Question

Let \( X \) be a Poisson distributed random variable with parameter \( \lambda (> 0) \) such that it satisfies the equation \[ P(X = 1) = 3P(X = 3) - P(X = 2) {. Then, the value of } \lambda { is:} \]

Hint:

In Poisson distribution problems, always use the probability mass function to relate given conditions to find the unknown parameter, in this case, \( \lambda \).

Answer 1

The probability mass function for a Poisson distribution is given by: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \] We are given the equation: \[ P(X = 1) = 3P(X = 3) - P(X = 2) \] Substitute the Poisson distribution formula for \(P(X = 1)\), \(P(X = 2)\), and \(P(X = 3)\): \[ \frac{\lambda^1 e^{-\lambda}}{1!} = 3 \cdot \frac{\lambda^3 e^{-\lambda}}{3!} - \frac{\lambda^2 e^{-\lambda}}{2!} \] Simplifying: \[ \lambda e^{-\lambda} = 3 \cdot \frac{\lambda^3 e^{-\lambda}}{6} - \frac{\lambda^2 e^{-\lambda}}{2} \] \[ \lambda e^{-\lambda} = \frac{\lambda^3 e^{-\lambda}}{2} - \frac{\lambda^2 e^{-\lambda}}{2} \] Factor out \( e^{-\lambda} \) (since it's non-zero) from both sides: \[ \lambda = \frac{\lambda^3}{2} - \frac{\lambda^2}{2} \] Multiply through by 2 to eliminate the denominators: \[ 2\lambda = \lambda^3 - \lambda^2 \] Rearranging the terms: \[ \lambda^3 - \lambda^2 - 2\lambda = 0 \] Factor out \( \lambda \): \[ \lambda(\lambda^2 - \lambda - 2) = 0 \] Thus, we have two possible solutions for \( \lambda \): \[ \lambda = 0 \quad {or} \quad \lambda^2 - \lambda - 2 = 0 \] Since \( \lambda>0 \), we solve the quadratic equation: \[ \lambda^2 - \lambda - 2 = 0 \] Using the quadratic formula: \[ \lambda = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)} = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm \sqrt{9}}{2} = \frac{1 \pm 3}{2} \] Thus, \( \lambda = 2 \) (since \( \lambda>0 \)). 
Correct Answer:} \( \lambda = 2 \)