Problem 16 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

The probability density function \(f\) of the random variable \(X\) used in Figure \(7.2\) is given by \(f(x)=0\) outside \((0,1)\) and \(f(x)=-4 x \ln (x)\) for \(0

Step-by-Step Solution

Answer

The position of the balancing point is \(\frac{4}{9}\).

Step 1: Definition of Expectation

The expectation of a continuous random variable \(X\) is given by the integral of the product of \(x\) and its probability density function \(f(x)\). For the given function, the expectation \(E(X)\) is given by \[ E(X) = \int_{0}^{1} x f(x) \, dx = \int_{0}^{1} x (-4x \ln(x)) \, dx. \]

Step 2: Simplify the Integral

Substitute \(f(x)\) into the expectation formula: \[ E(X) = \int_{0}^{1} -4x^2 \ln(x) \, dx. \] This is our expectation integral that needs to be evaluated.

Step 3: Integration by Parts Setup

Use integration by parts to solve the integral \(\int x^2 \ln(x) \, dx\). Choose \(u = \ln(x)\) and \(dv = x^2 \, dx\), which gives \(du = \frac{1}{x} \, dx\) and \(v = \frac{x^3}{3}\).

Step 4: Apply Integration by Parts Formula

Using the integration by parts formula \(\int u \, dv = uv - \int v \, du\), substitute the chosen \(u\) and \(v\): \[ \int x^2 \ln(x) \, dx = \frac{x^3}{3} \ln(x) \Big|_0^1 - \int_0^1 \frac{x^3}{3} \cdot \frac{1}{x} \, dx. \] The boundary term \(\frac{x^3}{3} \ln(x) \Big|_0^1\) evaluates to 0 as both the terms are zero for \(x = 1\) and the behavior at \(x = 0\) cancels.

Step 5: Simplify Remaining Integral

Evaluate the remaining integral: \[ -\frac{1}{3} \int_0^1 x^2 \, dx = -\frac{1}{3} \left[ \frac{x^3}{3} \right]_0^1 = -\frac{1}{3} \left( \frac{1}{3} - 0 \right) = -\frac{1}{9}. \]

Step 6: Multiply and Find Expectation

Substitute the result back into the expectation by considering the negative factor: \[ E(X) = -4 \times \left( -\frac{1}{9} \right) = \frac{4}{9}. \]

Step 7: Conclusion

Thus, the position of the balancing point (expectation of \(X\)) is \(\frac{4}{9}\).

Key Concepts

Integration by Parts

Integration by parts is a useful technique for solving integrals involving a product of functions. It simplifies complex integrals into more manageable components. The process can be summarized with the formula:

• \(\int u \, dv = uv - \int v \, du\)

In our exercise, we aimed to solve the integral \(\int x^2 \ln(x) \, dx\) using integration by parts. Here, we chose \(u = \ln(x)\) because differentiating \(\ln(x)\) simplifies it significantly, giving us \(du = \frac{1}{x} \, dx\). Such choices are typical as they simplify the \(du\) term. Meanwhile, choosing \(dv = x^2 \, dx\) provided us with a simple polynomial which when integrated, gives \(v = \frac{x^3}{3}\). This is a strategic way to handle the components of the integral and return a solvable expression. By applying integration by parts, we transformed our integral into more straightforward terms, ultimately achieving our goal of calculating the expectation.

Probability Density Function

The probability density function (PDF) is essential in statistics, especially for continuous random variables. This function describes the likelihood of a random variable to take on a specific value. For continuous distributions, the PDF is used to find probabilities within a certain range, as individual values have a probability of zero.

• In the context of our problem, the PDF is given by \(f(x) = -4x \ln(x)\) for the interval \(0
• Outside this interval, the PDF is zero.

An important property of PDFs is that the total area under the curve (over its domain) sums to 1, ensuring it coherently represents a probability. In this exercise, understanding the PDF allowed us to integrate the function over its domain and compute the expectation, which involved systematically using the function’s behavior within its defined limits.

Balancing Point in Probability

Expectation in probability is often regarded as the 'balancing point' or the mean of a distribution. It provides a long-term average of random variable outcomes and is defined for continuous variables using integration.

• In our problem, this meant finding the value where the distribution "balances" on the number line, computed through integration of the product \(x \cdot f(x)\).
• This integration reflects the weighted averages, each \(x\) being weighted by its probability density \(f(x)\).

The expectation is calculated via \(E(X) = \int x f(x)\, dx\), representing the theoretical center of mass. Specifically for our situation, given the calculated expectation, the balancing point was \(\frac{4}{9}\). This means if you imagine the probability distribution as a physical object, it balances at \(\frac{4}{9}\) on the interval \(0,1\). The concept helps in understanding distributions not only as abstract mathematical objects but also in terms of real-world expectation and approximation.