Question

Let \(X\) have the pdf \(f(x)=4 x^{3}, 0

Short answer

The cdf of \(X\) is \(x^{4}\), for \(0 < x < 1\). The cdf of \(Y\) is \(1 - e^{-y}\), and the pdf of \(Y\) is \(e^{-y}\), for \(y > 0.

Step-by-step solution

Find the CDF of X

First, find the cumulative distribution function (CDF) of \(X\) by integrating its probability density function (pdf).The CDF \(F(x)\) for \(0 < x < 1\) is given by the following integral,\[ F(x) = \int_{0}^{x} 4t^{3}dt = x^{4} \]for \(0 < x < 1\), and zero elsewhere.

Determine the transformation

The next task is to find the cumulative distribution function (CDF) of \(Y\). To do this, we need to determine the relationship between \(X\) and \(Y\). From the problem, \(Y = -\ln X^{4} = -4 \ln X\). Solve this equation for \(X\) to get \(X = e^{-Y/4}\).

Find the CDF and PDF of Y

First, find the cdf of \(Y\) using \(F_Y(y) = P(Y \leq y) = P(X > e^{-y/4}) = 1 - P(X \leq e^{-y/4}) = 1 - F_X(e^{-y/4}).\)Plugging in the cdf of X from Step 1 and simplifying, we get \(F_Y(y) = 1 - e^{-y}.\)To find the pdf of \(Y\), take the derivative of the cdf of \(Y\), \(f_Y(y) = \frac{df_Y(y)}{dy}.\)The derivative of \(1 - e^{-y}\) is \(e^{-y}\). Therefore, the pdf of \(Y\) is \(e^{-y}\), for \(y>0\).

Key concepts

Probability Density Function

The probability density function, or PDF, is a fundamental concept in probability theory and serves as a cornerstone for understanding continuous random variables. Imagine holding a lump of clay representing the total probability, which is always 1, and you're shaping it along an axis that represents all possible outcomes of a random variable. The height of the clay at any point indicates how densely the probability is packed around that outcome.

A PDF is defined only for continuous random variables and indicates the likelihood of the variable taking on a specific value. In the provided exercise, the function given, f(x) = 4x³ for 0 < x < 1, describes how the probability is distributed across the values between 0 and 1. Importantly, the area under the PDF curve between two points gives the probability that the random variable falls within that interval.

Integrating the PDF over its entire range, from negative infinity to positive infinity, must equal 1 because the random variable must take on some value within that range. However, for a defined range, like in the exercise where the PDF is only defined between 0 and 1, the integration bounds are limited accordingly.

Random Variable Transformation

When dealing with random variables, often we find it beneficial to transform them into new random variables which might be easier to work with or interpret. This is known as random variable transformation.

To transform a random variable, a function is applied to it, essentially reshaping the original variable's probability distribution into a new one that fits the transformed variable. In the original exercise, the transformation applies the function Y = -4ln(X) to the variable X. This results in a new random variable, Y, with its own PDF and CDF that differ from X.

The transformation function must be chosen carefully. It should be a one-to-one function—meaning every value of Y should come from exactly one value of X. This ensures that probabilities are correctly transferred from the original variable to the new one. In the context of the exercise, the natural logarithm function is both continuous and one-to-one in the interval (0,1), which are important properties to preserve the integrity of the probability distribution during the transformation.

Integration in Probability

Integration plays a pivotal role in probability, especially in the computation of cumulative distribution functions (CDFs) for continuous random variables. To understand this concept, picture integration as a tool for adding up tiny slices of probability over a range to get a total probability.

For a continuous random variable X with a probability density function (PDF) f(x), calculating the CDF requires integrating the PDF from the lowest value in the range to a variable upper limit x. This is represented by the integral F(x) = ∫_a^x f(t)dt, where a is the lower bound of the variable's domain and t is a dummy variable of integration.

In the context of the exercise, integration is used first to find the CDF of X, and then later to derive the CDF of the transformed variable Y. The idea is that since the CDF represents the probability that a random variable is less than or equal to a particular value, calculating this cumulative probability involves integrating the PDF up to that value. This is why understanding integration is vital to solving probability problems that involve continuous distributions.