Question

Let X ? Poisson (?).

a. Show that the Poisson probabilities for x = 0, 1, 2, . . . can be computed recursively by
$p_0 = e -\lambda$ and $p_k=\frac{\lambda}{k}p_{k-1}$ for k = 1, 2, . . ..

P= P(X = )

b. Use the scheme to find P(X ? 4) for ? = 4.5.

Answer 1

a)

We know that

a)����� We know that^p_x = P (X = x) = lambda^x e^-lambda/x! p_0 = e^-lambda Now, p_1 = lambda/1 p_0 = lambda e^-lambda and p_2 = lambda/2 p_1 = lambda/2 (lambda e^-lambda) = lambda^2 e^- lambda/2! And so on... So, we conclude that Poisson probabilities can be computed recursively. b) Now, p_0 = e^-lambda = e^- 4.5 p_1 = lambda/1 p_0 = 4.5 e^-4.5 p_2 = lambda/2 p_1 = lambda/2 (lambda e^-lambda) = 4.5^2 e^- 4.5/2! P_3 = lambda^3 e^- lambda/3! = 4.5^3 e^- 4.5/3! P_4 = lambda^4 e^-lambda/4! = 4.5^4 e^-4.5/4! P(X lessthanorequalto 4) = p_0 + p_1 + p_2 + p_3 + p_4 = e^-4.5 + 4.5^1 e^-4.5/1! + 4.5^2 e^-4.5/2! + 4.5^3 e^-4.5/3! + 4.5^4 e^-4.5/4! = 0.5321

$p_{0}=e^{-\lambda}$

Now, $p_{1}=\frac{\lambda}{1}p_{0}=\lambda e^{-\lambda}$

and $p_{2}=\frac{\lambda}{2}p_{1}=\frac{\lambda}{2}(\lambda e^{-\lambda})=\frac{\lambda^{2}e^{-\lambda}}{2!}$ and so on...

So, we conclude that Poisson probabilities can be computed recursively.

b)

Now, $p_{0}=e^{-\lambda}=e^{-4.5}$

$p_{1}=\frac{\lambda}{1}p_{0}=4.5 e^{-4.5}$

$p_{2}=\frac{\lambda}{2}p_{1}=\frac{\lambda}{2}(\lambda e^{-\lambda})=\frac{4.5^{2}e^{-4.5}}{2!}$

$p_{3}=\frac{\lambda^{3}e^{-\lambda}}{3!}=\frac{4.5^{3}e^{-4.5}}{3!}$

$p_{4}=\frac{\lambda^{4}e^{-\lambda}}{4!}=\frac{4.5^{4}e^{-4.5}}{4!}$

$P(X\leq 4)=p_{0}+p_{1}+p_{2}+p_{3}+p_{4}$

$=e^{-4.5}+\frac{4.5^{1}e^{-4.5}}{1!}+\frac{4.5^{2}e^{-4.5}}{2!}+\frac{4.5^{3}e^{-4.5}}{3!}+\frac{4.5^{4}e^{-4.5}}{4!}=0.5321$