1.3 Sum of Independent Binomial

\(X\thicksim B(n,P)\) and \(Y\thicksim B(m,P)\). \(X\) and \(Y\) are independent

Let \(Z=X+Y\) \begin {align*} P(Z=z) &= P(X+Y=z) =\sum ^z_{x=0}P(X=x, Y=z-x)\\ &=\sum ^z_{x=0}P(X=x)P(Y=z-x)\\ &=\sum ^z_{x=0} \begin {pmatrix} n\\x \end {pmatrix} P^x(1-P)^{n-x} \begin {pmatrix} m\\ z-x\\ \end {pmatrix} P^{z-x}(1-P)^{m-z+x}\\ &=P^z(1-P)^{n+m-z}\sum ^z_{x=0} \begin {pmatrix} n\\x\\ \end {pmatrix} \begin {pmatrix} m\\z-x\\ \end {pmatrix}\\ &=P^z(1-P)^{n+m-z} \begin {pmatrix} n+m\\ z\\ \end {pmatrix} \end {align*}

\[\text {Thus}\hspace {0.4cm} P(Z=z)= \begin {pmatrix} n+m\\z\\ \end {pmatrix} P^z(1-P)^{n+m-z},\hspace {0.5cm} Z\thicksim B(n+m,P) \]

If \(X\) and \(Y\) are independent Poisson with parameters \(\mu _1\) and \(\mu _2\) respectively. Find the conditional distribution of \(Y\) given \(X+Y=K\).

\begin {align*} P(Y=y \mid X+Y=K) &=P(Y=y, X+Y=K) =\frac {P(Y=y) P(X=K-y}{P(X+Y=K)}\\\\ &={\frac {\mu _2^ye^{-\mu _2}}{y!}\hspace {0.2cm}\frac {\mu _1^{k-y}e^{-\mu _1}}{(K-y)!}}\div {\frac {e^{-(\mu _1+\mu _2)}\hspace {0.2cm}(\mu _1+\mu _2)^K}{K!}}\\\\ &=\frac {K!}{y!(K-y)!}\hspace {0.1cm}\frac {\mu _2^y \mu _1^{K-y}}{(\mu _1+\mu _2)^K} = \begin {pmatrix} K\\y\\ \end {pmatrix} \frac {\mu _2^y}{(\mu _1+\mu _2)^y}\hspace {0.1cm}\frac {\mu _1^{K-y}}{(\mu _1+\mu _2)^{K-y}}\\\\ &= \begin {pmatrix} K\\y\\ \end {pmatrix} \Bigg (\frac {\mu _2}{\mu _1+\mu _2}\Bigg )^K \Bigg (\frac {\mu _1}{\mu _1+\mu _2}\Bigg )^{K-y}\\\\ &= \begin {pmatrix} K\\y\\ \end {pmatrix} P^y (1-P)^{K-y},\hspace {0.5cm} \text {where}\hspace {0.3cm} P=\frac {\mu _2}{\mu _1+\mu _2}\\ \end {align*}

Let \(X\) and \(Y\) be independent Binomial random variables with parameters \((n,P)\) and \((m,P)\) respectively. Find the conditional distribution of \(X\) given \(X+Y=K\) \begin {align*} P(X=x \mid X+Y=K) &=\frac {P(X=x, X+Y=K)}{P(X+Y=K)}\\\\ &=\frac {P(X=x, Y=K-x)}{ \begin {pmatrix} n+m\\K\\ \end {pmatrix} P^K(1-P)^{n+m-K}}\\\\ &=\frac {P(X=x) P(Y=K-x)}{ \begin {pmatrix} n+m\\K\\ \end {pmatrix} P^K(1-P)^{n+m-K}}\\\\ &=\frac { \begin {pmatrix} n\\x\\ \end {pmatrix} P^x(1-P)^{n-x} \begin {pmatrix} m\\K-x\\ \end {pmatrix} P^{K-x}(1-P)^{m-K+x}}{ \begin {pmatrix} n+m\\K\\ \end {pmatrix} P^K(1-P)^{n+m-K}}\\\\ &=\frac { \begin {pmatrix} n\\x\\ \end {pmatrix} \begin {pmatrix} m\\K-x\\ \end {pmatrix} }{ \begin {pmatrix} n+m\\ K\\ \end {pmatrix} }\\ \end {align*}

\[\therefore \hspace {0.4cm} P(X=x \mid X+Y=K)=\frac { \begin {pmatrix} n\\x\\ \end {pmatrix} \begin {pmatrix} m\\K-x\\ \end {pmatrix} }{ \begin {pmatrix} n+m\\ K\\ \end {pmatrix} },\hspace {0.5cm} x=0,1,2,\dots ,..\min (K,n)\]

\[\text {Thus}\hspace {0.5cm} P(X=x \mid X+Y=K)=\frac { \begin {pmatrix} n\\x\\ \end {pmatrix} \begin {pmatrix} N-n\\K-x\\ \end {pmatrix} }{ \begin {pmatrix} N\\K\\ \end {pmatrix} },\hspace {0.5cm} N=n+m\]

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