$. However, it is true that $$Pleft(X_1 = 9, Bigg{lvert},sumlimits_{i=1}^{5}X_i = 45right) =color{blue}{frac{Pleft(X_1 = 9, sumlimits_{i=2}^{5}X_i = 36 right)}{Pleft(sumlimits_{i=1}^{5}X_i = 45right)}}$$ (this is different to $Pleft(X_1 = 9,Bigg{lvert} ,sumlimits_{i=2}^{5}X_i = 36right)$). What did you get when you tried to calculate that ratio of probabilities?
$endgroup$
– Minus One-Twelfth
Mar 19 at 23:58






$begingroup$
You can't change the condition to $sumlimits_{i=2}^{5}X_i = 36$. However, it is true that $$Pleft(X_1 = 9, Bigg{lvert},sumlimits_{i=1}^{5}X_i = 45right) =color{blue}{frac{Pleft(X_1 = 9, sumlimits_{i=2}^{5}X_i = 36 right)}{Pleft(sumlimits_{i=1}^{5}X_i = 45right)}}$$ (this is different to $Pleft(X_1 = 9,Bigg{lvert} ,sumlimits_{i=2}^{5}X_i = 36right)$). What did you get when you tried to calculate that ratio of probabilities?
$endgroup$
– Minus One-Twelfth
Mar 19 at 23:58














$begingroup$
Thanks for the answer! I got $ 0.0427 $
$endgroup$
– bm1125
Mar 20 at 0:05




$begingroup$
Thanks for the answer! I got $ 0.0427 $
$endgroup$
– bm1125
Mar 20 at 0:05




1




1




$begingroup$
Maybe show us your working out that led to that answer? That way we can see if there was anything wrong with it.
$endgroup$
– Minus One-Twelfth
Mar 20 at 0:06




$begingroup$
Maybe show us your working out that led to that answer? That way we can see if there was anything wrong with it.
$endgroup$
– Minus One-Twelfth
Mar 20 at 0:06




1




1




$begingroup$
Note that $5times 5 =25$, not $50$.
$endgroup$
– Minus One-Twelfth
Mar 20 at 0:14






$begingroup$
Note that $5times 5 =25$, not $50$.
$endgroup$
– Minus One-Twelfth
Mar 20 at 0:14






1




1




$begingroup$
OK, well, that changes your calculations, doesn't it?
$endgroup$
– Gerry Myerson
Mar 20 at 0:16




$begingroup$
OK, well, that changes your calculations, doesn't it?
$endgroup$
– Gerry Myerson
Mar 20 at 0:16










2 Answers
2






active

oldest

votes


















1












$begingroup$

It is because the sum of independent Poisson random variables is a Poisson random variables whose mean is the sum of their means, and all $X_i$ are independent Poisson random variables whose mean is $10$, that we have the following.
$$X_1simmathcal{Pois}(10)\sum_{i=2}^5X_isimmathcal{Pois}(40)\sum_{i=1}^5X_isimmathcal{Pois}(50)$$
(Emphasis because the fact of independence is crucial to this solution.)



Also, if we define $Y_lambdasimmathcal {Pois}(lambda)$ then we have $~mathsf P(Y_lambda{=}y)~=~dfrac{lambda^ymathsf e^lambda}{y!}mathbf 1_{yinBbb N}$, and we can use this to ensure we are substituting the correct values.



So, therefore, we get the following.



$$begin{align}mathsf P(X_1{=}9mid sum_{i=1}^5X_i{=}45)&=dfrac{mathsf P(X_1{=}9)cdotmathsf P(sum_{i=2}^5 X_i{=}36)}{mathsf P(sum_{i=1}^5 X_i{=}45)}\[1ex]&=dfrac{mathsf P(Y_{10}{=}9)cdotmathsf P(Y_{40}{=}36)}{mathsf P(Y_{50}{=}45)}\[1ex]&=dfrac{dfrac{10^9mathsf e^{10}}{9!}cdotdfrac{40^{36}mathsf e^{40}}{36!}}{dfrac{50^{45}mathsf e^{50}}{45!}}\[1ex]&=dbinom{45}9 dfrac{10^9cdot 40^{36}}{50^{45}}end{align}$$



Which is where the binomial distribution comes from.



[The count of point events which occur in one among several independent Poisson processes has a conditionally binomial distribution when given the total count of events that occur among all of them, with success rate equal to the ratio of the sums of the Poisson rate parameters (favored versus total).]






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
    newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
    newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
    newcommand{dd}{mathrm{d}}
    newcommand{ds}[1]{displaystyle{#1}}
    newcommand{expo}[1]{,mathrm{e}^{#1},}
    newcommand{ic}{mathrm{i}}
    newcommand{mc}[1]{mathcal{#1}}
    newcommand{mrm}[1]{mathrm{#1}}
    newcommand{pars}[1]{left(,{#1},right)}
    newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
    newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
    newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
    newcommand{verts}[1]{leftvert,{#1},rightvert}$




    Let $ds{mc{P}pars{lambda,n} = {expo{-lambda}lambda^{n} over n!}}$.




    begin{align}
    &bbox[10px,#ffd]{sum_{x_{1} = 0}^{infty}mc{P}pars{lambda,x_{1}}
    !!sum_{x_{2} = 0}^{infty}mc{P}pars{lambda,x_{2}}
    !!sum_{x_{3} = 0}^{infty}mc{P}pars{lambda,x_{3}}
    !!sum_{x_{4} = 0}^{infty}mc{P}pars{lambda,x_{4}}
    !!sum_{x_{5} = 0}^{infty}mc{P}pars{lambda,x_{5}}}
    times
    \[2mm] & bracks{x_{1} = 9}
    bracks{z^{45}}! z^{sum_{i = 1}^{5}x_{i}}
    \[5mm] = &
    mc{P}pars{lambda,9}bracks{z^{45}}z^{9}
    bracks{sum_{x = 0}^{infty}mc{P}pars{x}z^{x}}^{4} =
    mc{P}pars{lambda,9}bracks{z^{36}}
    bracks{sum_{x = 0}^{infty}{expo{-lambda}lambda^{x} over x!}z^{x}}^{4}
    \[5mm] = &
    mc{P}pars{lambda,9}expo{-4lambda}bracks{z^{36}}
    bracks{sum_{x = 0}^{infty}{pars{lambda z}^{x} over x!}}^{4} =
    mc{P}pars{lambda,9}expo{-4lambda}bracks{z^{36}}
    expo{4lambda z}
    \[5mm] = &
    mc{P}pars{lambda,9}expo{-4lambda}{pars{4lambda}^{36} over 36!} =
    {expo{-lambda}lambda^{9} over 9!}expo{-4lambda}{4^{36}lambda^{36} over 36!}
    \[5mm] = &
    bbx{{4^{36} over 9!, 36!}
    ,expo{-5lambda}lambda^{45}}
    end{align}




    $ds{lambda = 10 implies approx 6.7474 times 10^{-3}}$.







    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Unfortunately, OP just edited the question, and wants Pois(10), not Pois(5).
      $endgroup$
      – Gerry Myerson
      Mar 20 at 0:17






    • 2




      $begingroup$
      $Hugeleft)smileright($
      $endgroup$
      – Felix Marin
      Mar 20 at 0:20














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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    1












    $begingroup$

    It is because the sum of independent Poisson random variables is a Poisson random variables whose mean is the sum of their means, and all $X_i$ are independent Poisson random variables whose mean is $10$, that we have the following.
    $$X_1simmathcal{Pois}(10)\sum_{i=2}^5X_isimmathcal{Pois}(40)\sum_{i=1}^5X_isimmathcal{Pois}(50)$$
    (Emphasis because the fact of independence is crucial to this solution.)



    Also, if we define $Y_lambdasimmathcal {Pois}(lambda)$ then we have $~mathsf P(Y_lambda{=}y)~=~dfrac{lambda^ymathsf e^lambda}{y!}mathbf 1_{yinBbb N}$, and we can use this to ensure we are substituting the correct values.



    So, therefore, we get the following.



    $$begin{align}mathsf P(X_1{=}9mid sum_{i=1}^5X_i{=}45)&=dfrac{mathsf P(X_1{=}9)cdotmathsf P(sum_{i=2}^5 X_i{=}36)}{mathsf P(sum_{i=1}^5 X_i{=}45)}\[1ex]&=dfrac{mathsf P(Y_{10}{=}9)cdotmathsf P(Y_{40}{=}36)}{mathsf P(Y_{50}{=}45)}\[1ex]&=dfrac{dfrac{10^9mathsf e^{10}}{9!}cdotdfrac{40^{36}mathsf e^{40}}{36!}}{dfrac{50^{45}mathsf e^{50}}{45!}}\[1ex]&=dbinom{45}9 dfrac{10^9cdot 40^{36}}{50^{45}}end{align}$$



    Which is where the binomial distribution comes from.



    [The count of point events which occur in one among several independent Poisson processes has a conditionally binomial distribution when given the total count of events that occur among all of them, with success rate equal to the ratio of the sums of the Poisson rate parameters (favored versus total).]






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      It is because the sum of independent Poisson random variables is a Poisson random variables whose mean is the sum of their means, and all $X_i$ are independent Poisson random variables whose mean is $10$, that we have the following.
      $$X_1simmathcal{Pois}(10)\sum_{i=2}^5X_isimmathcal{Pois}(40)\sum_{i=1}^5X_isimmathcal{Pois}(50)$$
      (Emphasis because the fact of independence is crucial to this solution.)



      Also, if we define $Y_lambdasimmathcal {Pois}(lambda)$ then we have $~mathsf P(Y_lambda{=}y)~=~dfrac{lambda^ymathsf e^lambda}{y!}mathbf 1_{yinBbb N}$, and we can use this to ensure we are substituting the correct values.



      So, therefore, we get the following.



      $$begin{align}mathsf P(X_1{=}9mid sum_{i=1}^5X_i{=}45)&=dfrac{mathsf P(X_1{=}9)cdotmathsf P(sum_{i=2}^5 X_i{=}36)}{mathsf P(sum_{i=1}^5 X_i{=}45)}\[1ex]&=dfrac{mathsf P(Y_{10}{=}9)cdotmathsf P(Y_{40}{=}36)}{mathsf P(Y_{50}{=}45)}\[1ex]&=dfrac{dfrac{10^9mathsf e^{10}}{9!}cdotdfrac{40^{36}mathsf e^{40}}{36!}}{dfrac{50^{45}mathsf e^{50}}{45!}}\[1ex]&=dbinom{45}9 dfrac{10^9cdot 40^{36}}{50^{45}}end{align}$$



      Which is where the binomial distribution comes from.



      [The count of point events which occur in one among several independent Poisson processes has a conditionally binomial distribution when given the total count of events that occur among all of them, with success rate equal to the ratio of the sums of the Poisson rate parameters (favored versus total).]






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        It is because the sum of independent Poisson random variables is a Poisson random variables whose mean is the sum of their means, and all $X_i$ are independent Poisson random variables whose mean is $10$, that we have the following.
        $$X_1simmathcal{Pois}(10)\sum_{i=2}^5X_isimmathcal{Pois}(40)\sum_{i=1}^5X_isimmathcal{Pois}(50)$$
        (Emphasis because the fact of independence is crucial to this solution.)



        Also, if we define $Y_lambdasimmathcal {Pois}(lambda)$ then we have $~mathsf P(Y_lambda{=}y)~=~dfrac{lambda^ymathsf e^lambda}{y!}mathbf 1_{yinBbb N}$, and we can use this to ensure we are substituting the correct values.



        So, therefore, we get the following.



        $$begin{align}mathsf P(X_1{=}9mid sum_{i=1}^5X_i{=}45)&=dfrac{mathsf P(X_1{=}9)cdotmathsf P(sum_{i=2}^5 X_i{=}36)}{mathsf P(sum_{i=1}^5 X_i{=}45)}\[1ex]&=dfrac{mathsf P(Y_{10}{=}9)cdotmathsf P(Y_{40}{=}36)}{mathsf P(Y_{50}{=}45)}\[1ex]&=dfrac{dfrac{10^9mathsf e^{10}}{9!}cdotdfrac{40^{36}mathsf e^{40}}{36!}}{dfrac{50^{45}mathsf e^{50}}{45!}}\[1ex]&=dbinom{45}9 dfrac{10^9cdot 40^{36}}{50^{45}}end{align}$$



        Which is where the binomial distribution comes from.



        [The count of point events which occur in one among several independent Poisson processes has a conditionally binomial distribution when given the total count of events that occur among all of them, with success rate equal to the ratio of the sums of the Poisson rate parameters (favored versus total).]






        share|cite|improve this answer









        $endgroup$



        It is because the sum of independent Poisson random variables is a Poisson random variables whose mean is the sum of their means, and all $X_i$ are independent Poisson random variables whose mean is $10$, that we have the following.
        $$X_1simmathcal{Pois}(10)\sum_{i=2}^5X_isimmathcal{Pois}(40)\sum_{i=1}^5X_isimmathcal{Pois}(50)$$
        (Emphasis because the fact of independence is crucial to this solution.)



        Also, if we define $Y_lambdasimmathcal {Pois}(lambda)$ then we have $~mathsf P(Y_lambda{=}y)~=~dfrac{lambda^ymathsf e^lambda}{y!}mathbf 1_{yinBbb N}$, and we can use this to ensure we are substituting the correct values.



        So, therefore, we get the following.



        $$begin{align}mathsf P(X_1{=}9mid sum_{i=1}^5X_i{=}45)&=dfrac{mathsf P(X_1{=}9)cdotmathsf P(sum_{i=2}^5 X_i{=}36)}{mathsf P(sum_{i=1}^5 X_i{=}45)}\[1ex]&=dfrac{mathsf P(Y_{10}{=}9)cdotmathsf P(Y_{40}{=}36)}{mathsf P(Y_{50}{=}45)}\[1ex]&=dfrac{dfrac{10^9mathsf e^{10}}{9!}cdotdfrac{40^{36}mathsf e^{40}}{36!}}{dfrac{50^{45}mathsf e^{50}}{45!}}\[1ex]&=dbinom{45}9 dfrac{10^9cdot 40^{36}}{50^{45}}end{align}$$



        Which is where the binomial distribution comes from.



        [The count of point events which occur in one among several independent Poisson processes has a conditionally binomial distribution when given the total count of events that occur among all of them, with success rate equal to the ratio of the sums of the Poisson rate parameters (favored versus total).]







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 20 at 0:45









        Graham KempGraham Kemp

        87.8k43578




        87.8k43578























            1












            $begingroup$

            $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
            newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
            newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
            newcommand{dd}{mathrm{d}}
            newcommand{ds}[1]{displaystyle{#1}}
            newcommand{expo}[1]{,mathrm{e}^{#1},}
            newcommand{ic}{mathrm{i}}
            newcommand{mc}[1]{mathcal{#1}}
            newcommand{mrm}[1]{mathrm{#1}}
            newcommand{pars}[1]{left(,{#1},right)}
            newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
            newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
            newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
            newcommand{verts}[1]{leftvert,{#1},rightvert}$




            Let $ds{mc{P}pars{lambda,n} = {expo{-lambda}lambda^{n} over n!}}$.




            begin{align}
            &bbox[10px,#ffd]{sum_{x_{1} = 0}^{infty}mc{P}pars{lambda,x_{1}}
            !!sum_{x_{2} = 0}^{infty}mc{P}pars{lambda,x_{2}}
            !!sum_{x_{3} = 0}^{infty}mc{P}pars{lambda,x_{3}}
            !!sum_{x_{4} = 0}^{infty}mc{P}pars{lambda,x_{4}}
            !!sum_{x_{5} = 0}^{infty}mc{P}pars{lambda,x_{5}}}
            times
            \[2mm] & bracks{x_{1} = 9}
            bracks{z^{45}}! z^{sum_{i = 1}^{5}x_{i}}
            \[5mm] = &
            mc{P}pars{lambda,9}bracks{z^{45}}z^{9}
            bracks{sum_{x = 0}^{infty}mc{P}pars{x}z^{x}}^{4} =
            mc{P}pars{lambda,9}bracks{z^{36}}
            bracks{sum_{x = 0}^{infty}{expo{-lambda}lambda^{x} over x!}z^{x}}^{4}
            \[5mm] = &
            mc{P}pars{lambda,9}expo{-4lambda}bracks{z^{36}}
            bracks{sum_{x = 0}^{infty}{pars{lambda z}^{x} over x!}}^{4} =
            mc{P}pars{lambda,9}expo{-4lambda}bracks{z^{36}}
            expo{4lambda z}
            \[5mm] = &
            mc{P}pars{lambda,9}expo{-4lambda}{pars{4lambda}^{36} over 36!} =
            {expo{-lambda}lambda^{9} over 9!}expo{-4lambda}{4^{36}lambda^{36} over 36!}
            \[5mm] = &
            bbx{{4^{36} over 9!, 36!}
            ,expo{-5lambda}lambda^{45}}
            end{align}




            $ds{lambda = 10 implies approx 6.7474 times 10^{-3}}$.







            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Unfortunately, OP just edited the question, and wants Pois(10), not Pois(5).
              $endgroup$
              – Gerry Myerson
              Mar 20 at 0:17






            • 2




              $begingroup$
              $Hugeleft)smileright($
              $endgroup$
              – Felix Marin
              Mar 20 at 0:20


















            1












            $begingroup$

            $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
            newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
            newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
            newcommand{dd}{mathrm{d}}
            newcommand{ds}[1]{displaystyle{#1}}
            newcommand{expo}[1]{,mathrm{e}^{#1},}
            newcommand{ic}{mathrm{i}}
            newcommand{mc}[1]{mathcal{#1}}
            newcommand{mrm}[1]{mathrm{#1}}
            newcommand{pars}[1]{left(,{#1},right)}
            newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
            newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
            newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
            newcommand{verts}[1]{leftvert,{#1},rightvert}$




            Let $ds{mc{P}pars{lambda,n} = {expo{-lambda}lambda^{n} over n!}}$.




            begin{align}
            &bbox[10px,#ffd]{sum_{x_{1} = 0}^{infty}mc{P}pars{lambda,x_{1}}
            !!sum_{x_{2} = 0}^{infty}mc{P}pars{lambda,x_{2}}
            !!sum_{x_{3} = 0}^{infty}mc{P}pars{lambda,x_{3}}
            !!sum_{x_{4} = 0}^{infty}mc{P}pars{lambda,x_{4}}
            !!sum_{x_{5} = 0}^{infty}mc{P}pars{lambda,x_{5}}}
            times
            \[2mm] & bracks{x_{1} = 9}
            bracks{z^{45}}! z^{sum_{i = 1}^{5}x_{i}}
            \[5mm] = &
            mc{P}pars{lambda,9}bracks{z^{45}}z^{9}
            bracks{sum_{x = 0}^{infty}mc{P}pars{x}z^{x}}^{4} =
            mc{P}pars{lambda,9}bracks{z^{36}}
            bracks{sum_{x = 0}^{infty}{expo{-lambda}lambda^{x} over x!}z^{x}}^{4}
            \[5mm] = &
            mc{P}pars{lambda,9}expo{-4lambda}bracks{z^{36}}
            bracks{sum_{x = 0}^{infty}{pars{lambda z}^{x} over x!}}^{4} =
            mc{P}pars{lambda,9}expo{-4lambda}bracks{z^{36}}
            expo{4lambda z}
            \[5mm] = &
            mc{P}pars{lambda,9}expo{-4lambda}{pars{4lambda}^{36} over 36!} =
            {expo{-lambda}lambda^{9} over 9!}expo{-4lambda}{4^{36}lambda^{36} over 36!}
            \[5mm] = &
            bbx{{4^{36} over 9!, 36!}
            ,expo{-5lambda}lambda^{45}}
            end{align}




            $ds{lambda = 10 implies approx 6.7474 times 10^{-3}}$.







            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Unfortunately, OP just edited the question, and wants Pois(10), not Pois(5).
              $endgroup$
              – Gerry Myerson
              Mar 20 at 0:17






            • 2




              $begingroup$
              $Hugeleft)smileright($
              $endgroup$
              – Felix Marin
              Mar 20 at 0:20
















            1












            1








            1





            $begingroup$

            $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
            newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
            newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
            newcommand{dd}{mathrm{d}}
            newcommand{ds}[1]{displaystyle{#1}}
            newcommand{expo}[1]{,mathrm{e}^{#1},}
            newcommand{ic}{mathrm{i}}
            newcommand{mc}[1]{mathcal{#1}}
            newcommand{mrm}[1]{mathrm{#1}}
            newcommand{pars}[1]{left(,{#1},right)}
            newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
            newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
            newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
            newcommand{verts}[1]{leftvert,{#1},rightvert}$




            Let $ds{mc{P}pars{lambda,n} = {expo{-lambda}lambda^{n} over n!}}$.




            begin{align}
            &bbox[10px,#ffd]{sum_{x_{1} = 0}^{infty}mc{P}pars{lambda,x_{1}}
            !!sum_{x_{2} = 0}^{infty}mc{P}pars{lambda,x_{2}}
            !!sum_{x_{3} = 0}^{infty}mc{P}pars{lambda,x_{3}}
            !!sum_{x_{4} = 0}^{infty}mc{P}pars{lambda,x_{4}}
            !!sum_{x_{5} = 0}^{infty}mc{P}pars{lambda,x_{5}}}
            times
            \[2mm] & bracks{x_{1} = 9}
            bracks{z^{45}}! z^{sum_{i = 1}^{5}x_{i}}
            \[5mm] = &
            mc{P}pars{lambda,9}bracks{z^{45}}z^{9}
            bracks{sum_{x = 0}^{infty}mc{P}pars{x}z^{x}}^{4} =
            mc{P}pars{lambda,9}bracks{z^{36}}
            bracks{sum_{x = 0}^{infty}{expo{-lambda}lambda^{x} over x!}z^{x}}^{4}
            \[5mm] = &
            mc{P}pars{lambda,9}expo{-4lambda}bracks{z^{36}}
            bracks{sum_{x = 0}^{infty}{pars{lambda z}^{x} over x!}}^{4} =
            mc{P}pars{lambda,9}expo{-4lambda}bracks{z^{36}}
            expo{4lambda z}
            \[5mm] = &
            mc{P}pars{lambda,9}expo{-4lambda}{pars{4lambda}^{36} over 36!} =
            {expo{-lambda}lambda^{9} over 9!}expo{-4lambda}{4^{36}lambda^{36} over 36!}
            \[5mm] = &
            bbx{{4^{36} over 9!, 36!}
            ,expo{-5lambda}lambda^{45}}
            end{align}




            $ds{lambda = 10 implies approx 6.7474 times 10^{-3}}$.







            share|cite|improve this answer











            $endgroup$



            $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
            newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
            newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
            newcommand{dd}{mathrm{d}}
            newcommand{ds}[1]{displaystyle{#1}}
            newcommand{expo}[1]{,mathrm{e}^{#1},}
            newcommand{ic}{mathrm{i}}
            newcommand{mc}[1]{mathcal{#1}}
            newcommand{mrm}[1]{mathrm{#1}}
            newcommand{pars}[1]{left(,{#1},right)}
            newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
            newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
            newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
            newcommand{verts}[1]{leftvert,{#1},rightvert}$




            Let $ds{mc{P}pars{lambda,n} = {expo{-lambda}lambda^{n} over n!}}$.




            begin{align}
            &bbox[10px,#ffd]{sum_{x_{1} = 0}^{infty}mc{P}pars{lambda,x_{1}}
            !!sum_{x_{2} = 0}^{infty}mc{P}pars{lambda,x_{2}}
            !!sum_{x_{3} = 0}^{infty}mc{P}pars{lambda,x_{3}}
            !!sum_{x_{4} = 0}^{infty}mc{P}pars{lambda,x_{4}}
            !!sum_{x_{5} = 0}^{infty}mc{P}pars{lambda,x_{5}}}
            times
            \[2mm] & bracks{x_{1} = 9}
            bracks{z^{45}}! z^{sum_{i = 1}^{5}x_{i}}
            \[5mm] = &
            mc{P}pars{lambda,9}bracks{z^{45}}z^{9}
            bracks{sum_{x = 0}^{infty}mc{P}pars{x}z^{x}}^{4} =
            mc{P}pars{lambda,9}bracks{z^{36}}
            bracks{sum_{x = 0}^{infty}{expo{-lambda}lambda^{x} over x!}z^{x}}^{4}
            \[5mm] = &
            mc{P}pars{lambda,9}expo{-4lambda}bracks{z^{36}}
            bracks{sum_{x = 0}^{infty}{pars{lambda z}^{x} over x!}}^{4} =
            mc{P}pars{lambda,9}expo{-4lambda}bracks{z^{36}}
            expo{4lambda z}
            \[5mm] = &
            mc{P}pars{lambda,9}expo{-4lambda}{pars{4lambda}^{36} over 36!} =
            {expo{-lambda}lambda^{9} over 9!}expo{-4lambda}{4^{36}lambda^{36} over 36!}
            \[5mm] = &
            bbx{{4^{36} over 9!, 36!}
            ,expo{-5lambda}lambda^{45}}
            end{align}




            $ds{lambda = 10 implies approx 6.7474 times 10^{-3}}$.








            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 20 at 0:40

























            answered Mar 20 at 0:12









            Felix MarinFelix Marin

            68.9k7110147




            68.9k7110147








            • 1




              $begingroup$
              Unfortunately, OP just edited the question, and wants Pois(10), not Pois(5).
              $endgroup$
              – Gerry Myerson
              Mar 20 at 0:17






            • 2




              $begingroup$
              $Hugeleft)smileright($
              $endgroup$
              – Felix Marin
              Mar 20 at 0:20
















            • 1




              $begingroup$
              Unfortunately, OP just edited the question, and wants Pois(10), not Pois(5).
              $endgroup$
              – Gerry Myerson
              Mar 20 at 0:17






            • 2




              $begingroup$
              $Hugeleft)smileright($
              $endgroup$
              – Felix Marin
              Mar 20 at 0:20










            1




            1




            $begingroup$
            Unfortunately, OP just edited the question, and wants Pois(10), not Pois(5).
            $endgroup$
            – Gerry Myerson
            Mar 20 at 0:17




            $begingroup$
            Unfortunately, OP just edited the question, and wants Pois(10), not Pois(5).
            $endgroup$
            – Gerry Myerson
            Mar 20 at 0:17




            2




            2




            $begingroup$
            $Hugeleft)smileright($
            $endgroup$
            – Felix Marin
            Mar 20 at 0:20






            $begingroup$
            $Hugeleft)smileright($
            $endgroup$
            – Felix Marin
            Mar 20 at 0:20




















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