4F (Probability)

Solution

λ > 0

P (Z = k) = e^{- λ} \frac{λ ^{k}}{k !}, k = 0, 1, 2, \dots .

The exponential series makes these probabilities sum to one. Shifting the indices in the absolutely convergent sums gives

EZ = e^{- λ} \sum_{k \geq 1} \frac{k λ ^{k}}{k !} = λ, E [Z (Z - 1)] = λ^{2} .

Hence

E Z^{2} = λ^{2} + λ

and

EZ = Var Z = λ .

Z_{1}, \dots, Z_{n}

of parameter one, let

S_{n} = \sum_{j} Z_{j}

. Summing their product probability mass function over all nonnegative

k_{1} + \dots + k_{n} = k

gives

P (S_{n} = k) = e^{- n} \sum_{k_{1} + \dots + k_{n} = k} \frac{1}{k _{1} ! \dots k _{n} !} = e^{- n} \frac{n ^{k}}{k !} .

The final equality follows from the multinomial theorem applied to

(1 + \dots + 1)^{k}

. Thus $S_{n}$ has the Poisson distribution of parameter $n$ .

Each

Z_{j}

\frac{S _{n} - n}{n} ⟹ N (0, 1) .

e^{- n} k = 0 \sum n \frac{n ^{k}}{k !} = P (S_{n} \leq n) = P (\frac{S _{n} - n}{n} \leq 0) ⟶ \frac{1}{2} .

S_{n} = n

causes no difficulty: the central limit theorem applies to the entire event with threshold zero.

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