Finding Mean and Variance for an Exponential Distribution

Math Problem Statement

A continuous random variable has the probability density function f(x) = 2e^(-2x) for x > 0. Find the mean and variance.

The image shows a problem involving a continuous random variable with a given probability density function (PDF):

\begin{cases} 2e^{-2x}, & x > 0 \\ 0, & x \leq 0 \end{cases}$$ The task is to find the **mean** and **variance** of this random variable. ### 1. Mean ($$\mu$$) The mean of a continuous random variable $$X$$ with probability density function $$f(x)$$ is given by: $$\mu = \int_{-\infty}^{\infty} x f(x) \, dx$$ For this specific PDF, the integral becomes: $$\mu = \int_0^\infty x \cdot 2e^{-2x} \, dx$$ We can solve this using integration by parts. ### 2. Variance ($$\sigma^2$$) The variance is defined as: \[ \sigma^2 = E[X^2] - (E[X])^2 \] First, we compute $$E[X^2]$$: $$E[X^2] = \int_{0}^{\infty} x^2 \cdot 2e^{-2x} \, dx$$ Once we find $$E[X^2]$$, we can calculate the variance using: $$\sigma^2 = E[X^2] - \mu^2$$ Would you like the detailed step-by-step solution for these integrals? --- **Related questions:** 1. How do we use integration by parts in finding the mean of a continuous distribution? 2. What is the interpretation of the mean and variance for a probability distribution? 3. How does this PDF resemble an exponential distribution? 4. Can we find the moment generating function for this distribution? 5. How do transformations affect the mean and variance of a random variable? **Tip:** For many distributions, like the exponential, the mean is the reciprocal of the rate parameter!

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Probability
Continuous Random Variables
Probability Density Functions
Exponential Distribution

Mean: μ = ∫[x * f(x)] dx
Variance: σ² = E[X²] - (E[X])²
E[X²] = ∫[x² * f(x)] dx

Properties of continuous probability distributions
Mean and variance of exponential distributions

Undergraduate or Advanced High School

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