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Chapter 1: Q 6E (page 1)

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

Short Answer

Expert verified

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

Step by step solution

01

Given information

\({X_1}\)and \({X_2}\)are independent random variables.

\(\begin{align}{X_1} \sim Bin\left( {{n_1},p} \right)\\{X_2} \sim Bin\left( {{n_2},p} \right)\end{align}\)

02

State and proof

For\(x = 0,1,...,k\),

\(\begin{align}P\left( {{X_1} = x|{X_1} + {X_2} = k} \right) &= \frac{{P\left( {{X_1} = x,{X_1} + {X_2} = k} \right)}}{{P\left( {{X_1} + {X_2} = k} \right)}}\\ &= \frac{{P\left( {{X_1} = x,{X_2} = k - x} \right)}}{{P\left( {{X_1} + {X_2} = k} \right)}}\end{align}\)

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,

\(P\left( {{X_1} = x} \right) = \left( {\begin{align}{}{{n_1}}\\x\end{align}} \right){p^x}{\left( {1 - p} \right)^{{n_1} - x}}\).

\(P\left( {{X_2} = k - x} \right) = \left( {\begin{align}{}{{n_2}}\\{k - x}\end{align}} \right){p^{k - x}}{\left( {1 - p} \right)^{{n_2} - k + x}}\)

\(P\left( {{X_1} + {X_2} = k} \right) = \left( {\begin{align}{}{{n_1} + {n_2}}\\k\end{align}} \right){p^k}{\left( {1 - p} \right)^{{n_1} + {n_2} - k}}\)

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

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

Hence, proved.

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