Question

Suppose thatX₁andX₂are independent random variables, thatX₁has the binomial distribution with parametersn1 andp, and thatX₂has the binomial distributionwith parametersn₂andp, wherepis the same for bothX₁andX₂. For each fixed value ofk (k=1,2,…, n₁+n₂),prove that the conditional distribution ofX₁given thatX₁+X₂=kis hypergeometric with parametersn₁,n₂,andk.

Short answer

\[P\left( {{X_1} = x|{X_1} + {X_2} = k} \right) = \frac{{\left( {\begin{array}{*{20}{c}}{{n_1}}\\x\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{n_2}}\\{k - x}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{n_1} + {n_2}}\\k\end{array}} \right)}}\] [Proved]

Step-by-step solution

Given information

\[{X_1}\]and\[{X_2}\]are independent random variables.

\[\begin{array}{l}{X_1} \sim Bin\left( {{n_1},p} \right)\\{X_2} \sim Bin\left( {{n_2},p} \right)\end{array}\]

State and proof

For\[x = 0,1,...,k\],

\[\begin{array}{c}P\left( {{X_1} = x|{X_1} + {X_2} = k} \right) = \frac{{P\left( {{X_1} = x,{X_1} + {X_2} = k} \right)}}{{P\left( {{X_1} + {X_2} = k} \right)}}\\ = \frac{{P\left( {{X_1} = x,{X_2} = k - x} \right)}}{{P\left( {{X_1} + {X_2} = k} \right)}}\end{array}\]

Since,\[{X_1}\]and\[{X_2}\]are independent,

\[P\left( {{X_1} = x,{X_2} = k - x} \right) = P\left( {{X_1} = x} \right)P\left( {{X_2} = k - x} \right)\]

Here,

\[{X_1} + {X_2} \sim Bin\left( {{n_1} + {n_2},p} \right)\]

So,

\[P\left( {{X_1} = x} \right) = \left( {\begin{array}{*{20}{c}}{{n_1}}\\x\end{array}} \right){p^x}{\left( {1 - p} \right)^{{n_1} - x}}\].

\[P\left( {{X_2} = k - x} \right) = \left( {\begin{array}{*{20}{c}}{{n_2}}\\{k - x}\end{array}} \right){p^{k - x}}{\left( {1 - p} \right)^{{n_2} - k + x}}\]

\[P\left( {{X_1} + {X_2} = k} \right) = \left( {\begin{array}{*{20}{c}}{{n_1} + {n_2}}\\k\end{array}} \right){p^k}{\left( {1 - p} \right)^{{n_1} + {n_2} - k}}\]

\[\begin{array}{c}P\left( {{X_1} = x|{X_1} + {X_2} = k} \right) = \frac{{\left( {\begin{array}{*{20}{c}}{{n_1}}\\x\end{array}} \right){p^x}{{\left( {1 - p} \right)}^{{n_1} - x}}\left( {\begin{array}{*{20}{c}}{{n_2}}\\{k - x}\end{array}} \right){p^{k - x}}{{\left( {1 - p} \right)}^{{n_2} - k + x}}}}{{\left( {\begin{array}{*{20}{c}}{{n_1} + {n_2}}\\k\end{array}} \right){p^k}{{\left( {1 - p} \right)}^{{n_1} + {n_2} - k}}}}\\ = \frac{{\left( {\begin{array}{*{20}{c}}{{n_1}}\\x\end{array}} \right){p^k}{{\left( {1 - p} \right)}^{{n_1} + {n_2} - k}}\left( {\begin{array}{*{20}{c}}{{n_2}}\\{k - x}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{n_1} + {n_2}}\\k\end{array}} \right){p^k}{{\left( {1 - p} \right)}^{{n_1} + {n_2} - k}}}}\\ = \frac{{\left( {\begin{array}{*{20}{c}}{{n_1}}\\x\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{n_2}}\\{k - x}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{n_1} + {n_2}}\\k\end{array}} \right)}}\end{array}\]

Therefore, the conditional distribution is a hypergeometric distribution with parameters\[{n_1}\],\[{n_2}\]and\[k\].

Hence, proved.