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

How to say job offer in Mandarin/Cantonese?

What are these boxed doors outside store fronts in New York?

Email Account under attack (really) - anything I can do?

Minkowski space

Mathematical cryptic clues

Why can't I see bouncing of a switch on an oscilloscope?

Today is the Center

How can I prevent hyper evolved versions of regular creatures from wiping out their cousins?

Test if tikzmark exists on same page

Have astronauts in space suits ever taken selfies? If so, how?

How to format long polynomial?

What is the offset in a seaplane's hull?

Why was the small council so happy for Tyrion to become the Master of Coin?

Is it legal for company to use my work email to pretend I still work there?

Fencing style for blades that can attack from a distance

Python: next in for loop

Is it possible to do 50 km distance without any previous training?

Why doesn't H₄O²⁺ exist?

What typically incentivizes a professor to change jobs to a lower ranking university?

How much RAM could one put in a typical 80386 setup?

Can an x86 CPU running in real mode be considered to be basically an 8086 CPU?

The use of multiple foreign keys on same column in SQL Server

Show that if two triangles built on parallel lines, with equal bases have the same perimeter only if they are congruent.

How do I create uniquely male characters?



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














    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3154806%2fcompute-px-i-sum-x-i-45-where-x-i-sim-mathrmpois%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    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




















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3154806%2fcompute-px-i-sum-x-i-45-where-x-i-sim-mathrmpois%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Nidaros erkebispedøme

            Image
            For det lutherske bispedømet, sjå Nidaros bispedøme. Nidaros erkebispedøme var eit katolsk erkebispedøme med sete i Nidaros. Det vart oppretta av kong Olav Kyrre mellom 1070 og 1080 som bispedøme. Bispedømet vart opphøgd til erkebispedøme i 1153. Erkebispedømet vart innstifta av pave Anastasius IV i 1154, og var i funksjon fram til reformasjonen i 1537. Erkebiskopen sitt hovudsete var Erkebispegarden, rett ved domkyrkja Nidarosdomen. Frå 1152 til 1154 var Nicholas Breakspear i Skandinavia som paveleg utsending, og organiserte opprettinga av erkesetet. Innhaldsliste 1 Utbreiing og jurisdiksjon 2 Overhyrdar 2.1 Lydbiskopar 3 Biskopar 3.1 Biskopar 3.2 Erkebiskopar Utbreiing og jurisdiksjon | Kart over den nordiske kyrkjeprovinsen (1153-1387) Erkebispedømet omfatta i tillegg til Noreg øyane i vest i Noregsveldet. Nidaros var eit metropolitanerkebispedøme med fylgjande underliggjande bispedøme suffraganar: Bjørgvin bispedøme Oslo bispedøme Hamar bispedøme Stavanger bispedøme Kirkjub
            Read more

            Birsay

            Image
            Tettstaden Birsay på Orknøyane. Ruinane av jarleplasset om lag midt i biletet. (Fotografi frå 2012) Utsikt mot Birsay og Brough of Brisay (fotografi frå 2009) Birsay er ein tettstad i eit sokn (parish) med same namn, nordvest på øya Mainland [Note 1] på Orknøyane, om lag 31 km frå Kirkwall. Staden er omtala fleire stader i Orknøyingasoga [1] , der han er kalla «Byrgesherad» (norrønt: Birgisherað ). Namnet skriv seg truleg frå elvenamnet Birgisá [2] . Innhaldsliste 1 Den gamle hovudstaden 2 Bispesetet 3 Jarlepalasset 4 Kjelder og utfyllande opplysningar 4.1 Notar 4.2 Fotnotar 4.3 Bakgrunnsstoff Den gamle hovudstaden | Hovudartikkelen for dette emnet er Torfinn Sigurdsson . Soknet er i dag i hovudsak ei jordbrukskommune men er mest kjend for dei mange kulturminna i området. Tettstaden Birsay blir gjerne omtala som «den gamle hovudstaden på Orknøyane» [3] , og ein viser då gjerne til Orknøyingasoga (kap.31) som fortel at orknøyjarlen Torfinn Sigurdsson den mektige «hadde jamleg
            Read more

            Where did Arya get these scars? Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar Manara Favourite questions and answers from the 1st quarter of 2019Why did Arya refuse to end it?Has the pronunciation of Arya Stark's name changed?Has Arya forgiven people?Why did Arya Stark lose her vision?Why can Arya still use the faces?Has the Narrow Sea become narrower?Does Arya Stark know how to make poisons outside of the House of Black and White?Why did Nymeria leave Arya?Why did Arya not kill the Lannister soldiers she encountered in the Riverlands?What is the current canonical age of Sansa, Bran and Arya Stark?

            Image
            How to translate "red flag" into Spanish? France's Public Holidays' Puzzle How did Elite on the NES work? Where can I find how to tex symbols for different fonts? In search of the origins of term censor, I hit a dead end stuck with the greek term, to censor, λογοκρίνω How to check if string is entirely made of same substring? What's called a person who works as someone who puts products on shelves in stores? What helicopter has the most rotor blades? Did war bonds have better investment alternatives during WWII? Page Layouts : 1 column , 2 columns-left , 2 columns-right , 3 column Marquee sign letters Why did Israel vote against lifting the American embargo on Cuba? What *exactly* is electrical current, voltage, and resistance? Mechanism of the formation of peracetic acid Does Prince Arnaud cause someone holding the Princess to lose? Bright yellow or light yellow? What is /etc/mtab in Linux? What is the ongoing value of the Kanban
            Read more