Question #322952

The random variable X_{1} , X 2 ,.......,Xn , are independent and each has a poisson distribution with mean 1. Let Y=X 1 +X 2 +........+Xn . find the conditional distribution of X mathcal N given Y.

1
Expert's answer
2022-04-04T16:11:49-0400

X1+X2+...+Xn1Poiss(n1)X1+X2+...+XnPoiss(n)E(XnY=y)=k=0ykP(Xn=kY=y)==k=0ykP(Xn=k,X1+X2+...+Xn1=yk)P(X1+X2+...+Xn=y)==k=0yk1ke1k!(n1)yke(n1)(yk)!nyeny!=k=0ykCyk(1n)k(11n)yk==[Cyk(1n)k(11n)ykBin(y,1n)]=ynE(XnY)=YnX_1+X_2+...+X_{n-1}\sim Poiss\left( n-1 \right) \\X_1+X_2+...+X_n\sim Poiss\left( n \right) \\E\left( X_n|Y=y \right) =\sum_{k=0}^y{kP\left( X_n=k|Y=y \right)}=\\=\sum_{k=0}^y{k\frac{P\left( X_n=k,X_1+X_2+...+X_{n-1}=y-k \right)}{P\left( X_1+X_2+...+X_n=y \right)}}=\\=\sum_{k=0}^y{k\frac{\frac{1^ke^{-1}}{k!}\cdot \frac{\left( n-1 \right) ^{y-k}e^{-\left( n-1 \right)}}{\left( y-k \right) !}}{\frac{n^ye^{-n}}{y!}}}=\sum_{k=0}^y{kC_{y}^{k}\left( \frac{1}{n} \right) ^k\left( 1-\frac{1}{n} \right) ^{y-k}}=\\=\left[ C_{y}^{k}\left( \frac{1}{n} \right) ^k\left( 1-\frac{1}{n} \right) ^{y-k}\sim Bin\left( y,\frac{1}{n} \right) \right] =\frac{y}{n}\\E\left( X_n|Y \right) =\frac{Y}{n}


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