<Dd> P (x) = ∏ i = 1 n λ x i e − λ x i! = 1 ∏ i = 1 n x i! × λ ∑ i = 1 n x i e − n λ (\ displaystyle P (\ mathbf (x)) = \ prod _ (i = 1) ^ (n) (\ frac (\ lambda ^ (x_ (i)) e ^ (- \ lambda)) (x_ (i)!)) = (\ frac (1) (\ prod _ (i = 1) ^ (n) x_ (i)!)) \ times \ lambda ^ (\ sum _ (i = 1) ^ (n) x_ (i)) e ^ (- n \ lambda)) </Dd> <P> Note that the first term, h (x) (\ displaystyle h (\ mathbf (x))), depends only on x (\ displaystyle \ mathbf (x)). The second term, g (T (x) λ) (\ displaystyle g (T (\ mathbf (x)) \ lambda)), depends on the sample only through T (x) = ∑ i = 1 n x i (\ displaystyle T (\ mathbf (x)) = \ sum _ (i = 1) ^ (n) x_ (i)). Thus, T (x) (\ displaystyle T (\ mathbf (x))) is sufficient . </P> <P> To find the parameter λ that maximizes the probability function for the Poisson population, we can use the logarithm of the likelihood function: </P> <Dl> <Dd> l (λ) = ln ⁡ ∏ i = 1 n f (k i ∣ λ) = ∑ i = 1 n ln (e − λ λ k i k i!) = − n λ + (∑ i = 1 n k i) ln ⁡ (λ) − ∑ i = 1 n ln ⁡ (k i!). (\ displaystyle (\ begin (aligned) \ ell (\ lambda) & = \ ln \ prod _ (i = 1) ^ (n) f (k_ (i) \ mid \ lambda) \ \ & = \ sum _ (i = 1) ^ (n) \ ln \! \ left ((\ frac (e ^ (- \ lambda) \ lambda ^ (k_ (i))) (k_ (i)!)) \ right) \ \ & = - n \ lambda + \ left (\ sum _ (i = 1) ^ (n) k_ (i) \ right) \ ln (\ lambda) - \ sum _ (i = 1) ^ (n) \ ln (k_ (i)!). \ end (aligned))) </Dd> </Dl>

How to find the mean and standard deviation of a poisson distribution