The maximum likelihood estimate for λ
of a Poisson distribution P(X=x)=f(x,λ)
∑ilnf(xi,λ)→maxl=∑ilnλxie−λxi!=(x1+x2+...+xn)lnλ−nλ−lnx1!x2!...xn!→maxdldλ=x1+x2+...+xnλ−ndldλ=0⇒λ=xˉIt is maxbecause l(0)=−∞,l(+∞)=−∞Thusλ^=xˉ\sum_i{\ln f\left( x_i,\lambda \right)}\rightarrow \max \\l=\sum_i{\ln \frac{\lambda ^{x_i}e^{-\lambda}}{x_i!}}=\left( x_1+x_2+...+x_n \right) \ln \lambda -n\lambda -\ln x_1!x_2!...x_n!\rightarrow \max \\\frac{dl}{d\lambda}=\frac{x_1+x_2+...+x_n}{\lambda}-n\\\frac{dl}{d\lambda}=0\Rightarrow \lambda =\bar{x}\\It\,\,is\,\,\max because\,\,l\left( 0 \right) =-\infty ,l\left( +\infty \right) =-\infty \\Thus\\\hat{\lambda}=\bar{x}∑ilnf(xi,λ)→maxl=∑ilnxi!λxie−λ=(x1+x2+...+xn)lnλ−nλ−lnx1!x2!...xn!→maxdλdl=λx1+x2+...+xn−ndλdl=0⇒λ=xˉItismaxbecausel(0)=−∞,l(+∞)=−∞Thusλ^=xˉ
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