Question

Problem 2 Let X is a random variable with Poisson distribution X ~ Poisson(λ), (a) Find E(X1X2 i). λ > 0. 、 (b) Find E(xIx2). (c)Prove that λ>2-2a-ka for λ>0.

a b and c please

and thankyou

Answer 1

The Poisson distribution has the probability distribution $P(X=x) = \frac{\lambda^x e^{-\lambda}}{x!}$ .

(a) $E(X|X \geq 1) = \sum_{x \geq 1} xP(X=x)$ or $E(X|X \geq 1) = \sum_{x \geq 1} x\frac{\lambda^x e^{-\lambda}}{x!}$ or $E(X|X \geq 1) = 1*\frac{\lambda^1 e^{-\lambda}}{1!} + 2*\frac{\lambda^2 e^{-\lambda}}{2!} + 3*\frac{\lambda^3 e^{-\lambda}}{3!} + 4*\frac{\lambda^1 e^{-\lambda}}{4!} + ...$ or $E(X|X \geq 1) = 0*\frac{\lambda^0 e^{-\lambda}}{0!} + 1*\frac{\lambda^1 e^{-\lambda}}{1!} + 2*\frac{\lambda^2 e^{-\lambda}}{2!} + 3*\frac{\lambda^3 e^{-\lambda}}{3!} + ... - 0*\frac{\lambda^0 e^{-\lambda}}{0!}$ or $E(X|X \geq 1) = \sum_{x \geq 0} x\frac{\lambda^x e^{-\lambda}}{x!} - 0*\frac{\lambda^0 e^{-\lambda}}{0!}$ or $E(X|X \geq 1) = E(X) - 0*\frac{\lambda^0 e^{-\lambda}}{0!}$ or $E(X|X \geq 1) = \lambda - 0$ or $E(X|X \geq 1) = \lambda$ .

(b) $E(X|X \geq 2) = \sum_{x \geq 2} xP(X=x)$ or $E(X|X \geq 2) = \sum_{x \geq 2} x\frac{\lambda^x e^{-\lambda}}{x!}$ or $E(X|X \geq 2) = 2*\frac{\lambda^2 e^{-\lambda}}{2!} + 3*\frac{\lambda^3 e^{-\lambda}}{3!} + 4*\frac{\lambda^1 e^{-\lambda}}{4!} + 5*\frac{\lambda^5 e^{-\lambda}}{5!} + ...$ or $E(X|X \geq 2) = 0*\frac{\lambda^0 e^{-\lambda}}{0!} + 1*\frac{\lambda^1 e^{-\lambda}}{1!} + 2*\frac{\lambda^2 e^{-\lambda}}{2!} + 3*\frac{\lambda^3 e^{-\lambda}}{3!} + 4*\frac{\lambda^4 e^{-\lambda}}{4!} + 5*\frac{\lambda^5 e^{-\lambda}}{5!} ... - 0*\frac{\lambda^0 e^{-\lambda}}{0!} - 1*\frac{\lambda^1 e^{-\lambda}}{1!}$ or $E(X|X \geq 2) = \sum_{x \geq 0} x\frac{\lambda^x e^{-\lambda}}{x!} - 0*\frac{\lambda^0 e^{-\lambda}}{0!} - 1*\frac{\lambda^1 e^{-\lambda}}{1!}$ or $E(X|X \geq 2) = E(X) - 0*\frac{\lambda^0 e^{-\lambda}}{0!} - 1*\frac{\lambda^1 e^{-\lambda}}{1!}$ or $E(X|X \geq 2) = \lambda - 0 - \lambda e^{-\lambda}$ or $E(X|X \geq 2) = \lambda - \lambda e^{-\lambda}$ .

(c) We have to prove that $\lambda > 2 - 2e^{-\lambda} - \lambda e^{-\lambda}$ or $\lambda + 2e^{-\lambda} + \lambda e^{-\lambda} > 2$ for $\lambda > 0$ .

We have $\lambda + 2e^{-\lambda} + \lambda e^{-\lambda} = \lambda + e^{-\lambda} (2 + \lambda)$ , and taking the limit that $\lambda \rightarrow 0$ , we have $\underset{\lambda \rightarrow 0}{lim} [\lambda + 2e^{-\lambda} + \lambda e^{-\lambda}] = \underset{\lambda \rightarrow 0}{lim} [\lambda + e^{-\lambda} (2 + \lambda)]$ or $\underset{\lambda \rightarrow 0}{lim} [\lambda + 2e^{-\lambda} + \lambda e^{-\lambda}] = \underset{\lambda \rightarrow 0}{lim} \lambda + \underset{\lambda \rightarrow 0}{lim} [e^{-\lambda} (2 + \lambda)]$ or $\underset{\lambda \rightarrow 0}{lim} [\lambda + 2e^{-\lambda} + \lambda e^{-\lambda}] = 0 + 2 * \underset{\lambda \rightarrow 0}{lim} [e^{-\lambda}]$ or $\underset{\lambda \rightarrow 0}{lim} [\lambda + 2e^{-\lambda} + \lambda e^{-\lambda}] = 0 + 2 * 1$ or $\underset{\lambda \rightarrow 0}{lim} [\lambda + 2e^{-\lambda} + \lambda e^{-\lambda}] = 2$ .

The reason being that for any constant c>0, as x(>0) decreases, $c^x$ tends to 1, and so does $c^{-x}$ . The difference is that for c>0 and x>0, and it decreases to 1, while , and it increases to 1.

The worst case where lambda is zero, is where $\lambda + 2e^{-\lambda} + \lambda e^{-\lambda} = 2$ . For any $\lambda > 0$ , $\lambda + 2e^{-\lambda} + \lambda e^{-\lambda} > 2$ .

Further, as   $\frac{\partial }{\partial \lambda}(\lambda + 2e^{-\lambda} + \lambda e^{-\lambda}) = 1 - 2e^{-\lambda} + e^{-\lambda} - \lambda e^{-\lambda}$ or $\frac{\partial }{\partial \lambda}(\lambda + 2e^{-\lambda} + \lambda e^{-\lambda}) = 1 - e^{-\lambda} - \lambda e^{-\lambda}$ and $e^{-\lambda} - \lambda e^{-\lambda}$ is always less than one for $0 < \lambda < 1$ , meaning that $\frac{\partial }{\partial \lambda}(\lambda + 2e^{-\lambda} + \lambda e^{-\lambda}) > 0$ for all lambda, which funrther means that the $\lambda + 2e^{-\lambda} + \lambda e^{-\lambda}$ is increasing function in lambda, and never decreases as lambda increases.

Hence, we can say that for $\lambda > 0$ , $\lambda + 2e^{-\lambda} + \lambda e^{-\lambda} > 2$ or $\lambda > 2 - 2e^{-\lambda} - \lambda e^{-\lambda}$ .

A graph for lambda even greater than 1 is as below.