Question

Suppose that the random variables \({{\bf{X}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{X}}_{\bf{k}}}\) are independent and that \({{\bf{X}}_{\bf{i}}}\) has the Poisson distribution with mean \({{\bf{\lambda }}_{\bf{i}}}\left( {{\bf{i = 1, \ldots ,k}}} \right)\). Show that for each fixed positive integer n, the conditional distribution of the random Vector \({\bf{X = }}\left( {{{\bf{X}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{X}}_{\bf{k}}}} \right)\), given that \(\sum\limits_{{\bf{i = 1}}}^{\bf{k}} {{{\bf{X}}_{\bf{i}}}{\bf{ = n}}} \) it is the multinomial distribution with parameters n and

\(\begin{array}{l}{\bf{p = }}\left( {{{\bf{p}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{p}}_{\bf{k}}}} \right){\bf{,}}\,{\bf{where}}\\{{\bf{p}}_{\bf{i}}}{\bf{ = }}\frac{{{{\bf{\lambda }}_{\bf{i}}}}}{{\sum\limits_{{\bf{j = 1}}}^{\bf{k}} {{{\bf{\lambda }}_{\bf{j}}}} }}\,{\bf{for}}\,{\bf{i = 1, \ldots ,k}}{\bf{.}}\end{array}\)

Short answer

The proof has been established

Step-by-step solution

Given information

It is given that \({X_1} \ldots {X_k}\)it follows a Poisson distribution with parameters \({\lambda _i}\left( {i = 1, \ldots ,k} \right)\) respectively.

The proof

\(\begin{array}{l} = P\left[ {{X_1} \cap \ldots \cap {X_k}|\sum\limits_{i = 1}^k {{X_i} = n} } \right]\\ = \frac{{P\left[ {{X_1} = {r_1} \cap \ldots \cap {X_k} = {r_k} \cap \sum\limits_{i = 1}^k {{X_i} = n} } \right]}}{{P\left( {\sum\limits_{i = 1}^k {{X_i} = n} } \right)}}\\\frac{{ = \left( {{X_1} = {r_1} \cap \ldots \cap {X_k} = n - {r_1} - \ldots - {r_{k - 1}}} \right)}}{{P\left( {\sum\limits_{i = 1}^k {{X_i} = n} } \right)}}\\ = \frac{{P\left( {{X_1} = {r_1}} \right) \ldots P\left( {{X_k} = n - {r_1} - \ldots - {r_{k - 1}}} \right)}}{{P\left( {\sum\limits_{i = 1}^k {{X_i} = n} } \right)}},\,\,{\rm{Since}}\,{{\rm{X}}_{\rm{i}}}\,\,{\rm{are}}\,\,{\rm{independent}}\end{array}\)

Further, since \({X_1} \ldots {X_k}\) they are independent, Poisson variates with parameters \({\lambda _i}\left( {i = 1, \ldots ,k} \right)\) \(X = \sum\limits_{i = 1}^k {{X_i}} \) is also a Poisson variate with parameters \(\lambda = \sum\limits_{i = 1}^k {{\lambda _i}} \).

Now,

\(\begin{array}{l} = \frac{{\frac{{{e^{ - {\lambda _1}}}{\lambda _1}^{{r_1}}}}{{{r_1}!}} \ldots \frac{{{e^{ - {\lambda _k}}}{\lambda _k}^{{r_{n - {r_1} - \ldots - {r_{k - 1}}}}}}}{{\left( {n - {r_1} - \ldots - {r_{k - 1}}} \right)!}}}}{{\frac{{{e^{ - \lambda }}{\lambda ^r}}}{{r!}}}}\\ = \left\{ {\frac{{n!}}{{{r_1}! \ldots \left( {n - {r_1} - \ldots - {r_{k - 1}}} \right)!}}} \right\}\left\{ {{{\left( {\frac{{{\lambda _1}}}{\lambda }} \right)}^{{r_1}}} \ldots {{\left( {\frac{{{\lambda _k}}}{\lambda }} \right)}^{n - {r_1} - \ldots - {r_{k - 1}}}}} \right\}\\ = \frac{{n!}}{{{r_1}! \ldots {r_k}!}}p_1^{{r_1}} \ldots p_k^{{r_k}}\\Where,\,\\\sum\limits_{i = 1}^k {{r_i} = n\,\,and\,\,\sum\limits_{i = 1}^k {{p_i} = \sum\limits_{i = 1}^k {\left( {\frac{{{\lambda _i}}}{\lambda }} \right)} } } = \frac{1}{\lambda }\sum\limits_{i = 1}^k {\left( {{\lambda _i}} \right) = 1} \end{array}\)

Therefore, the conditional distribution of the random Vector\({\bf{X = }}\left( {{{\bf{X}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{X}}_{\bf{k}}}} \right)\), given that \(\sum\limits_{{\bf{i = 1}}}^{\bf{k}} {{{\bf{X}}_{\bf{i}}}{\bf{ = n}}} \)it is the multinomial distribution with parameters n and

\(\begin{array}{l}{\bf{p = }}\left( {{{\bf{p}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{p}}_{\bf{k}}}} \right){\bf{,}}\,{\bf{where}}\\{{\bf{p}}_{\bf{i}}}{\bf{ = }}\frac{{{{\bf{\lambda }}_{\bf{i}}}}}{{\sum\limits_{{\bf{j = 1}}}^{\bf{k}} {{{\bf{\lambda }}_{\bf{j}}}} }}\,{\bf{for}}\,{\bf{i = 1, \ldots ,k}}{\bf{.}}\end{array}\)