Question

If the second moment of a Poisson distribution is 6 , find the probability P(X ≥ 2).

Answer 1

Answer on Question #66776 – Math – Statistics and Probability

Question

If the second moment of a Poisson distribution is 6, find the probability $P(X \geq 2)$.

Solution

If $X$ has a Poisson distribution, then

$$P(X = k) = \frac{\lambda^k}{k!} e^{-\lambda}, \quad \lambda > 0, k = 0, 1, 2, \dots$$

(see https://en.wikipedia.org/wiki/Poisson_distribution#Definition).

From the definition of the second moment (see https://en.wikipedia.org/wiki/Moment_(mathematics)) we get

$$\mathbb{E}(X^2) = 6.$$

On the other hand,

$$\mathbb{E}(X^2) = \operatorname{Var}(X) + [\mathbb{E}(X)]^2$$

(see https://en.wikipedia.org/wiki/Variance#Definition).

Since $X$ has a Poisson distribution then

$$\mathbb{E}(X) = \operatorname{Var}(X) = \lambda$$

(see https://en.wikipedia.org/wiki/Poisson_distribution#Definition).

We have

$$6 = \lambda + \lambda^2 \Rightarrow \lambda^2 + \lambda - 6 = 0.$$

Let us solve the latter equation. Using the quadratic formula (see https://en.wikipedia.org/wiki/Quadratic_formula) we obtain

$$\lambda = \frac{-1 \pm \sqrt{1 + 24}}{2} \Rightarrow \begin{bmatrix} \lambda_1 = 2 \\ \lambda_2 = -3 \end{bmatrix}.$$

Since $\lambda_2 = -3 < 0$ we conclude that $\lambda = 2$ and

$$P(X = k) = \frac{2^k}{k!} e^{-2}.$$

Then required probability is

$$\begin{array}{l} P (X \geq 2) = 1 - P (X < 2) = \\ = 1 - \left[ P (X = 0) + P (X = 1) \right] = 1 - \left[ e ^ {- 2} + 2 e ^ {- 2} \right] = 1 - 3 e ^ {- 2} = 1 - \frac {3}{e ^ {2}}. \end{array}$$

Answer: $1 - \frac{3}{e^2}$

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