Math Problem Statement
Solution
The problem states that is a continuous random variable with the probability density function (pdf) given by:
2e^{-2x}, & x > 0 \\ 0, & \text{otherwise} \end{cases}$$ We are asked to find $$E[X^2]$$ and $$\text{Var}(X)$$. ### Step 1: Finding $$E[X^2]$$ The expectation $$E[X^2]$$ is defined as: $$E[X^2] = \int_{-\infty}^{\infty} x^2 f(x) \, dx$$ Given the pdf, we only need to integrate from 0 to $$\infty$$ because $$f(x) = 0$$ for $$x \leq 0$$. Thus: $$E[X^2] = \int_0^{\infty} x^2 \cdot 2e^{-2x} \, dx$$ We can use integration by parts to evaluate this integral. Alternatively, the integral of the form $$\int_0^{\infty} x^n e^{-ax} \, dx = \frac{n!}{a^{n+1}}$$ can be applied here. For $$n = 2$$ and $$a = 2$$, we get: $$E[X^2] = \frac{2!}{2^3} = \frac{2}{8} = \frac{1}{4}$$ ### Step 2: Finding $$E[X]$$ To calculate $$\text{Var}(X)$$, we also need $$E[X]$$, which is defined as: $$E[X] = \int_{0}^{\infty} x \cdot 2e^{-2x} \, dx$$ Using the same rule for integrals $$\int_0^{\infty} x^n e^{-ax} \, dx = \frac{n!}{a^{n+1}}$$, with $$n = 1$$ and $$a = 2$$, we get: $$E[X] = \frac{1!}{2^2} = \frac{1}{4}$$ ### Step 3: Finding $$\text{Var}(X)$$ The variance of $$X$$, $$\text{Var}(X)$$, is given by: \[ \text{Var}(X) = E[X^2] - (E[X])^2 \] Substituting the values we found: $$\text{Var}(X) = \frac{1}{4} - \left( \frac{1}{4} \right)^2 = \frac{1}{4} - \frac{1}{16} = \frac{4}{16} - \frac{1}{16} = \frac{3}{16}$$ ### Final Answers: - $$E[X^2] = \frac{1}{4}$$ - $$\text{Var}(X) = \frac{3}{16}$$ Let me know if you'd like any further details or clarification! --- Here are 5 follow-up questions related to this topic: 1. What are the conditions for a function to be a valid probability density function (pdf)? 2. How is the expected value of a continuous random variable related to the concept of center of mass in physics? 3. Can you compute $$E[X^3]$$ for the same pdf? What does this represent? 4. How would the variance change if the pdf was scaled by a factor? 5. What is the physical interpretation of variance in terms of probability distributions? **Tip:** When dealing with exponential functions in probability, knowing key integrals of the form $$\int_0^\infty x^n e^{-ax} \, dx$$ can save a lot of time!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability Density Function (PDF)
Expected Value
Variance
Integration
Formulas
E[X^2] = ∫ x^2 f(x) dx
E[X] = ∫ x f(x) dx
Var(X) = E[X^2] - (E[X])^2
Theorems
Integration by parts
Exponential Distribution Properties
Suitable Grade Level
Undergraduate (Probability and Statistics)
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