/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 A continuous random variable \(X... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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
Y = F(X) is uniformly distributed over [0, 1] because P(Y ≤ y) = y.

Step by step solution

01

Understand the Problem

We need to show that if we apply the cumulative distribution function (CDF) of a continuous random variable to itself, the result follows a uniform distribution between 0 and 1.
02

Define the Functions

Let the continuous random variable be denoted as \(X\) with probability density function (PDF) \(f(x)\) and cumulative distribution function (CDF) \(F(x)\). The new variable is defined as \(Y = F(X)\).
03

Express the Cumulative Distribution Function of Y

To show \(Y\) is uniformly distributed, we need to find its CDF. Define the CDF of \(Y\) as \(G(y)\). We need to compute \(G(y) = P(Y \leq y)\).
04

Compute \(P(Y \leq y)\)

By the definition of \(Y\), this is equivalent to computing \(P(F(X) \leq y)\). Since \(F(X)\) is the CDF of \(X\), it is a non-decreasing function that maps to \([0, 1]\).
05

Invert the CDF

We need \(P(F(X) \leq y)\). Since \(F\) is a CDF, it is invertible where \(F^{-1}(y)\) exists. Hence, \(P(F(X) \leq y) = P(X \leq F^{-1}(y))\).
06

Substitute the CDF of X

Utilizing the definition of the CDF, \(P(X \leq F^{-1}(y)) = F(F^{-1}(y))\). Since the CDF evaluated at its inverse yields \(y\), we get \(F(F^{-1}(y)) = y\).
07

Conclusion

We have shown that \(P(Y \leq y) = y\). Thus, the CDF of \(Y\) is \(G(y) = y\), demonstrating that \(Y\) is uniformly distributed over \([0, 1]\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Cumulative Distribution Function
A cumulative distribution function (CDF) is a fundamental concept in probability theory.
It describes the probability that a random variable takes on a value less than or equal to a given number.
For a continuous random variable \(X\) with a probability density function \(f(x)\), the CDF \(F(x)\) is defined as:
\[ F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t) \ dt \] This means that \(F(x)\) accumulates the probabilities from the leftmost point of the distribution up to \(x\).
The CDF always ranges from 0 to 1, because the probability of \(X\) falling within its entire range is 1.
This cumulative property makes the CDF very useful in various applications, such as determining percentiles and constructing other distributions.
Probability Density Function
A probability density function (PDF) describes how the probabilities of a continuous random variable are distributed.
For a variable \(X\), the PDF is denoted by \(f(x)\).
The PDF has several key properties:
  • The area under the curve \(f(x)\) over all possible values of \(X\) is 1.
  • \(f(x)\) is always non-negative: \(f(x) \geq 0\) for all \(x\).
  • The probability that \(X\) falls in an interval \([a, b]\) is the integral of \(f(x)\) from \(a\) to \(b\): \(P(a \leq X \leq b) = \int_{a}^{b} f(x) \ dx\).
Understanding the shape and properties of a PDF helps in interpreting the behavior of a random variable.
In the exercise, the PDF \(f(x)\) and its associated CDF \(F(x)\) are used to transform the random variable \(X\) into another variable \(Y\) to demonstrate uniform distribution.
Inverse Function
An inverse function essentially reverses the operation of the original function.
For a given function \(f\), its inverse \(f^{-1}\) satisfies \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\).
In probability theory, the inverse of the CDF \(F\) is crucial for transforming random variables.
When dealing with continuous random variables, the CDF \(F(x)\) is typically a monotonically increasing function, which means it is invertible.
Here's how the concept is used in the exercise:
To prove that the variable \(Y = F(X)\) is uniformly distributed between 0 and 1, we need to find its CDF \(G(y)\).
By expressing \(G(y)\) as \(P(Y \leq y)\), we identify that:
\(P(Y \leq y) = P(F(X) \leq y)\).
Since \(F\) is a CDF, it is invertible, hence:
\(P(F(X) \leq y) = P(X \leq F^{-1}(y))\).
Finally, substituting this into the definition of the CDF, we see that \(P(X \leq F^{-1}(y)) = y\).
Therefore, the CDF of \(Y\) is \(G(y) = y\), which confirms that \(Y\) is uniformly distributed.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

As every student of probability theory will know, Bayesylvania is awash with natives, not all of whom can be trusted to tell the truth, and lost, and apparently somewhat deaf, travellers who ask the same question several times in an attempt to get directions to the nearest village. One such traveller finds himself at a T-junction in an area populated by the Asciis and Bisciis in the ratio 11 to \(5 .\) As is well known, the Biscii always lie, but the Ascii tell the truth three quarters of the time, giving independent answers to all questions, even to immediately repeated ones. (a) The traveller asks one particular native twice whether he should go to the left or to the right to reach the local village. Each time he is told 'left'. Should he take this advice, and, if he does, what are his chances of reaching the village? (b) The traveller then asks the same native the same question a third time, and for a third time receives the answer 'left'. What should the traveller do now? Have his chances of finding the village been altered by asking the third question?

A particle is confined to the one-dimensional space \(0 \leq x \leq a\), and classically it can be in any small interval \(d x\) with equal probability. However, quantum mechanics gives the result that the probability distribution is proportional to \(\sin ^{2}(n \pi x / a)\), where \(n\) is an integer. Find the variance in the particle's position in both the classical and quantum- mechanical pictures, and show that, although they differ, the latter tends to the former in the limit of large \(n\), in agreement with the correspondence principle of physics.

A discrete random variable \(X\) takes integer values \(n=0,1, \ldots, N\) with probabilities \(p_{n}\). A second random variable \(Y\) is defined as \(Y=(X-\mu)^{2}\), where \(\mu\) is the expectation value of \(X\). Prove that the covariance of \(X\) and \(Y\) is given by $$ \operatorname{Cov}[X, Y]=\sum_{n=0}^{N} n^{3} p_{n}-3 \mu \sum_{n=0}^{N} n^{2} p_{n}+2 \mu^{3} $$ Now suppose that \(X\) takes all of its possible values with equal probability, and hence demonstrate that two random variables can be uncorrelated, even though one is defined in terms of the other.

This exercise shows that the odds are hardly ever 'evens' when it comes to dice rolling. (a) Gamblers \(A\) and \(B\) each roll a fair six-faced die, and \(B\) wins if his score is strictly greater than \(A\) 's. Show that the odds are 7 to 5 in A's favour. (b) Calculate the probabilities of scoring a total \(T\) from two rolls of a fair die for \(T=2,3, \ldots, 12 .\) Gamblers \(C\) and \(D\) each roll a fair die twice and score respective totals \(T_{C}\) and \(T_{D}, D\) winning if \(T_{D}>T_{C} .\) Realising that the odds are not equal, \(D\) insists that \(C\) should increase her stake for each game. \(C\) agrees to stake \(£ 1.10\) per game, as compared to D's \(£ 1.00\) stake. Who will show a profit?

A husband and wife decide that their family will be complete when it includes two boys and two girls - but that this would then be enough! The probability that a new baby will be a girl is \(p\). Ignoring the possibility of identical twins, show that the expected size of their family is $$ 2\left(\frac{1}{p q}-1-p q\right) $$ where \(q=1-p\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.