Problem 181 - An Introduction to Mathematical Statistics and Its Applications

Question

Suppose that \(X_{1}, X_{2}, X_{3}\), and \(X_{4}\) are independent random variables, each with pdf \(f_{X_{i}}\left(x_{i}\right)=4 x_{i}^{3}, 0 \leq x_{i} \leq 1\). Find (a) \(P\left(X_{1}<\frac{1}{2}\right)\). (b) \(P\left(\right.\) exactly one \(\left.X_{i}<\frac{1}{2}\right)\). (c) \(f_{X_{1}, X_{2}, X_{3}, X_{4}}\left(x_{1}, x_{2}, x_{3}, x_{4}\right)\). (d) \(F_{X_{2}, X_{3}}\left(x_{2}, x_{3}\right)\).

Step-by-Step Solution

Answer

The answers for the respective parts are: (a) 0.0625, (b) 0.316, (c) \(256{x_{1}}^{3}{x_{2}}^{3}{x_{3}}^{3}{x_{4}}^{3}\), (d) \(x_{2}^4 * x_{3}^4 \).

Step 1: Solving Part (a)

In order to find \(P\left(X_{1}<\frac{1}{2}\right)\), one needs to integrate the pdf from 0 to 1/2. The required probability is given as \[\int_{0}^{\frac{1}{2}} f_{X_{1}}(x_{1}) dx_{1}\] = \[\int_{0}^{\frac{1}{2}} 4{x_{1}}^{3} dx_{1}\] = \( \frac{1}{2^4} \) = 0.0625

Step 2: Solving Part (b)

For \(P\left(\right.\) exactly one \(\left.X_{i}<\frac{1}{2}\right)\), it involves the case where exactly one \(X_{i}\) takes a value less than 1/2 and the rest are 1/2 or greater. This can be approached as: 4 * \( \int_{0}^{\frac{1}{2}} f_{X_{i}}(x_{i}) dx_{i}\) * \( \int_{\frac{1}{2}}^{1} f_{X_{i}}(x_{i}) dx_{i}\)^3 = 4 * 0.0625 * (1-0.0625)^3 = 0.316

Step 3: Solving Part (c)

To find the joint probability density function \(f_{X_{1}, X_{2}, X_{3}, X_{4}}\left(x_{1}, x_{2}, x_{3}, x_{4}\right)\), one needs to recognize that the random variables are independent. This implies the multiplication of their marginal pdfs: \( f_{X_{1}}(x_{1}) * f_{X_{2}}(x_{2}) * f_{X_{3}}(x_{3}) * f_{X_{4}}(x_{4})\), which equals \(4{x_{1}}^{3} * 4{x_{2}}^{3} * 4{x_{3}}^{3} * 4{x_{4}}^{3} = 256{x_{1}}^{3}{x_{2}}^{3}{x_{3}}^{3}{x_{4}}^{3}\)

Step 4: Solving Part (d)

The cumulative distribution funciton (cdf) \(F_{X_{2}, X_{3}}\left(x_{2}, x_{3}\right)\) is the probability that \(X_{2}\) is less than \(x_{2}\) and \(X_{3}\) is less than \(x_{3}\). This means we need to integrate the joint pdf within the range of the values, yielding \(\int_{0}^{x_{2}} f_{X_{2}}(x_{2}) dx_{2} \int_{0}^{x_{3}} f_{X_{3}}(x_{3}) dx_{3} = \int_{0}^{x_{2}} 4x_{2}^{3} dx_{2} \int_{0}^{x_{3}} 4x_{3}^{3} dx_{3} = x_{2}^4 * x_{3}^4\)

Key Concepts

Probability Density Function (PDF)

Understanding the probability density function (PDF) is key when dealing with continuous random variables. It's a function that describes the likelihood of a random variable taking on a certain value. The PDF can be thought of as a curve where the area under the curve between two points corresponds to the probability of the random variable falling within that range.

For example, the PDF given in the exercise is defined as \(f_{X_{i}}\left(x_{i}\right)=4 x_{i}^{3}\), where \(0 \leq x_{i} \leq 1\). To find the probability that the random variable \(X_1\) is less than \(1/2\), you would calculate the area under the PDF curve from \(0\) to \(1/2\). This integration process gives us the probability in question, and for our exercise, the probability that \(X_{1}<\frac{1}{2}\) is as calculated, \(0.0625\).

Why is the PDF important?

• It helps us to calculate the probabilities for continuous random variables within specific ranges.
• Allows the evaluation of how random variables are likely to distribute across their potential values.
• Is foundational in fields such as engineering, economics, and even social sciences.

Cumulative Distribution Function (CDF)

The cumulative distribution function (CDF) provides the cumulative probability associated with a random variable and gives the probability that a random variable will have a value less than or equal to a certain value. The CDF is integral to understanding the overall distribution of probabilities for a random variable.

In our exercise, we look at the CDF for two independent variables, \(X_2\) and \(X_3\). To find this, we integrate their individual PDFs up to \(x_2\) and \(x_3\), respectively. The result, as seen in the final step of our solution, is a product of their individual integrated PDFs, given by \(x_{2}^4 * x_{3}^4\).

Here's why understanding CDFs is important:

• CDPs allow us to find probabilities for intervals and marginal probabilities easily.
• They provide a complete description of the random variable's distribution.
• If known, the CDF can be used to derive other statistical measures like median, percentile, etc.

Independent Random Variables

The concept of independent random variables is crucial in the field of probability and statistics. Two or more random variables are independent if knowing the occurence of one provides no information about the occurrence of the others. In other words, the outcome of one random variable has no effect on the outcome of another.

In our exercise, the random variables \(X_{1}, X_{2}, X_{3},\) and \(X_{4}\) are considered independent. This independence is a pivotal piece of information since it allows us to calculate the joint probability density function by simply multiplying their individual PDFs together, resulting in \(256{x_{1}}^{3}{x_{2}}^{3}{x_{3}}^{3}{x_{4}}^{3}\).

Importance of Independent Variables

• Many statistical methods and tests assume independence between variables. Violating this assumption can lead to incorrect conclusions.
• Independence simplifies calculations and interpretations of random phenomena.
• Understanding the independence of variables is essential in modeling and simulations.