Problem 4 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Let \(F\) be the distribution function as given in Quick exercise \(6.3: F(x)\) is 0 for \(x<1\) and 1 for \(x>3\), and \(F(x)=\frac{1}{4}(x-1)^{2}\) if \(1 \leq x \leq 3\). In the answer it is claimed that \(X=1+2 \sqrt{U}\) has distribution function \(F\), where \(U\) is a \(U(0,1)\) random variable. Verify this by computing \(\mathrm{P}(X \leq a)\) and checking that this equals \(F(a)\), for any \(a\).

Step-by-Step Solution

Answer

The transformation \( X = 1+2\sqrt{U} \) matches the distribution function \( F \).

Step 1: Calculate F(x) for different intervals

The distribution function \( F(x) \) is defined piecewise:- For \( x < 1 \), \( F(x) = 0 \).- For \( 1 \leq x \leq 3 \), \( F(x) = \frac{1}{4}(x-1)^2 \).- For \( x > 3 \), \( F(x) = 1 \). This means that \( F(x) \) progressively increases from 0 to 1 as \( x \) goes from 1 to 3.

Step 2: Setup the transformation for X

Given that \( X = 1 + 2 \sqrt{U} \), where \( U \sim U(0,1) \), we note:- The value of \( X \) is restricted between 1 and 3, because \( \sqrt{U} \) ranges from 0 to 1.- Therefore, studying \( X \) is equivalent to studying how it transforms the probability from \( U \).

Step 3: Derive the expression for P(X ≤ a)

To find \( \mathrm{P}(X \leq a) \), solve the inequality: \[ X = 1 + 2 \sqrt{U} \leq a \]Simplify to find:\[ 2\sqrt{U} \leq a-1 \]\[ \sqrt{U} \leq \frac{a-1}{2} \]\[ U \leq \left(\frac{a-1}{2}\right)^2 \]Therefore, \( \mathrm{P}(X \leq a) = \mathrm{P}(U \leq (\frac{a-1}{2})^2) = (\frac{a-1}{2})^2 \) given \( U \sim U(0,1) \).

Step 4: Verify equivalence with F(a)

Check if the result from Step 3 matches \( F(a) \):- For \( 1 \leq a \leq 3 \), \( F(a) = \frac{1}{4}(a-1)^2 \), exactly equal to \( (\frac{a-1}{2})^2 \).- For \( a < 1 \), both \( F(a) = 0 \) and \( \mathrm{P}(X \leq a) = 0 \) since \( U \) is never less than 0.- For \( a > 3 \), both \( F(a) = 1 \) and \( \mathrm{P}(X \leq a) = 1 \) since all \( U \leq 1 \). Therefore, \( X = 1 + 2\sqrt{U} \) reproduces the distribution function \( F \) exactly.

Key Concepts

Uniform Distribution

The uniform distribution is one of the simplest forms of probability distributions. In the case of a continuous uniform distribution, every outcome in a specific range is equally likely. This is denoted as \(U(a, b)\), where \(a\) and \(b\) are the lower and upper bounds, respectively. For this exercise, we focus on the uniform distribution \(U(0, 1)\).
A practical example of a uniform distribution is the value of a rolled six-sided dice, which has equal probability for any number from 1 to 6. In mathematical problems, such a distribution helps describe random processes where each outcome within the interval has an equal chance of occurring.
It serves a key role in transformation techniques and generating random variables with different distributions by applying specific functions to uniform variables.

Transformation of Variables

Transformations are used to change a simple, well-understood random variable, like a uniform distribution, into a more complex one. In the original exercise, the transformation is expressed as \(X = 1 + 2 \sqrt{U}\), and \(U\) follows a \(U(0, 1)\) distribution.
This transformation rescales and shifts \(U\) to produce a variable \(X\) that fits the specific desired distribution \(F(x)\).

• The equation \(X = 1 + 2 \sqrt{U}\) modifies \(U\) first by taking its square root, compressing values closer to 1.
• It then scales by 2, expanding the compressed values into a wider range.
• Finally, it shifts all values by 1, translating the interval from \([0, 1]\) to \([1, 3]\).

This transformation is validated by matching \(X\) to the given distribution function through the derived cumulative distribution function (CDF). Such transformations are valuable when modeling different statistical properties.

Distribution Function

A distribution function describes the probability distribution of a random variable. It indicates how the total probability of 1 is distributed over different ranges of values. In this exercise, the function \(F(x)\) is piecewise:

• For \(x < 1\), \(F(x) = 0\).
• For \(1 \leq x \leq 3\), \(F(x) = \frac{1}{4}(x-1)^2\).
• For \(x > 3\), \(F(x) = 1\).

The purpose of \(F(x)\) is to assign probabilities to intervals of values of the random variable \(X\). As \(x\) increases from 1 to 3, \(F(x)\) increases from 0 to 1, showing that \(X\) is more likely to assume values towards the upper end of its range.
Understanding \(F(x)\) enables determination of the probability for \(X\) to fall below specific thresholds, forming the heart of probabilistic forecasting and statistical inference.

Cumulative Distribution Function

A cumulative distribution function (CDF) is crucial in probability theory, as it represents the probability that a random variable \(X\) will take a value less than or equal to some \(a\). For our exercise, verifying that the transformation \(X = 1 + 2 \sqrt{U}\) matches the desired distribution \(F(a)\) involves calculating this probability:

• Start with the inequality \(X = 1 + 2 \sqrt{U} \leq a\)
• Solve this to get \(U \leq \left(\frac{a-1}{2}\right)^2\)
• Since \(U\) is uniformly distributed, \( P(U \leq \left(\frac{a-1}{2}\right)^2) = \left(\frac{a-1}{2}\right)^2\)

Thus, the CDF confirms that \(\mathrm{P}(X \leq a)\) exactly equals \(F(a)\) for the same intervals. As a continuous function, the CDF reaches 0 for values less than the minimum and 1 beyond the maximum, making it a key tool in statistical modeling and analysis.