Math Problem Statement
express this using the fonction de repartition
Solution
The exercise requires using the cumulative distribution function (CDF), often denoted , to calculate the probabilities for a continuous random variable with a given probability density function (PDF).
The PDF provided for is as follows:
x & \text{if } 0 \leq x \leq 2, \\ 0 & \text{elsewhere}. \end{cases}$$ To solve this using the CDF $$F_X(x)$$, let's go through each part: ### Step 1: Find the CDF $$F_X(x)$$ The CDF $$F_X(x)$$ is obtained by integrating the PDF from $$-\infty$$ up to $$x$$: $$F_X(x) = \int_{-\infty}^{x} f_X(t) \, dt.$$ Since $$f_X(x) = 0$$ for $$x < 0$$ and $$x > 2$$, we only need to integrate from 0 to $$x$$ within the interval $$0 \leq x \leq 2$$: 1. **For $$x < 0$$:** $$F_X(x) = 0$$, since $$X$$ cannot take values outside $$[0, 2]$$. 2. **For $$0 \leq x \leq 2$$:** $$F_X(x) = \int_{0}^{x} t \, dt = \left[ \frac{t^2}{2} \right]_{0}^{x} = \frac{x^2}{2}.$$ 3. **For $$x \geq 2$$:** $$F_X(x) = 1$$, since the total probability must equal 1. Thus, the CDF is: $$F_X(x) = \begin{cases} 0 & \text{if } x < 0, \\ \frac{x^2}{2} & \text{if } 0 \leq x \leq 2, \\ 1 & \text{if } x \geq 2. \end{cases}$$ ### Step 2: Compute the Probabilities Using the CDF Now, let's solve each part of the question: #### i) $$P(X > 1)$$ Using the complement rule, $$P(X > 1) = 1 - P(X \leq 1)$$: $$P(X > 1) = 1 - F_X(1).$$ Since $$F_X(1) = \frac{1^2}{2} = \frac{1}{2}$$, $$P(X > 1) = 1 - \frac{1}{2} = \frac{1}{2}.$$ #### ii) $$P(X < 1)$$ This is simply $$F_X(1)$$: $$P(X < 1) = F_X(1) = \frac{1}{2}.$$ #### iii) $$P\left( \frac{1}{2} < X < \frac{3}{2} \right)$$ We calculate this as $$P\left( X < \frac{3}{2} \right) - P\left( X \leq \frac{1}{2} \right)$$: $$P\left( \frac{1}{2} < X < \frac{3}{2} \right) = F_X\left( \frac{3}{2} \right) - F_X\left( \frac{1}{2} \right).$$ 1. **Calculate $$F_X\left( \frac{3}{2} \right)$$:** $$F_X\left( \frac{3}{2} \right) = \frac{\left( \frac{3}{2} \right)^2}{2} = \frac{\frac{9}{4}}{2} = \frac{9}{8}.$$ 2. **Calculate $$F_X\left( \frac{1}{2} \right)$$:** $$F_X\left( \frac{1}{2} \right) = \frac{\left( \frac{1}{2} \right)^2}{2} = \frac{\frac{1}{4}}{2} = \frac{1}{8}.$$ So, $$P\left( \frac{1}{2} < X < \frac{3}{2} \right) = \frac{9}{8} - \frac{1}{8} = \frac{8}{8} = 1.$$ ### Final Answers: 1. $$P(X > 1) = \frac{1}{2}$$. 2. $$P(X < 1) = \frac{1}{2}$$. 3. $$P\left( \frac{1}{2} < X < \frac{3}{2} \right) = 1$$. Would you like more details or have any questions? ### Additional Questions: 1. How is the CDF derived from the PDF in other cases? 2. What are some properties of the cumulative distribution function? 3. How does integration help in finding probabilities for continuous random variables? 4. Can the probability of a continuous random variable at a single point be non-zero? 5. How does the complement rule work for calculating probabilities? ### Tip: Always verify the boundaries of the random variable's support when calculating probabilities to avoid errors.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability
Cumulative Distribution Function
Continuous Random Variables
Formulas
Cumulative distribution function F_X(x) = ∫_{-∞}^x f_X(t) dt
P(X > a) = 1 - F_X(a)
P(a < X < b) = F_X(b) - F_X(a)
Theorems
Definition of CDF from PDF
Complement Rule for Probabilities
Suitable Grade Level
College-level Probability and Statistics
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