Problem 144 - An Introduction to Mathematical Statistics and Its Applications

Question

Suppose that five independent observations are drawn from the continuous pdf \(f_{T}(t)=2 t, 0 \leq t \leq 1\). Let \(X\) denote the number of \(t\) 's that fall in the interval \(0 \leq t<\frac{1}{3}\) and let \(Y\) denote the number of \(t\) 's that fall in the interval \(\frac{1}{3} \leq t<\frac{2}{3}\). Find \(p_{X, Y}(1,2)\).

Step-by-Step Solution

Answer

The value of \(p_{X, Y}(1,2)\) is \(\frac{175}{6561}\)

Step 1: Calculate PDF for given intervals

The probability density function \(f_{T}(t)=2 t\) is given for the interval \(0 \leq t \leq 1\). To calculate the probability \(p_{X, Y}(1,2)\), we need to calculate probability for the specified intervals: \(0 \leq t<\frac{1}{3}\) and \(\frac{1}{3} \leq t<\frac{2}{3}\). By integrating the PDF over these intervals, we get \(\int_{0}^{\frac{1}{3}} 2t dt = \frac{1}{9}\) and \(\int_{\frac{1}{3}}^{\frac{2}{3}} 2t dt = \frac{7}{27}\) respectively.

Step 2: Apply definition of X and Y

Now, based on the definition, \(X\) denotes the number of \(t\) 's that fall in the interval \(0 \leq t<\frac{1}{3}\), and \(Y\) denotes the number of \(t\) 's that fall in the interval \(\frac{1}{3} \leq t<\frac{2}{3}\). To get \(p_{X, Y}(1,2)\), we want 1 t to fall in the first interval and 2 in the second.

Step 3: Calculate joint probability

We now use the calculated probabilities for both the intervals and the number of event occurences to calculate the joint probability. This can be done using the formula for joint probability of two independent events. You also need to remember here that since we have 5 independent observations, the remaining 2 would fall in the interval \(\frac{2}{3} \leq t \leq 1\) by default. So, \(p_{X, Y}(1,2) = ^5C_1*^4C_2 * (\frac{1}{9})^1 * (\frac{7}{27})^2 * (1-\frac{1}{9}-\frac{7}{27})^2 = \frac{175}{6561}\)

Key Concepts

Probability Density Function

A probability density function (PDF) is a function that describes the relative likelihood for a continuous random variable to take on a particular value. Unlike discrete probabilities, which are straightforward, PDFs require an understanding of probability regarding intervals. Essentially, the PDF provides the "density" of probability at any point, meaning how probable it is for the variable to be near that point. For any continuous variable:

• The total area under the curve of the PDF over all possible values of the variable is 1.
• The probability of the variable taking an exact value is actually 0, it's the area under the curve for a range that's meaningful.

In our exercise, the PDF is given as \( f_{T}(t) = 2t \) for the interval \( 0 \leq t \leq 1 \). This implies the probability changes across the interval, being zero at \( t = 0 \) and peaking at \( t = 1 \). To find probabilities over specific ranges, we must integrate this function over those intervals.

Independent Observations

Independent observations in probability mean that the outcome of one observation has no effect on another. In simple terms, knowing the result of one observation provides no information about others. This concept is foundational because it allows us to model and compute probabilities of combined outcomes easily. In scenarios involving independent observations:

• The joint probability of two independent events A and B occurring is the product of their marginal probabilities: \( P(A \cap B) = P(A) \times P(B) \).
• Each observation or trial is considered separate and does not affect the results of any other observation.

In our exercise, understanding independence is crucial as we draw five observations from a given PDF. By assuming independence, we model each draw separately but solve jointly for probabilities like \( p_{X,Y}(1,2) \), which describes specific counts falling within defined intervals.

Integrals in Probability

Integrals are significant tools used in probability, especially when working with continuous random variables. Calculating the probability that a variable falls within a certain range involves finding the area under the PDF curve over that range through integration.Here's why integrals are useful in probability:

• Integrals calculate the total probability over an interval, providing the cumulative distribution function (CDF) from the PDF.
• They allow us to handle variable changes and calculate probabilities over continuous ranges rather than just specific points.

For example, in the given exercise, we had to find the probability of \( t \) values falling within the intervals \( 0 \leq t < \frac{1}{3} \) and \( \frac{1}{3} \leq t < \frac{2}{3} \) by integrating the PDF over these intervals:\[\int_{0}^{\frac{1}{3}} 2t \, dt = \frac{1}{9} \, \text{and} \, \int_{\frac{1}{3}}^{\frac{2}{3}} 2t \, dt = \frac{7}{27}\]These integrations provide the necessary probabilities for further calculations, like finding the joint probability of specific outcomes.