我有一個大學的機率數學題 ><!急
when calculating all the values of a Poisson distribution the work can often be simplified by first calculating p(0,λ) and then using the recursion formula
p(x+1;λ)=λ/(x+1)*p(x;λ)
verify this formula and use it and e^-2=0.1353 to verify the values given in table II for λ=2
please steps by steps !
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Poission Distributionp(x;λ)=e^-λ*λ^x/x!p(x+1;λ)=e^-λ*λ^(x+1)/(x+1)!p(x+1;λ)/ p(x;λ)=[ e^-λ*λ^(x+1)/(x+1)!]/[ e^-λ*λ^x/x!]=λ/(x+1)p(x+1;λ)=λ/(x+1)*p(x;λ) p(0;λ)=e^-λ*λ^0/0!= e^-λp(0;2)=e^-2=0.1353p(1;2)=2/1* p(0;2)=0.2706p(2;2)=2/2* p(1;2)=0.2706p(3;2)=2/3* p(2;2)=0.1804p(4;2)=2/4* p(3;2)=0.0902p(5;2)=2/5* p(4;2)=0.0361p(6;2)=2/6* p(5;2)=0.0120p(7;2)=2/7* p(6;2)=0.0034p(8;2)=2/8* p(7;2)=0.0009p(9;2)=2/9* p(8;2)=0.0002……