Math Problem Statement
Solution
Let's solve the problem step by step based on the information given in the image:
Problem Statement:
We are given a cumulative distribution function (CDF) defined as:
\begin{cases} 0, & x \leq 0 \\ \frac{x^2}{C + x^2}, & x > 0 \end{cases}$$ We are tasked with solving the following: 1. Find the parameter $$C$$. 2. Derive the probability density function (PDF), $$f(x)$$. 3. Plot the graphs of $$f(x)$$ and $$F(x)$$. 4. Find the main characteristics (mean, variance, etc.). 5. Find $$P(1 \leq x \leq 3)$$. --- ### Solution: #### **Step 1: Find the parameter $$C$$** The cumulative distribution function $$F(x)$$ must satisfy the condition: $$\lim_{x \to \infty} F(x) = 1.$$ From $$F(x) = \frac{x^2}{C + x^2}$$, as $$x \to \infty$$: $$\frac{x^2}{C + x^2} \to 1 \implies C = 1.$$ Thus, $$C = 1$$. --- #### **Step 2: Derive the PDF, $$f(x)$$** The probability density function is the derivative of the CDF: $$f(x) = \frac{d}{dx} F(x).$$ For $$x > 0$$, using $$F(x) = \frac{x^2}{1 + x^2}$$: $$f(x) = \frac{d}{dx} \left( \frac{x^2}{1 + x^2} \right).$$ Using the quotient rule, $$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$$, where $$u = x^2$$ and $$v = 1 + x^2$$: $$f(x) = \frac{2x(1 + x^2) - x^2(2x)}{(1 + x^2)^2}.$$ Simplify: $$f(x) = \frac{2x + 2x^3 - 2x^3}{(1 + x^2)^2} = \frac{2x}{(1 + x^2)^2}.$$ Thus, the PDF is: $$f(x) = \begin{cases} 0, & x \leq 0 \\ \frac{2x}{(1 + x^2)^2}, & x > 0 \end{cases}.$$ --- #### **Step 3: Plot the graphs of $$f(x)$$ and $$F(x)$$** The graphs of $$F(x)$$ and $$f(x)$$ can be plotted. I can generate them for you if you'd like. --- #### **Step 4: Main characteristics** To find the main characteristics such as the mean $$\mathbb{E}[X]$$ and variance $$\text{Var}(X)$$, we need to evaluate the following: 1. **Mean:** $$\mathbb{E}[X] = \int_{0}^{\infty} x f(x) dx = \int_{0}^{\infty} x \cdot \frac{2x}{(1 + x^2)^2} dx.$$ Simplify: $$\mathbb{E}[X] = \int_{0}^{\infty} \frac{2x^2}{(1 + x^2)^2} dx.$$ Let $$u = 1 + x^2$$, so $$du = 2x dx$$. When $$x = 0, u = 1$$, and as $$x \to \infty, u \to \infty$$. The integral becomes: \[ \mathbb{E}[X] = \int_{1}^{\infty} \frac{1}{u^2} du = \left[ -\frac{1}{u} \right]_{1}^{\infty}. \] Evaluate: $$\mathbb{E}[X] = 0 - (-1) = 1.$$ Thus, the mean is $$\mathbb{E}[X] = 1$$. 2. **Variance:** The variance is: \[ \text{Var}(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2. \] First, find $$\mathbb{E}[X^2]$$: $$\mathbb{E}[X^2] = \int_{0}^{\infty} x^2 f(x) dx = \int_{0}^{\infty} x^2 \cdot \frac{2x}{(1 + x^2)^2} dx.$$ Simplify: $$\mathbb{E}[X^2] = \int_{0}^{\infty} \frac{2x^3}{(1 + x^2)^2} dx.$$ Using the same substitution $$u = 1 + x^2$$, $$du = 2x dx$$, and $$x^2 = u - 1$$, the integral becomes: $$\mathbb{E}[X^2] = \int_{1}^{\infty} \frac{u - 1}{u^2} du.$$ Split into two terms: $$\mathbb{E}[X^2] = \int_{1}^{\infty} \frac{1}{u} du - \int_{1}^{\infty} \frac{1}{u^2} du.$$ Evaluate each term: \[ \int_{1}^{\infty} \frac{1}{u} du = \ln(u) \Big|_{1}^{\infty} = \infty \quad \text{(Correct Integrals needed adjustAsk a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability Theory
Cumulative Distribution Function (CDF)
Probability Density Function (PDF)
Integration
Differentiation
Formulas
F(x) = 0 for x ≤ 0; F(x) = x^2 / (C + x^2) for x > 0
f(x) = dF(x)/dx = derivative of F(x)
P(a ≤ X ≤ b) = ∫[a, b] f(x) dx
Theorems
Properties of CDFs
Normalization condition for PDFs
Basic integration and substitution rules
Suitable Grade Level
Undergraduate (Mathematics or Statistics)
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