Math Problem Statement

If a random variable x is Gamma distribution with parameter λ = 3, compute the value of

  1. P(x ≤ 1) 2. P(1 ≤ x ≤ 2)

Solution

Given that xx is a random variable that follows a Gamma distribution with parameter λ=3\lambda = 3, we need to compute the following probabilities:

  1. P(x1)P(x \leq 1)
  2. P(1x2)P(1 \leq x \leq 2)

Let's proceed step-by-step:

Gamma Distribution

The Gamma distribution's probability density function (PDF) for a random variable xx is given by:

f(x;α,λ)=λαxα1eλxΓ(α)f(x; \alpha, \lambda) = \frac{\lambda^\alpha x^{\alpha - 1} e^{-\lambda x}}{\Gamma(\alpha)}

where:

  • α\alpha is the shape parameter,
  • λ\lambda is the rate parameter,
  • Γ(α)\Gamma(\alpha) is the Gamma function, which generalizes the factorial function.

Interpreting the Problem

The problem provides us with λ=3\lambda = 3, but we do not have an explicit value for α\alpha (shape parameter). If only λ\lambda is given and no α\alpha is specified, a typical interpretation (or the most straightforward) would be to assume the standard exponential distribution, where α=1\alpha = 1. This is because the exponential distribution is a special case of the Gamma distribution with α=1\alpha = 1.

Hence, we assume:

  • Shape parameter α=1\alpha = 1
  • Rate parameter λ=3\lambda = 3

Exponential Distribution PDF

With α=1\alpha = 1, the Gamma distribution simplifies to an exponential distribution with PDF:

f(x;λ)=λeλxf(x; \lambda) = \lambda e^{-\lambda x}

For our problem, this becomes:

f(x;3)=3e3xf(x; 3) = 3 e^{-3x}

Computing the Probabilities

  1. Probability P(x1)P(x \leq 1)

P(x1)=013e3xdxP(x \leq 1) = \int_0^1 3 e^{-3x} \, dx

Using integration by substitution, let u=3xu = -3x, hence du=3dxdu = -3dx, which gives dx=13dudx = -\frac{1}{3}du. The integral becomes:

P(x1)=013e3xdx=[e3x]01=1e3P(x \leq 1) = \int_0^1 3 e^{-3x} \, dx = \left[-e^{-3x}\right]_0^1 = 1 - e^{-3}

P(x1)1e310.04980.9502P(x \leq 1) \approx 1 - e^{-3} \approx 1 - 0.0498 \approx 0.9502

  1. Probability P(1x2)P(1 \leq x \leq 2)

P(1x2)=123e3xdxP(1 \leq x \leq 2) = \int_1^2 3 e^{-3x} \, dx

Following a similar method:

P(1x2)=[e3x]12=e3e6P(1 \leq x \leq 2) = \left[-e^{-3x}\right]_1^2 = e^{-3} - e^{-6}

P(1x2)0.04980.00250.0473P(1 \leq x \leq 2) \approx 0.0498 - 0.0025 \approx 0.0473

Final Answers

  1. P(x1)0.9502P(x \leq 1) \approx 0.9502
  2. P(1x2)0.0473P(1 \leq x \leq 2) \approx 0.0473

Would you like more details, or do you have any questions?

Related Questions

  1. How does the Gamma distribution differ from the exponential distribution?
  2. What is the cumulative distribution function (CDF) of the exponential distribution?
  3. How can we interpret the shape and rate parameters of the Gamma distribution?
  4. How would the probability change if the shape parameter α\alpha were different?
  5. What are the mean and variance of a Gamma-distributed random variable?

Tip

The exponential distribution is a special case of the Gamma distribution and is commonly used to model the time until the next event in a Poisson process.

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Math Problem Analysis

Mathematical Concepts

Gamma Distribution
Exponential Distribution
Probability Theory

Formulas

Gamma distribution PDF
Exponential distribution PDF
Integral calculus

Theorems

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Suitable Grade Level

Advanced Mathematics