The second moment of a Poisson-distributed random variable is 2. The mean of the random variable is _______.
Correct Answer
Solution & Step-by-step Explanation
In the range: 1 - 1
Explanation:
The Poisson distribution is characterized by the property that its mean (μ) is equal to its variance (σ²). In other words, if X is a Poisson-distributed random variable:
E[X] = μ = λ,
Var[X] = σ² = λ,
where λ is the rate parameter for the Poisson distribution.
Furthermore, the second moment of a random variable X, E[X²], is related to its mean and variance by:
E[X²] = Var[X] + (E[X])².
Substituting the expressions for the mean and variance of a Poisson distribution, we get:
E[X²] = λ + λ².
In your case, you're given that the second moment E[X²] is 2. Equating this to λ + λ² and solving for λ, you get a quadratic equation:
λ² + λ - 2 = 0.
Solving this quadratic equation gives two solutions, λ = 1 and λ = -2. Since the rate parameter λ of a Poisson distribution must be positive (it represents a rate or number of occurrences), we discard the negative solution, so:
λ = 1.
Therefore, the mean of the random variable, which is equal to λ for a Poisson distribution, is 1.
Explanation:
The Poisson distribution is characterized by the property that its mean (μ) is equal to its variance (σ²). In other words, if X is a Poisson-distributed random variable:
E[X] = μ = λ,
Var[X] = σ² = λ,
where λ is the rate parameter for the Poisson distribution.
Furthermore, the second moment of a random variable X, E[X²], is related to its mean and variance by:
E[X²] = Var[X] + (E[X])².
Substituting the expressions for the mean and variance of a Poisson distribution, we get:
E[X²] = λ + λ².
In your case, you're given that the second moment E[X²] is 2. Equating this to λ + λ² and solving for λ, you get a quadratic equation:
λ² + λ - 2 = 0.
Solving this quadratic equation gives two solutions, λ = 1 and λ = -2. Since the rate parameter λ of a Poisson distribution must be positive (it represents a rate or number of occurrences), we discard the negative solution, so:
λ = 1.
Therefore, the mean of the random variable, which is equal to λ for a Poisson distribution, is 1.