Math Problem Statement

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Solution

Let's address each problem one by one.

Problem 3:

Given:

  • XX is a continuous random variable with the probability density function (pdf) fX(x)=8x3f_X(x) = \frac{8}{x^3} for x>2x > 2.
  • We need to find E[W]E[W], where W=X3W = \frac{X}{3}.

Solution:

  1. Since W=X3W = \frac{X}{3}, we can express E[W]E[W] as: [ E[W] = E\left[\frac{X}{3}\right] = \frac{1}{3}E[X]. ]

  2. To find E[X]E[X], we use: E[X]=2xfX(x)dx=2x8x3dx.E[X] = \int_{2}^{\infty} x \cdot f_X(x) dx = \int_{2}^{\infty} x \cdot \frac{8}{x^3} dx.

  3. Simplify the integral: [ E[X] = 8 \int_{2}^{\infty} \frac{1}{x^2} dx = 8 \left[ -\frac{1}{x} \right]_2^{\infty} = 8 \left(0 + \frac{1}{2}\right) = 4. ]

  4. Therefore, E[W]=134=43.E[W] = \frac{1}{3} \cdot 4 = \frac{4}{3}.

Problem 4:

Given:

  • We need to find the expected value of the function g(x)=x2g(x) = x^2 where XX is a random variable with pdf fX(x)=aeaxu(x)f_X(x) = ae^{-ax}u(x), where aa is a constant, and u(x)u(x) is the unit step function.

Solution:

  1. E[g(X)]=E[X2]E[g(X)] = E[X^2] can be found by: E[X2]=0x2aeaxdx.E[X^2] = \int_{0}^{\infty} x^2 \cdot ae^{-ax} dx.

  2. This is a standard expectation for an exponential distribution. We know: E[X2]=2a2.E[X^2] = \frac{2}{a^2}.

Problem 5:

Given:

  • We need to find the mean and variance of the random variable whose pdf is represented by a triangular distribution (as per the graph).

Let's assume the graph indicates a triangular distribution on the interval [0,4][0, 4] with a peak at x=2x = 2. For this type of distribution:

  • fX(x)=12xf_X(x) = \frac{1}{2}x for 0x20 \leq x \leq 2,
  • fX(x)=12(4x)f_X(x) = \frac{1}{2}(4 - x) for 2<x42 < x \leq 4.

Solution:

  1. Mean E[X]E[X] for a triangular distribution can be calculated as: E[X]=02x22dx+24x(4x)2dx=2.E[X] = \int_{0}^{2} \frac{x^2}{2} dx + \int_{2}^{4} \frac{x(4-x)}{2} dx = 2.

  2. Variance is found by: [ \text{Var}(X) = E[X^2] - (E[X])^2 = \int_{0}^{2} \frac{x^3}{2} dx + \int_{2}^{4} \frac{x^2(4-x)}{2} dx - 4 = \frac{4}{3}. ]

Problem 6:

Given:

  • The pdf fX(x)f_X(x) is defined as: \begin{cases} x, & 0 < x < 1, \\ 2 - x, & 1 < x < 2, \\ 0, & \text{elsewhere}. \end{cases}$$
  • We need to find the moment generating function (MGF) MX(t)M_X(t).

Solution:

  1. The MGF is defined as: MX(t)=E[etX]=etxfX(x)dx=01xetxdx+12(2x)etxdx.M_X(t) = E[e^{tX}] = \int_{-\infty}^{\infty} e^{tx}f_X(x) dx = \int_{0}^{1} xe^{tx} dx + \int_{1}^{2} (2 - x)e^{tx} dx.

  2. Evaluate these integrals to find the complete MGF.


Would you like me to provide more detailed solutions to any of these problems, or any clarifications? Here are some related questions to explore:

  1. How do you determine the expected value and variance for other types of continuous distributions?
  2. What is the significance of the moment generating function (MGF) in probability theory?
  3. Can you generalize the process for finding the mean of piecewise functions like in Problem 5?
  4. How do you interpret the unit step function u(x)u(x) when defining pdfs?
  5. What are some other examples of real-world problems that use triangular distributions?

Tip: Understanding how to manipulate integrals is crucial in solving problems involving continuous random variables, especially when calculating moments or expected values.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Expected Value
Variance
Moment Generating Functions
Triangular Distribution

Formulas

Expected value: E[X] = ∫x * f_X(x) dx
Variance: Var(X) = E[X^2] - (E[X])^2
Moment Generating Function: M_X(t) = E[e^{tX}]

Theorems

Properties of continuous probability distributions
Basic integration techniques
Moment generating functions

Suitable Grade Level

College / Undergraduate Level