Confusion with a multivariate Poisson distribution












1












$begingroup$


I am looking at a multinomial Poisson distribution. Failing to obtain the denominator.



We have that $X_1, ..., X_n sim mathbb{P}(lambda)$, where $theta = exp^{-lambda}$



We know that $$mathbb{P}(X=x) = frac{lambda^x theta}{x!}$$



Since the variables are IID, we know that:



$$mathbb{P}(mathbf{X}) = prod_{i=1}^n frac{lambda^{x_i} theta}{x_i!} $$



this when simplified, gives me:



$$mathbb{P}(mathbf{X}) = left( theta^n lambda^{sum_i x_i} right)left( frac{1}{prod_{i=1}^n x_i!} right) $$



If denote the sufficient statistic $T$ as $sum_i x_i = t$, then we have:



$$mathbb{P}(mathbf{X}) = left( theta^n lambda^t right) left( frac{1}{prod_{i=1}^n x_i!} right)$$



However, Cambridge stats notes, page 15 example 3.4 (b) last equation line, implies that $prod_{i=1}^n x_i! = t!$. Actually, I think I am mixing something up... Isn't the distribution $mathbb{P}(sum_i X_i = t)$ the one that I have been deriving above?






probability statistics







share|cite|improve this question











$endgroup$












  • $begingroup$
    $frac{1}{prod_{i=1}^n x_i!}$ is in the expression for $mathbb P(mathbf{X}=mathbf{x})$ while $frac{1}{t!}$ in in the expression for the rather larger $mathbb Pleft(sum X_i =tright)$
    $endgroup$
    – Henry
    Dec 19 '18 at 16:50












  • $begingroup$
    thanks! Can you give me a tip on how to derive the equation for $mathbb{P} left( sum_i X_i = t right)$
    $endgroup$
    – i squared - Keep it Real
    Dec 19 '18 at 16:53


















1












$begingroup$


I am looking at a multinomial Poisson distribution. Failing to obtain the denominator.



We have that $X_1, ..., X_n sim mathbb{P}(lambda)$, where $theta = exp^{-lambda}$



We know that $$mathbb{P}(X=x) = frac{lambda^x theta}{x!}$$



Since the variables are IID, we know that:



$$mathbb{P}(mathbf{X}) = prod_{i=1}^n frac{lambda^{x_i} theta}{x_i!} $$



this when simplified, gives me:



$$mathbb{P}(mathbf{X}) = left( theta^n lambda^{sum_i x_i} right)left( frac{1}{prod_{i=1}^n x_i!} right) $$



If denote the sufficient statistic $T$ as $sum_i x_i = t$, then we have:



$$mathbb{P}(mathbf{X}) = left( theta^n lambda^t right) left( frac{1}{prod_{i=1}^n x_i!} right)$$



However, Cambridge stats notes, page 15 example 3.4 (b) last equation line, implies that $prod_{i=1}^n x_i! = t!$. Actually, I think I am mixing something up... Isn't the distribution $mathbb{P}(sum_i X_i = t)$ the one that I have been deriving above?






probability statistics







share|cite|improve this question











$endgroup$












  • $begingroup$
    $frac{1}{prod_{i=1}^n x_i!}$ is in the expression for $mathbb P(mathbf{X}=mathbf{x})$ while $frac{1}{t!}$ in in the expression for the rather larger $mathbb Pleft(sum X_i =tright)$
    $endgroup$
    – Henry
    Dec 19 '18 at 16:50












  • $begingroup$
    thanks! Can you give me a tip on how to derive the equation for $mathbb{P} left( sum_i X_i = t right)$
    $endgroup$
    – i squared - Keep it Real
    Dec 19 '18 at 16:53
















1












1








1





$begingroup$


I am looking at a multinomial Poisson distribution. Failing to obtain the denominator.



We have that $X_1, ..., X_n sim mathbb{P}(lambda)$, where $theta = exp^{-lambda}$



We know that $$mathbb{P}(X=x) = frac{lambda^x theta}{x!}$$



Since the variables are IID, we know that:



$$mathbb{P}(mathbf{X}) = prod_{i=1}^n frac{lambda^{x_i} theta}{x_i!} $$



this when simplified, gives me:



$$mathbb{P}(mathbf{X}) = left( theta^n lambda^{sum_i x_i} right)left( frac{1}{prod_{i=1}^n x_i!} right) $$



If denote the sufficient statistic $T$ as $sum_i x_i = t$, then we have:



$$mathbb{P}(mathbf{X}) = left( theta^n lambda^t right) left( frac{1}{prod_{i=1}^n x_i!} right)$$



However, Cambridge stats notes, page 15 example 3.4 (b) last equation line, implies that $prod_{i=1}^n x_i! = t!$. Actually, I think I am mixing something up... Isn't the distribution $mathbb{P}(sum_i X_i = t)$ the one that I have been deriving above?






probability statistics







share|cite|improve this question











$endgroup$




I am looking at a multinomial Poisson distribution. Failing to obtain the denominator.



We have that $X_1, ..., X_n sim mathbb{P}(lambda)$, where $theta = exp^{-lambda}$



We know that $$mathbb{P}(X=x) = frac{lambda^x theta}{x!}$$



Since the variables are IID, we know that:



$$mathbb{P}(mathbf{X}) = prod_{i=1}^n frac{lambda^{x_i} theta}{x_i!} $$



this when simplified, gives me:



$$mathbb{P}(mathbf{X}) = left( theta^n lambda^{sum_i x_i} right)left( frac{1}{prod_{i=1}^n x_i!} right) $$



If denote the sufficient statistic $T$ as $sum_i x_i = t$, then we have:



$$mathbb{P}(mathbf{X}) = left( theta^n lambda^t right) left( frac{1}{prod_{i=1}^n x_i!} right)$$



However, Cambridge stats notes, page 15 example 3.4 (b) last equation line, implies that $prod_{i=1}^n x_i! = t!$. Actually, I think I am mixing something up... Isn't the distribution $mathbb{P}(sum_i X_i = t)$ the one that I have been deriving above?






probability statistics




probability statistics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 19 '18 at 16:28







i squared - Keep it Real

















asked Dec 19 '18 at 16:15









i squared - Keep it Reali squared - Keep it Real

1,5871927




1,5871927












  • $begingroup$
    $frac{1}{prod_{i=1}^n x_i!}$ is in the expression for $mathbb P(mathbf{X}=mathbf{x})$ while $frac{1}{t!}$ in in the expression for the rather larger $mathbb Pleft(sum X_i =tright)$
    $endgroup$
    – Henry
    Dec 19 '18 at 16:50












  • $begingroup$
    thanks! Can you give me a tip on how to derive the equation for $mathbb{P} left( sum_i X_i = t right)$
    $endgroup$
    – i squared - Keep it Real
    Dec 19 '18 at 16:53




















  • $begingroup$
    $frac{1}{prod_{i=1}^n x_i!}$ is in the expression for $mathbb P(mathbf{X}=mathbf{x})$ while $frac{1}{t!}$ in in the expression for the rather larger $mathbb Pleft(sum X_i =tright)$
    $endgroup$
    – Henry
    Dec 19 '18 at 16:50












  • $begingroup$
    thanks! Can you give me a tip on how to derive the equation for $mathbb{P} left( sum_i X_i = t right)$
    $endgroup$
    – i squared - Keep it Real
    Dec 19 '18 at 16:53


















$begingroup$
$frac{1}{prod_{i=1}^n x_i!}$ is in the expression for $mathbb P(mathbf{X}=mathbf{x})$ while $frac{1}{t!}$ in in the expression for the rather larger $mathbb Pleft(sum X_i =tright)$
$endgroup$
– Henry
Dec 19 '18 at 16:50






$begingroup$
$frac{1}{prod_{i=1}^n x_i!}$ is in the expression for $mathbb P(mathbf{X}=mathbf{x})$ while $frac{1}{t!}$ in in the expression for the rather larger $mathbb Pleft(sum X_i =tright)$
$endgroup$
– Henry
Dec 19 '18 at 16:50














$begingroup$
thanks! Can you give me a tip on how to derive the equation for $mathbb{P} left( sum_i X_i = t right)$
$endgroup$
– i squared - Keep it Real
Dec 19 '18 at 16:53






$begingroup$
thanks! Can you give me a tip on how to derive the equation for $mathbb{P} left( sum_i X_i = t right)$
$endgroup$
– i squared - Keep it Real
Dec 19 '18 at 16:53












1 Answer
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$begingroup$

The key point is that the $X_i$´s are poisson distributed. Then the sum of poisson distributed random variables can be derived. For instance here (two variables). For $n$ variables we get



$$Pleft( sum_{i=1}^{n} X_i=t right)=frac{left( ncdot lambdaright)^t}{t!}cdot e^{-ncdot lambda}$$



Now it is straigtforward to caclculate



$$frac{mathbb P(X_1=0 cap sum_{i=2}^n X_i=t)}{mathbb P(sum_{i=1}^n X_i=t)}=frac{mathbb P(X_1=0) cdot mathbb P( sum_{i=2}^n X_i=t)}{mathbb P(sum_{i=1}^n X_i=t)}$$






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    $begingroup$

    The key point is that the $X_i$´s are poisson distributed. Then the sum of poisson distributed random variables can be derived. For instance here (two variables). For $n$ variables we get



    $$Pleft( sum_{i=1}^{n} X_i=t right)=frac{left( ncdot lambdaright)^t}{t!}cdot e^{-ncdot lambda}$$



    Now it is straigtforward to caclculate



    $$frac{mathbb P(X_1=0 cap sum_{i=2}^n X_i=t)}{mathbb P(sum_{i=1}^n X_i=t)}=frac{mathbb P(X_1=0) cdot mathbb P( sum_{i=2}^n X_i=t)}{mathbb P(sum_{i=1}^n X_i=t)}$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The key point is that the $X_i$´s are poisson distributed. Then the sum of poisson distributed random variables can be derived. For instance here (two variables). For $n$ variables we get



      $$Pleft( sum_{i=1}^{n} X_i=t right)=frac{left( ncdot lambdaright)^t}{t!}cdot e^{-ncdot lambda}$$



      Now it is straigtforward to caclculate



      $$frac{mathbb P(X_1=0 cap sum_{i=2}^n X_i=t)}{mathbb P(sum_{i=1}^n X_i=t)}=frac{mathbb P(X_1=0) cdot mathbb P( sum_{i=2}^n X_i=t)}{mathbb P(sum_{i=1}^n X_i=t)}$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The key point is that the $X_i$´s are poisson distributed. Then the sum of poisson distributed random variables can be derived. For instance here (two variables). For $n$ variables we get



        $$Pleft( sum_{i=1}^{n} X_i=t right)=frac{left( ncdot lambdaright)^t}{t!}cdot e^{-ncdot lambda}$$



        Now it is straigtforward to caclculate



        $$frac{mathbb P(X_1=0 cap sum_{i=2}^n X_i=t)}{mathbb P(sum_{i=1}^n X_i=t)}=frac{mathbb P(X_1=0) cdot mathbb P( sum_{i=2}^n X_i=t)}{mathbb P(sum_{i=1}^n X_i=t)}$$






        share|cite|improve this answer









        $endgroup$



        The key point is that the $X_i$´s are poisson distributed. Then the sum of poisson distributed random variables can be derived. For instance here (two variables). For $n$ variables we get



        $$Pleft( sum_{i=1}^{n} X_i=t right)=frac{left( ncdot lambdaright)^t}{t!}cdot e^{-ncdot lambda}$$



        Now it is straigtforward to caclculate



        $$frac{mathbb P(X_1=0 cap sum_{i=2}^n X_i=t)}{mathbb P(sum_{i=1}^n X_i=t)}=frac{mathbb P(X_1=0) cdot mathbb P( sum_{i=2}^n X_i=t)}{mathbb P(sum_{i=1}^n X_i=t)}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 17:37









        callculuscallculus

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        18.1k31427






























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