Problem 12
Question
Let \(X\) be a random variable with distribution function $$F(x)=\left\\{\begin{array}{ll} 0 & x<0 \\ 0.05 & 0 \leq x<1.3 \\ 0.30 & 1.3 \leq x<1.7 \\ 0.85 & 1.7 \leq x<1.9 \\ 0.90 & 1.9 \leq x<2 \\ 1.0 & x \geq 2 \end{array}\right.$$ Determine the probability mass function of \(X\).
Step-by-Step Solution
Verified Answer
The PMF of \(X\) is: 0.05 at \(x=0\), 0.25 at \(x=1.3\), 0.55 at \(x=1.7\), 0.05 at \(x=1.9\), and 0.10 at \(x=2\).
1Step 1: Identify the Discontinuities
Examine the distribution function and identify where it changes values. These points indicate the possible values that the random variable can take. Here, the changes occur at \(x = 0\), \(x = 1.3\), \(x = 1.7\), \(x = 1.9\), and \(x = 2\).
2Step 2: Calculate the Probability Mass for Each Value
To find the probability mass function (PMF), compute the difference in the distribution function at each identified point:- For \(x=0\), \(P(X=0) = F(0) - F(- ext{inf}) = 0.05 - 0 = 0.05\).- For \(x=1.3\), \(P(X=1.3) = F(1.3) - F(0) = 0.30 - 0.05 = 0.25\).- For \(x=1.7\), \(P(X=1.7) = F(1.7) - F(1.3) = 0.85 - 0.30 = 0.55\).- For \(x=1.9\), \(P(X=1.9) = F(1.9) - F(1.7) = 0.90 - 0.85 = 0.05\).- For \(x=2\), \(P(X=2) = F(2) - F(1.9) = 1.0 - 0.90 = 0.10\).
3Step 3: Write the PMF
Combine the calculated probabilities to write the probability mass function of the random variable \(X\):\[P(X=x) = \left\{\begin{array}{ll}0.05 & \text{if } x=0, \0.25 & \text{if } x=1.3, \0.55 & \text{if } x=1.7, \0.05 & \text{if } x=1.9, \0.10 & \text{if } x=2\end{array}\right.\]
Key Concepts
Understanding Random VariablesDistribution Function ExplainedPMF Calculation StepsUnderstanding Probability Concepts
Understanding Random Variables
In probability theory, a random variable is a variable that can take different values based on the outcome of a random event. This concept is fundamental in statistics, as it helps to quantify the results of different random processes.
The values that a random variable can take are determined by a probability distribution. This is essentially a mathematical function that provides the probabilities of occurrence of different possible outcomes. In our example, the random variable is denoted by \(X\), with specific values identified through a distribution function.
Remember:
The values that a random variable can take are determined by a probability distribution. This is essentially a mathematical function that provides the probabilities of occurrence of different possible outcomes. In our example, the random variable is denoted by \(X\), with specific values identified through a distribution function.
Remember:
- Random variables can be discrete or continuous. In this context, \(X\) is discrete, as it takes specific values.
- These values represent specific outcomes that are possible within a given scenario, governed by a distribution function.
Distribution Function Explained
A distribution function, or cumulative distribution function (CDF), is a function that describes the probability that a random variable \(X\) will take a value less than or equal to \(x\). It's an integral part of understanding how probabilities are distributed across possible outcomes.
In the exercise, we are given the CDF \(F(x)\) for different ranges of \(x\). The steps help us understand the behavior of the function at discontinuities, which are points where the probability changes.
Key points:
In the exercise, we are given the CDF \(F(x)\) for different ranges of \(x\). The steps help us understand the behavior of the function at discontinuities, which are points where the probability changes.
Key points:
- The CDF is non-decreasing and generally ranges from 0 to 1.
- Each value in the CDF indicates the probability of the random variable being less than or equal to that value.
- The difference in values at these points helps determine the probability mass function.
PMF Calculation Steps
To derive the Probability Mass Function (PMF) from a distribution function, you need to assess where the function changes. These points of change, known as discontinuities, signify the unique values that the random variable can take.
Here's how to determine the PMF:
Here's how to determine the PMF:
- Start by identifying all discontinuities in the distribution function. These indicate potential values for \(X\).
- Calculate the probability mass for each value by finding the difference between successive values of the distribution function.
- Combine these probabilities to express the PMF of \(X\).
Understanding Probability Concepts
Probability is the bedrock of statistical analysis and involves calculating the likelihood of different outcomes. In the context of random variables, probability helps to predict the behavior of different scenarios.
Crucial facets of probability in this context include:
Crucial facets of probability in this context include:
- The probability mass function (PMF) details the likelihood of a random variable taking specific discrete values.
- The cumulative nature of a distribution function (or CDF), which builds up probabilities over a range of values, helps us understand the lead-up to certain outcomes.
- The sum of probabilities for all possible values of \(X\) equals 1, ensuring a complete distribution of outcomes.
Other exercises in this chapter
Problem 11
Let \(\left(X_{1}, X_{2}, \ldots, X_{n}\right)\) denote a sample of size \(n\). Show that $$ \sum_{k=1}^{n}\left(X_{k}-\bar{X}\right)=0 $$ where \(\bar{X}\) is
View solution Problem 11
Suppose \(X_{1}, X_{2}, \ldots, X_{n}\) are independent random variables with density function $$ f(x)=\frac{1}{\pi\left(1+x^{2}\right)}, \quad x \in \mathbf{R}
View solution Problem 12
Suppose that \(f(x)\) is the density function of a normal distribution with mean \(\mu\) and standard deviation \(\sigma\). Show that $$ \mu=\int_{-\infty}^{\in
View solution Problem 12
In Problems \(9-12\), assume that $$ \Omega=\\{1,2,3,4,5\\} $$ \(P(1)=0.1, P(2)=0.2\), and \(P(3)=P(4)=0.05 .\) Furthermore, assume that \(A=\\{1,3,5\\}\) and \
View solution