91影视

X₁ and X₂ are random variables independent as well that have
a Poisson distribution with mean\({\lambda _1}\,{\rm{and}}\,{\lambda _2}\)respectively

We have to determine the conditional distribution \({X_1}\) given that \({X_1} + {X_2} = k\)

Let \(Y = {X_1} + {X_2}\)

\({X_1},{X_2}\)are independent and have Poisson distribution with parameters \({\lambda _1}\,{\rm{and}}\,{\lambda _2}\), respectively.

Thus, Y follows a Poisson distribution with parameters \({\lambda _1} + {\lambda _2}\)

\(P\left( {Y = k} \right) = \frac{{{{\left( {{\lambda _1} + {\lambda _2}} \right)}^k}}}{{k!}}{e^{ - {\lambda _1} - {\lambda _2}}}\)

The conditional probability is

\(\begin{array}{l}P\left( {{X_1} = m|\,Y = k} \right)\\ = \frac{{P\left( {{X_1} = m,\,Y = k} \right)}}{{P\left( {Y = k} \right)}}\\ = \frac{{P\left( {{X_1} = m,{X_1} + {X_2} = k} \right)}}{{P\left( {Y = k} \right)}}\\ = \frac{{P\left( {{X_1} = m,{X_2} = k - m} \right)}}{{P\left( {Y = k} \right)}}\\ = \frac{{P\left( {{X_1} = m} \right)P\left( {{X_2} = k - m} \right)}}{{P\left( {Y = k} \right)}}\end{array}\)

By the independence of property\({X_1}\,{\rm{and}}\,{X_2}\), we can write

\(\begin{array}{l}P\left( {{X_1} = m|\,Y = k} \right)\\\\ = \frac{{\frac{{{\lambda _1}^m}}{{m!}}{e^{ - {\lambda _1}}}\,\frac{{{\lambda _2}^{k - m}}}{{\left( {k - m} \right)!}}{e^{ - {\lambda _2}}}}}{{\frac{{{{\left( {{\lambda _1} + {\lambda _2}} \right)}^k}}}{{k!}}{e^{ - {\lambda _1} - {\lambda _2}}}}}\\\\ = \frac{{k!}}{{m!\left( {k - m} \right)!}}\,\frac{{{\lambda _1}^m{\lambda _2}^{k - m}}}{{{{\left( {{\lambda _1} + {\lambda _2}} \right)}^k}}}\\\\ = \left( \begin{array}{l}k\\m\end{array} \right){\left( {\frac{{{\lambda _1}}}{{{\lambda _1} + {\lambda _2}}}} \right)^m}{\left( {\frac{{{\lambda _2}}}{{{\lambda _1} + {\lambda _2}}}} \right)^{k - m}}\end{array}\)

\( = \left( \begin{array}{l}k\\m\end{array} \right){p^m}{\left( {1 - p} \right)^{k - m}}\,{\rm{where}}\,p = \frac{{{\lambda _1}}}{{{\lambda _1} + {\lambda _2}}}\)