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

            Magento 2 - Add success message with knockout Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Success / Error message on ajax request$.widget is not a function when loading a homepage after add custom jQuery on custom themeHow can bind jQuery to current document in Magento 2 When template load by ajaxRedirect page using plugin in Magento 2Magento 2 - Update quantity and totals of cart page without page reload?Magento 2: Quote data not loaded on knockout checkoutMagento 2 : I need to change add to cart success message after adding product into cart through pluginMagento 2.2.5 How to add additional products to cart from new checkout step?Magento 2 Add error/success message with knockoutCan't validate Post Code on checkout page

            Image
            What were wait-states, and why was it only an issue for PCs? “Since the train was delayed for more than an hour, passengers were given a full refund.” – Why is there no article before “passengers”? Will I be more secure with my own router behind my ISP's router? Short story about an alien named Ushtu(?) coming from a future Earth, when ours was destroyed by a nuclear explosion When does Bran Stark remember Jamie pushing him? How to charge percentage of transaction cost? Pointing to problems without suggesting solutions false 'Security alert' from Google - every login generates mails from 'no-reply@accounts.google.com' Why doesn't the university give past final exams' answers? Like totally amazing interchangeable sister outfit accessory swapping or whatever Does traveling In The United States require a passport or can I use my green card if not a US citizen? Who's this lady in the war room? Who can become a wight? "Destructi
            Read more

            Fil:Tokke komm.svg

            Image
            Skildring Skildring Tokke komm.svg Norsk nynorsk: Våpen for Tokke kommune i Telemark fylke, godkjent ved kongeleg resolusjon 6. februar 1987. Motiv: På gull grunn ein gåande svart bjørn. [1] Tokke ligg i eit gamalt bjørneområde og bjørnen er eit viktig innslag i fleire lokale historier. Bjørnen kan òg symbolisere naturen og dyrelivet i kommunen. English: Coat of arms of the Norwegian municipality of Tokke, Telemark. Dato 6. februar 1987 Kjelde Original by Stein Davidsen, based upon image from caplex.no. Opphavsperson Eige arbeid Vektorgrafikken vart laga med Inkscape. Løyve (Gjenbruk av denne fila) Public domain Public domain false false This official Norwegian coat of arms is ineligible for copyright, but its usage is restricted according to Norwegian penal law, independent of the copyright status of the depiction shown here. Dette biletet er sett saman av eit flagg, eit våpen, eit sigill eller eit anna offisielt insignium . Bruken av slike symbol er avgrensa i fleire land. Desse r
            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