Problem 17 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Let \(U_{1}\) and \(U_{2}\) be two independent random variables, both uniformly distributed over \([0, a] .\) Let \(V=\min \left\\{U_{1}, U_{2}\right\\}\) and \(Z=\max \left\\{U_{1}, U_{2}\right\\} .\) Show that the joint distribution function of \(V\) and \(Z\) is given by $$ F(s, t)=\mathrm{P}(V \leq s, Z \leq t)=\frac{t^{2}-(t-s)^{2}}{a^{2}} \quad \text { for } 0 \leq s \leq t \leq a . $$

Step-by-Step Solution

Answer

The joint distribution function is \(F(s, t) = \frac{t^2 - (t-s)^2}{a^2}.\)

Step 1: Identify the Condition

Since \(U_1\) and \(U_2\) are uniformly distributed over \([0, a]\), both can take any value in this interval with equal probability. We need to find \(\mathrm{P}(V \leq s, Z \leq t)\). Here, \(V = \min\{U_1, U_2\}\) and \(Z = \max\{U_1, U_2\}\). This probability means both \(U_1\) and \(U_2\) must be in the interval \([0, t]\), and at least one of them must be less than or equal to \(s\).

Step 2: Establish Joint Probability Requirements

For \(V \leq s\) and \(Z \leq t\), consider the scenarios: both \(U_1\) and \(U_2\) end up less than or equal to \(t\). In this region, the area of their possibilities is determined by the condition that at least one is less than or equal to \(s\). This is equivalent to the union of two events: one where \(U_1 \leq s\) and \(U_2 \leq t\), and the other where \(U_2 \leq s\) and \(U_1 \leq t\).

Step 3: Calculate Relevant Areas

The probability \(\mathrm{P}(U_1 \leq t, U_2 \leq t)\) can be calculated using the independence of \(U_1\) and \(U_2\) as \( (t/a)^2 \) because it represents the area of a square in a \(t \times t\) square. The unwanted area where both are greater than \(s\) can be given as \((t-s)^2/a^2\), the square formation from \((s,s)\) to \((t,t)\).

Step 4: Derive the Desired Probability Expression

By subtracting the area of the probabilities where both variables are greater than \(s\) from the area of where both are less than or equal to \(t\), we find the expression \[(t/a)^2 - ((t-s)/a)^2 = \frac{t^2 - (t-s)^2}{a^2}.\] Hence, the probability \(\mathrm{P}(V \leq s, Z \leq t)\) is \(\frac{t^2 - (t-s)^2}{a^2}\).

Key Concepts

Uniform Distribution

A uniform distribution is a fundamental concept in probability and statistics. It involves a probability distribution where all outcomes are equally likely to occur. In the context of the exercise, both random variables, \(U_1\) and \(U_2\), follow a uniform distribution over the interval \([0, a]\). This means any value within this range is just as probable as any other.

The formula for the probability density function (PDF) of a uniform distribution over \([0, a]\) is given by:

• \( f(x) = \frac{1}{a} \) for \(0 \leq x \leq a\)

This highlights the constant probability across the interval. In simpler terms, because \(U_1\) and \(U_2\) are uniformly distributed, finding probabilities is often about determining the area under a flat 'curve', reflecting equal likelihood.

Understanding this helps when expanding to joint distributions, where combinations of variables come into play, like finding probability expressions for \(V = \min\{U_1, U_2\}\) and \(Z = \max\{U_1, U_2\}\).

Independent Random Variables

Random variables are considered independent when the outcome of one variable does not affect the outcome of the other. This is crucial when you are dealing with joint distributions, as it simplifies probability calculations significantly.

For \(U_1\) and \(U_2\), being independent implies that:

• The probability of both events happening is the product of their probabilities: \(\mathrm{P}(U_1 \leq t) \times \mathrm{P}(U_2 \leq t)\).
• The joint probability of \(U_1\) and \(U_2\) lying within a certain region, like \([0, t]\), simplifies to their individual probabilities multiplied together.

In the given exercise, this characteristic allows us to calculate \(\mathrm{P}(U_1 \leq t, U_2 \leq t)\) as \(\left(\frac{t}{a}\right)^2\). This is because each variable independently can take values between \(0\) and \(t\), with genuine uniformity governed by the interval \([0, a]\).

Independence is a powerful simplifier. It helps by turning what could be a more complex joint probability problem into a sequence of simpler calculations that can be solved using multiplication.

Probability Calculation

Calculating probabilities for various events involves using the foundational concepts we've discussed. Specifically, with the exercise at hand, we need to understand how to compute the probability of combined events, particularly using the minimum and maximum of two random variables.

To find the probability \(\mathrm{P}(V \leq s, Z \leq t)\), we employ the understanding of joint probability spaces and subtract out the undesired area. For our particular function, this involves:

• First calculating the probability that both \(U_1\) and \(U_2\) are below or equal to \(t\), which was reasoned as \(\frac{t^2}{a^2}\).
• Then, understanding the portion where both exceed \(s\) is \(\frac{(t-s)^2}{a^2}\).
• Finally, subtracting the unwanted probability from the total, yielding the formula \[ \frac{t^2 - (t-s)^2}{a^2} \].

This subtraction reflects removing areas of the probability space where neither \(U_1\) nor \(U_2\) meet the condition \(V \leq s\). This mathematical maneuver echoes logical reasoning about the likelihood of events and reaffirms the power of independence and uniformity in calculating joint probabilities.