Chapter 1: Problem 74
Let \(X\) have the p.d.f. \(f(x)=(x+2) / 18,-2
Short Answer
Expert verified
The answers are the results of the integral calculations above for \(E(X)\), \(E\left[(X+2)^{3}\right]\), and \(E\left[6X-2(X+2)^{3}\right]\) respectively.
Step by step solution
01
Calculate \(E(X)\)
Calculate the expectation \(E(X)\) for a continuous random variable defined by the probability density function (pdf) \(f(x)\). The formula to calculate \(E(X)\) is: \(E(X) = \int_{-\infty}^{\infty}x \cdot f(x) dx\). Here, \(-2 < x < 4\) so the integral becomes \(E(X) = \int_{-2}^{4}x \cdot f(x) dx\). Plugging in \(f(x)\), we get \(E(X)= \int_{-2}^{4}(x) \cdot (x+2) / 18 dx \). Calculate this integral and simplify the solution.
02
Calculate \(E\left[(X+2)^{3}\right]\)
Similarly, we know that the expectation of a function \(g(X)\) of a random variable \(X\) is \(E[g(X)] = \int_{-\infty}^{\infty}g(x) \cdot f(x) dx\). Here, \(g(X) = (X + 2)^3\). So, \(E\left[(X+2)^{3}\right] = \int_{-2}^{4} (x+2)^3 \cdot f(x) dx = \int_{-2}^{4} (x+2)^3 \cdot (x+2)/18 dx\). Solve this integral and simplify the solution.
03
Calculate \(E\left[6X-2(X+2)^{3}\right]\)
Here, \(g(X) = 6X - 2(X+2)^3\). So, \(E\left[6X-2(X+2)^{3}\right]\) can be calculated as \(E\left[6X-2(X+2)^{3}\right] = \int_{-2}^{4} \left[6x - 2(x+2)^3 \right] \cdot f(x) dx = \int_{-2}^{4} \left[ 6x - 2(x+2)^3 \right] \cdot (x+2) /18 dx\). Compute this integral and simplify the result.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Probability Density Function
In probability theory, a Probability Density Function, or pdf, describes the likelihood of a continuous random variable to take on a particular value. Unlike discrete variables, which have exact probabilities, continuous variables are described over a range of values, requiring the use of integrals.
In our example, the pdf is given as \( f(x) = \frac{x+2}{18} \) for \(-2 < x < 4\), and zero elsewhere. This means the function's support—where it can actually assign probabilities—is between \(-2\) and \(4\). Outside this range, the probability is zero.
Key properties of a probability density function include:
In our example, the pdf is given as \( f(x) = \frac{x+2}{18} \) for \(-2 < x < 4\), and zero elsewhere. This means the function's support—where it can actually assign probabilities—is between \(-2\) and \(4\). Outside this range, the probability is zero.
Key properties of a probability density function include:
- The area under the pdf curve over its support equals 1, representing total certainty.
- The function must be non-negative for all values. Negative probability does not make sense.
Continuous Random Variables
Continuous Random Variables are variables that can take on any real value within a specified range. They differ from discrete random variables, which can only take on certain distinct values.
Consider the variable \(X\) with pdf \( f(x) = \frac{x+2}{18} \). Here, \(X\) can take any value between \(-2\) and \(4\). Notably, because \(X\) is continuous, we use integrals instead of sums to compute probabilities or expectations.
Some important aspects of continuous random variables include:
Consider the variable \(X\) with pdf \( f(x) = \frac{x+2}{18} \). Here, \(X\) can take any value between \(-2\) and \(4\). Notably, because \(X\) is continuous, we use integrals instead of sums to compute probabilities or expectations.
Some important aspects of continuous random variables include:
- The probability of \(X\) being exactly a specific value is zero because there are infinite possibilities.
- We often talk about the probability of \(X\) falling within an interval, which is computed using the area under the pdf over that interval.
- Instead of probabilities at points, the integral gives us the probabilities over a continuous span.
Expectation of a Function
In probability, the Expectation of a Function of a random variable is a critical concept that extends the idea of an average to functions of random variables. If \(g(X)\) is some function of the random variable \(X\), the expectation \(E[g(X)]\) is calculated by integrating \(g(x)\) multiplied by the pdf over its support: \(E[g(X)] = \int_{a}^{b} g(x) \cdot f(x) \, dx\).
This is a generalization beyond just finding the expected value of \(X\) itself. For our example, we compute:
This is a generalization beyond just finding the expected value of \(X\) itself. For our example, we compute:
- \(E(X)\), the mean of \(X\), by \(\int_{-2}^{4} x \cdot \frac{x+2}{18} \, dx\).
- \(E\left[(X+2)^3\right]\), which involves \(g(x) = (x+2)^3\) and accounts for how \((X+2)^3\) behaves with \(X\) distributed by the given pdf.
- \(E\left[6X-2(X+2)^{3}\right]\), combining linear and higher-order terms, showcases flexibility in expectation computations.