1.6.2 Example: Poisson distribution Let Y1,., Yn be independent random variables each with the Poisson distri- bution f(y; 0) Yi = 0, 1, 2, ... y!! with the same parameter 0. Their joint distribution is OYie-0 QYne-0 f (41,... ,Yn; 0) |[ S(4;;0) = ! Y2! Yn! i=1 . -no yn!y2!..yn! This is also the likelihood function L(0; y1,..., Yn). It is easier to use the log- likelihood function 1(0; Y1,. Yn) = log L(0; y1,..., Yn) = (E 4) log 0 no E(log y:!). ..... To find the maximum likelihood estimate 0, use dl . Yi - n. OP Equate this to zero to obtain the solution =E4i/n = J. Since de1/de? = -E4:/02 < 0, 1 has its maximum value when 0 = 6, con- firming that j is the maximum likelihood estimate.