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Chapter 1: Problem 62

Let \(F(x)\) be the distribution function of the random variable \(X .\) If \(m\) is a number such that \(F(m)=\frac{1}{2}\), show that \(m\) is a median of the distribution.

Short Answer

Expert verified
For \(m\) to be the median of a distribution: first, at least half of its probability mass should be less than or equal to \(m\). Given is \(F(m) = \frac{1}{2}\), which satisfies this condition. Second, at least half of its probability mass should be greater than or equal to \(m\). With total probability 1, we find \(P(X>m)= \frac{1}{2}\), satisfying the second condition. Therefore, \(m\) is indeed the median of the distribution.

Step by step solution

01

Definition of Median

The definition of the median \(m\) of a distribution is a number such that at least half of the probability mass of the distribution is less than or equal to \(m\) and at least half of the probability mass is greater than or equal to \(m\). Hence, for \(m\) to be a median, \(P(X \le m) \ge \frac{1}{2}\) and \(P(X \ge m) \ge \frac{1}{2}\)
02

Applying Distribution Function

The cumulative distribution function \(F(x)\) gives the probability that the random variable \(X\) will take a value less than or equal to \(x\). Given \(F(m) = \frac{1}{2}\), this implies that the probability \(P(X \le m) = F(m) = \frac{1}{2}\)
03

Proving the other half

We know that the the total probability must be 1. So, if \(P(X \le m) = \frac{1}{2}\), then \(P(X > m) = 1 - P(X \le m) = 1 - \frac{1}{2} = \frac{1}{2}\), which means that probability of \(X\) being greater than \(m\) is also at least half. Therefore, both conditions for being a median are satisfied.

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Key Concepts

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

Distribution Function
In probability and statistics, a distribution function is a crucial concept that describes the probability distribution of a random variable. Specifically, it tells us how probabilities are allocated across different possible values of a random variable. For a random variable like \(X\), the distribution function is often represented as \(F(x)\) or \(P(X \leq x)\). This function gives us the probability that the random variable will take on a value less than or equal to \(x\).

It's important to note:
  • The distribution function is always non-decreasing; this means it never goes down as \(x\) increases.
  • It ranges from 0 to 1. At the very start, when \(x\) is very small, \(F(x) = 0\). As \(x\) grows, \(F(x)\) climbs towards 1.
  • For continuous random variables, the distribution function is continuous, while for discrete random variables, it can jump at certain points.
Understanding the distribution function is essential for grasping other related concepts like the cumulative distribution function and the median.
Random Variable
A random variable is a mathematical device used in probability theory to model random phenomena. Although it sounds like a single random number, it actually represents a set of possible outcomes from a random process and their associated probabilities. Think of it as a link between a random experiment and its possible numerical results.

There are two main types of random variables:
  • Discrete: These take on a countable number of distinct values, like the roll of a die. Each outcome has a specific probability mass.
  • Continuous: These can take any value within a given interval, like the weight of an object. The probability is not associated with specific outcomes but rather intervals within the distribution function.
Random variables form the basis for probability distributions, helping to describe how likely different outcomes are from a random process.
Probability Mass
Probability mass is linked to the probabilities associated with the discrete outcomes of a random variable. For discrete random variables, probability mass tells you how likely each potential outcome is. As each outcome has a probability, the sum of all probabilities for all possible outcomes gives you 1, representing the completeness and certainty of all possible events.

Key points about probability mass:
  • It's only applicable to discrete random variables, not continuous ones, where we deal with probability density functions instead.
  • Each outcome or "state" that the random variable can assume has a specific probability mass value ranging from 0 to 1.
  • In the context of medians and probability, at least half of the probability mass needs to be on either side of a median value.
Proper understanding of probability mass gives insights into the distribution of outcomes for discrete random variables, essentially telling us how the probability "mass" spreads across different possible values.
Cumulative Distribution Function
The cumulative distribution function (CDF) is an extension of the distribution function and an integral concept in probability theory. It's typically represented as \(F(x)\) and provides an aggregate way to track probability.

The CDF gives the probability that a random variable \(X\) is less than or equal to some value \(x\), essentially summing up probabilities up to that point. It builds off the individual probability values into a cumulative measure.
  • The CDF starts at 0 and "accumulates" values up to 1 as \(x\) increases.
  • For a given value \(m\), the statement \(F(m) = \frac{1}{2}\) signifies \(m\) as a median if it has half the probability mass on either side.
  • For a continuous random variable, it is the integral of its probability density function. For discrete, its the sum across values less than or equal to \(x\).
Mastering the cumulative distribution function helps in understanding how probabilities "build up" over possible outcomes in a distribution.

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Most popular questions from this chapter

A mode of a distribution of one random variable \(X\) of the continuous or discrete type is a value of \(x\) that maximizes the p.d.f. \(f(x)\). If there is only one such \(x\), it is called the mode of the distribution. Find the mode of each of the following distributions: (a) \(f(x)=\left(\frac{1}{2}\right)^{x}, x=1,2,3, \ldots\), zero elsewhere. (b) \(f(x)=12 x^{2}(1-x), 0

Let \(\mathscr{A}\) denote the set of points that are interior to or on the boundary of a square with opposite vertices at the point \((0,0)\) and at the point \((1,1) . \operatorname{Let} Q(A)=\int_{A} \int d y d x .(a)\) If \(A \subset \mathscr{A}\) is the set \(\\{(x, y) ; 0

A bowl contains 10 chips, of which 8 are marked \(\$ 2\) each and 2 are marked \(\$ 5\) each. Let a person choose, at random and without replacement, 3 chips from this bowl. If the person is to receive the sum of the resulting amounts, find his expectation.

Let the probability set function of the random variable \(X\) be $$P(A)=\int_{A} e^{-x} d x, \quad \text { where } \mathscr{A}=\\{x ; 0

Let \(Q(A)=\int_{A} \int\left(x^{2}+y^{2}\right) d x d y\) for every two- dimensional set \(A\) for which the integral exists; otherwise, let \(Q(A)\) be undefined. If \(A_{1}=\) \(\\{(x, y) ;-1 \leq x \leq 1,-1 \leq y \leq 1\\}, A_{2}=\\{(x, y) ;-1 \leq x=y \leq 1\\}\), and \(A_{3}=\left\\{(x, y) ; x^{2}+y^{2} \leq 1\right\\}\), find \(Q\left(A_{1}\right), Q\left(A_{2}\right)\), and \(Q\left(A_{3}\right) .\) Hint. In evaluating \(Q\left(A_{2}\right)\), recall the definition of the double integral (or consider the volume under the surface \(z=x^{2}+y^{2}\) above the line segment \(-1 \leq x=y \leq 1\) in the \(x y\) -plane). Use polar coordinates in the calculation of \(Q\left(A_{3}\right)\).
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