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Chapter 4: Problem 19

If the distribution function of \(X\) is given by $$F(b)=\left\\{\begin{array}{ll} 0 & b<0 \\ \frac{1}{2} & 0 \leq b < 1 \\ \frac{3}{5} & 1 \leq b < 2 \\ \frac{4}{5} & 2 \leq b < 3 \\ \frac{9}{10} & 3 \leq b < 3.5 \\ 1 & b \geq 3.5 \end{array}\right.$$ calculate the probability mass function of \(X\)

Short Answer

Expert verified
The probability mass function of the discrete random variable X is given by: $$P(X=x)=\left\\{\begin{array}{ll} \frac{1}{2} & x=0 \\\ \frac{1}{10} & x=1 \\\ \frac{1}{5} & x=2 \\\ \frac{1}{10} & x=3 \\\ \frac{1}{10} & x=3.5 \\\ 0 & \text{otherwise} \end{array}\right.$$

Step by step solution

01

Identify the intervals and their corresponding probabilities

We are given the following distribution function F(b): $$F(b)=\left\\{\begin{array}{ll} 0 & b<0 \\\ \frac{1}{2} & 0 \leq b < 1 \\\ \frac{3}{5} & 1 \leq b < 2 \\\ \frac{4}{5} & 2 \leq b < 3 \\\ \frac{9}{10} & 3 \leq b < 3.5 \\\ 1 & b \geq 3.5 \end{array}\right.$$ From the distribution function, we have five intervals for b and their corresponding probabilities.
02

Calculate the probability mass for each value of X

We will find the probability of X taking each specific value by evaluating the differences of F(b) in the intervals. Let P(X=x) be the probability mass function of X. 1. For X=0: Since the value of the CDF jumps from 0 to 1/2 at b=0, the probability mass is: P(X=0) = F(0) - F(-) = \(\frac{1}{2} - 0 = \frac{1}{2}\) 2. For X=1: The probability mass at X=1 is found by taking the difference in the CDF at b=1 and b=0: P(X=1) = F(1) - F(0) = \(\frac{3}{5} - \frac{1}{2} = \frac{1}{10}\) 3. For X=2: The probability mass at X=2 is found by taking the difference in the CDF at b=2 and b=1: P(X=2) = F(2) - F(1) = \(\frac{4}{5} - \frac{3}{5} = \frac{1}{5}\) 4. For X=3: The probability mass at X=3 is found by taking the difference in the CDF at b=3 and b=2: P(X=3) = F(3) - F(2) = \(\frac{9}{10} - \frac{4}{5} = \frac{1}{10}\) 5. For X=3.5: The probability mass at X=3.5 is found by taking the difference in the CDF at b=3.5 and b=3: P(X=3.5) = F(3.5) - F(3) = \(1 - \frac{9}{10} = \frac{1}{10}\)
03

Write the probability mass function

Now we can write the probability mass function P(X=x) for the discrete random variable X: $$P(X=x)=\left\\{\begin{array}{ll} \frac{1}{2} & x=0 \\\ \frac{1}{10} & x=1 \\\ \frac{1}{5} & x=2 \\\ \frac{1}{10} & x=3 \\\ \frac{1}{10} & x=3.5 \\\ 0 & \text{otherwise} \end{array}\right.$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cumulative Distribution Function
The cumulative distribution function (CDF) is a cornerstone concept within probability theory, which offers much insight into the behavior of a random variable. A CDF, denoted as F(b), provides the probability that a random variable X will take a value less than or equal to a particular number b. It is a fundamental tool that helps to understand the distribution and likelihood of different outcomes.

When attempting to improve one鈥檚 grasp of CDFs, considering the context of the problem is key. For instance, in the given exercise, the CDF of random variable X is presented in intervals. When the CDF rises at a certain point, it suggests that there has been an increment in probability at that exact point, indicating the presence of a 'jump'. These jumps are very telling; they actually reveal the probability mass function of a discrete random variable.

It is also important to distinguish between discrete and continuous random variables. A discrete random variable has a countable number of possible values, and thus its CDF will have a step-like shape with jumps at each value it can take. The height of each jump precisely corresponds to the probability of the random variable taking that value.
Discrete Random Variable
Discrete random variables are those that can take on only a finite or countably infinite set of distinct values. Each of these values has a certain likelihood of occurring, quantified by probabilities. Probability mass functions (PMFs) are associated with discrete random variables and provide these probabilities explicitly for each possible value of the variable.

In our exercise, we focus on a particular type of discrete random variable that is defined by a given CDF. The PMF is derived from the CDF by computing the increase in probability at each value the random variable can take. A key tip for students is to meticulously analyze the intervals provided in the CDF. By noting the probabilities just before and just after a value that the variable can assume, one can calculate the probability mass. The PMF derived this way presents a nuanced, granular view of the distribution of our discrete random variable.
Probability Theory
Probability theory lays the foundation for quantifying uncertainty in various phenomena and processes. It allows us to make informed predictions and decisions based on the likelihood of different events. One of the primary components in this branch of mathematics is the probability mass function for discrete random variables, which we've discussed earlier.

Understanding the underlying principles of probability theory, such as the concepts of random variables, CDF, and PMF, is instrumental for students not just on paper, but in real-world applications as well. In the context of our exercise, the theory comes to life when transitioning from the CDF, which presents probability in a cumulative manner, to the PMF, which breaks down probability into individual occurrences. As we apply these theory-based steps to find the PMF of a given exercise, interpreting the results in terms of the likelihood of specific outcomes becomes an exercise in applying probability theory to dissect and understand the problem.

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Most popular questions from this chapter

A total of \(2 n\) people, consisting of \(n\) married couples, are randomly seated (all possible orderings being equally likely) at a round table. Let \(C_{i}\) denote the event that the members of couple \(i\) are seated next to each other, \(i=1, \ldots, n\) (a) Find \(P\left(C_{i}\right)\) (b) For \(j \neq i,\) find \(P\left(C_{j} | C_{i}\right)\) (c) Approximate the probability, for \(n\) large, that there are no married couples who are seated next to each other.

Suppose that a biased coin that lands on heads with probability \(p\) is flipped 10 times. Given that a total of 6 heads results, find the conditional probability that the first 3 outcomes are (a) \(h, t, t\) (meaning that the first flip results in heads, the second in tails, and the third in tails); (b) \(t, h, t\)

A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win \(\$ 1.10 ;\) if they are different colors, then you win \(-\$ 1.00 .\) (That is, you lose \(\$ 1.00 .\) ) Calculate (a) the expected value of the amount you win; (b) the variance of the amount you win.

Each of 500 soldiers in an army company independently has a certain disease with probability \(1 / 10^{3} .\) This disease will show up in a blood test, and to facilitate matters, blood samples from all 500 soldiers are pooled and tested. (a) What is the (approximate) probability that the blood test will be positive (that is, at least one person has the disease)? Suppose now that the blood test yields a positive result. (b) What is the probability, under this circumstance, that more than one person has the disease? Now, suppose one of the 500 people is Jones, who knows that he has the disease. (c) What does Jones think is the probability that more than one person has the disease? Because the pooled test was positive, the authorities have decided to test each individual separately. The first \(i-1\) of these tests were negative, and the \(i\) th one-which was on Jones-was positive. (d) Given the preceding scenario, what is the probability, as a function of \(i,\) that any of the remaining people have the disease?

\(A\) and \(B\) play the following game: \(A\) writes down either number 1 or number \(2,\) and \(B\) must guess which one. If the number that \(A\) has written down is \(i\) and \(B\) has guessed correctly, \(B\) receives \(i\) units from \(A\). If \(B\) makes a wrong guess, \(B\) pays \(\frac{3}{4}\) unit to \(A .\) If \(B\) randomizes his decision by guessing 1 with probability \(p\) and 2 with probability \(1-p,\) determine his expected gain if (a) \(A\) has written down number 1 and (b) \(A\) has written down number 2 What value of \(p\) maximizes the minimum possible value of \(B\) 's expected gain, and what is this maximin value? (Note that \(B\) 's expected gain depends not only on \(p,\) but also on what \(A\) does.) Consider now player \(A\). Suppose that she also randomizes her decision, writing down number 1 with probability q. What is \(A\) 's expected loss if (c) \(B\) chooses number 1 and (d) \(B\) chooses number \(2 ?\) What value of \(q\) minimizes \(A\) 's maximum expected loss? Show that the minimum of \(A\) 's maximum expected loss is equal to the maximum of \(B\) 's minimum expected gain. This result, known as the minimax theorem, was first established in generality by the mathematician John von Neumann and is the fundamental result in the mathematical discipline known as the theory of games. The common value is called the value of the game to player \(B\).
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