Solution:
We know that exponential family of distribution is of the form:
"P(x,\\theta)=h(x).\\exp(\\theta^TT(x)-A(\\theta))" ...(i)
Given, "P(x,a)=a^x(1-a)^{1-x}; 0<a<1;x=0,1"
"=\\exp[\\log (a^x(1-a)^{1-x}) ]\n\\\\=\\exp[\\log a^x+\\log(1-a)^{1-x} ]\n\\\\=\\exp[x\\log a+(1-x)\\log(1-a) ]\n\\\\=\\exp[x\\log a+\\log(1-a)-x\\log(1-a) ]\n\\\\=\\exp[x\\log \\dfrac{a}{1-a}+\\log(1-a) ]\n\\\\=\\exp[x\\theta-\\log(1+e^{\\theta})] ...(ii)"
where "T(x)=x,\\theta=\\log \\dfrac{a}{1-a},A(\\theta)=\\log(1+e^{\\theta})"
Hence, (ii) matches with (i), given probability density function is a member of an exponential family
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