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Chapter 26: Problem 19

A continuous random variable \(X\) has a probability density function \(f(x)\); the corresponding cumulative probability function is \(F(x) .\) Show that the random variable \(Y=F(X)\) is uniformly distributed between 0 and 1 .

Short Answer

Expert verified
The random variable Y = F(X) has its cdf F_Y(t)= t, which implies it is uniformly distributed on [0,1].

Step by step solution

01

Understand the Given Functions

Notice that we have two functions: the probability density function (pdf) of the random variable X, denoted as f(x), and the cumulative distribution function (cdf) of X, denoted as F(x). The cdf is the integral of the pdf.
02

Define the Random Variable Y

The exercise requires showing that the new random variable Y = F(X) is uniformly distributed between 0 and 1. Recall that a random variable Y is uniformly distributed on the interval [0,1] if its cdf is F_Y(y) = y for y in [0,1].
03

Find the Cumulative Distribution Function of Y

To find the cdf of Y, calculate the probability that Y is less than or equal to some value t in [0,1]. This can be written as P(Y 鈮� t) = P(F(X) 鈮� t).
04

Use the Inverse Relationship of the CDF

Since F(X) is a non-decreasing function, P(F(X) 鈮� t) equals the probability that X is less than or equal to F^{-1}(t). Thus, P(Y 鈮� t) = P(F(X) 鈮� t) = P(X 鈮� F^{-1}(t)).
05

Apply the Definition of Cumulative Probability Function

The cumulative distribution function F(x) is defined such that F(x) = P(X 鈮� x). Therefore, P(X 鈮� F^{-1}(t)) = t. This implies that F_Y(t) = t for t in [0,1].
06

Conclude the Proof

Since F_Y(t) = t for t in [0, 1], by definition, Y is uniformly distributed on [0,1].

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Key Concepts

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

Probability Density Function
A Probability Density Function (pdf) represents the likelihood of a continuous random variable taking on a specific value within its range. Unlike discrete variables, the probability at any single point for continuous variables is zero. Instead, the pdf helps to find the probability that the variable falls within a certain interval.

Key aspects of pdf:
  • The pdf is a non-negative function.
  • The area under the curve is equal to one, representing the total probability.
  • To find the probability that the variable falls within an interval \[a, b\], you integrate the pdf over that interval: \[ P(a \leq X \leq b) = \int_{a}^{b} f(x) dx \].
Cumulative Distribution Function
The Cumulative Distribution Function (cdf) of a continuous random variable X, denoted as F(x), describes the probability that X will take a value less than or equal to x. Essentially, the cdf accumulates the probabilities up to x.

Properties of the cdf:
  • It is a non-decreasing function.
  • The value ranges between 0 and 1.
  • It is calculated by integrating the pdf from negative infinity to x: \[ F(x) = \int_{-\infty}^{x} f(t) dt \].
  • If x is extremely small, F(x) approaches 0, and if x is extremely large, F(x) approaches 1.

To use a cdf to find the probability that X falls within the interval \[a, b\], subtract the cdf values: \[ P(a \leq X \leq b) = F(b) - F(a) \].
Uniformly Distributed Random Variable
A uniformly distributed random variable has all intervals of the same length on the distribution's support with equal probability. For a variable Y uniformly distributed over \[0, 1\], every outcome between 0 and 1 is equally likely.

Important characteristics:
  • The pdf of Y is uniform and denoted as \[ f_Y(y) = 1 \text{ for } 0 \leq y \leq 1 \].
  • The corresponding cdf is \[ F_Y(y) = y \text{ for } 0 \leq y \leq 1 \].

In our original exercise, we transform X into Y using Y = F(X). By showing that the cdf of Y is \[ t \text{ for } 0 \leq t \leq 1 \], it is evident that Y is uniformly distributed between 0 and 1. Thus, Y follows the properties of a uniformly distributed random variable.

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Most popular questions from this chapter

As assistant to a celebrated and imperious newspaper proprietor, you are given the job of running a lottery in which each of his five million readers will have an equal independent chance \(p\) of winning a million pounds; you have the job of choosing \(p .\) However, if nobody wins it will be bad for publicity whilst if more than two readers do so, the prize cost will more than offset the profit from extra circulation - in either case you will be sacked! Show that, however you choose \(p\), there is more than a \(40 \%\) chance you will soon be clearing your desk.

Villages \(A, B, C\) and \(D\) are connected by overhead telephone lines joining \(A B\), \(A C, B C, B D\) and \(C D .\) As a result of severe gales, there is a probability \(p\) (the same for each link) that any particular link is broken. (a) Show that the probability that a call can be made from \(A\) to \(B\) is $$ 1-2 p^{2}+p^{3} $$ (b) Show that the probability that a call can be made from \(D\) to \(A\) is $$ 1-2 p^{2}-2 p^{3}+5 p^{4}-2 p^{5} $$

A shopper buys 36 items at random in a supermarket where, because of the sales tax imposed, the final digit (the number of pence) in the price is uniformly and randomly distributed from 0 to \(9 .\) Instead of adding up the bill exactly she rounds each item to the nearest 10 pence, rounding up or down with equal probability if the price ends in a ' 5 '. Should she suspect a mistake if the cashier asks her for 23 pence more than she estimated?

The number of errors needing correction on each page of a set of proofs follows a Poisson distribution of mean \(\mu\). The cost of the first correction on any page is \(\alpha\) and that of each subsequent correction on the same page is \(\beta\). Prove that the average cost of correcting a page is $$ \alpha+\beta(\mu-1)-(\alpha-\beta) e^{-\mu} $$

\(26.3 A\) and \(B\) each have two unbiased four-faced dice, the four faces being numbered \(1,2,3,4\). Without looking, \(B\) tries to guess the sum \(x\) of the numbers on the bottom faces of \(A\) 's two dice after they have been thrown onto a table. If the guess is correct \(B\) receives \(x^{2}\) euros, but if not he loses \(x\) euros. Determine \(B\) 's expected gain per throw of \(A\) 's dice when he adopts each of the following strategies: (a) he selects \(x\) at random in the range \(2 \leq x \leq 8\); (b) he throws his own two dice and guesses \(x\) to be whatever they indicate; (c) he takes your advice and always chooses the same value for \(x\). Which number would you advise? 26.4 Use the method of induction to prove equation (26.16), the probability addition law for the union of \(n\) general events.
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