Density Function Of U = Y^2: A Step-by-Step Solution

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Hey guys! Today, we're diving into a fun probability problem: figuring out the probability density function (PDF) of a transformed random variable. Specifically, we'll tackle the case where we have a random variable Y, and we create a new random variable U by squaring Y. This kind of transformation is super common in statistics, so understanding how it works is a must. Let's break it down together!

Problem Setup: Understanding the Random Variables

First things first, let's clearly define what we're working with. We're given that Y is a random variable, which means it's a variable whose value is a numerical outcome of a random phenomenon. Y has a specific probability distribution, described by its probability density function (PDF). Think of the PDF as a function that tells us the relative likelihood of Y taking on a particular value. Our main keyword here is probability density function (PDF), which is crucial for understanding random variables. The PDF, denoted as f(y), is defined as follows:

f(y) = egin{cases} 2(1 - y) & \text{if } 0 	ext{ ≤ } y 	ext{ ≤ } 1 \\ 0 & \text{otherwise} 
\end{cases}

This formula tells us that Y can only take values between 0 and 1. For values within this range, the probability density decreases linearly as y increases. Outside the range of 0 to 1, the probability density is 0, meaning Y will never take those values. Now, we introduce a new random variable, U, which is simply the square of Y: U = Y². Our mission, should we choose to accept it, is to find the PDF of U. This means we want to find a function, let's call it g(u), that tells us the probability density for different values of U. The process of finding the PDF of U involves a transformation of variables, a common technique in probability and statistics. We need to carefully consider how the transformation U = Y² affects the probabilities associated with different values.

Breaking Down the Problem: Why This Matters

Understanding transformations of random variables is fundamental in many areas of statistics and probability. For instance, in signal processing, we might square a signal to analyze its power. In finance, we might look at the square of returns to understand volatility. Therefore, understanding PDF transformations is very essential. The core idea is that when we transform a random variable, we change its distribution. We need a way to figure out what the new distribution looks like. In our case, we're going from Y to U = Y². Since Y is between 0 and 1, U will also be between 0 and 1. However, the relationship isn't linear. Squaring a number changes the way probabilities are distributed. The original PDF of Y provides the foundation for deriving the PDF of U. We'll use the relationship U = Y² and the known f(y) to find g(u). It's like we're taking the probability mass and reshaping it according to the transformation. So, buckle up, and let's dive into the solution! We'll use a combination of calculus and probability concepts to get there. Remember, the key is to think about how the transformation affects the probabilities associated with different intervals.

Finding the Cumulative Distribution Function (CDF) of U

The first strategic move in our quest to find the PDF of U is to determine its cumulative distribution function (CDF). *The CDF, denoted as G(u), tells us the probability that the random variable U takes on a value less than or equal to a specific value u. * Think of it as the accumulated probability up to a certain point. This is a crucial step because it often simplifies the process of finding the PDF, which we'll do later by differentiating the CDF. Mathematically, the CDF of U, denoted as G(u), is defined as:

G(u) = P(U ≤ u)

Where P(U ≤ u) represents the probability that U is less than or equal to u. Now, remember that U = Y². So, we can rewrite the above expression in terms of Y:

G(u) = P(Y² ≤ u)

This is where things get interesting! We need to think about how the inequality Y² ≤ u translates into conditions on Y. Since Y is between 0 and 1, taking the square root of both sides is perfectly valid and helps us isolate Y. So, we have:

G(u) = P(-√u ≤ Y ≤ √u)

However, we know that Y is only defined between 0 and 1. Therefore, we only need to consider the positive square root and the interval [0, √u]. The key here is to remember the range of Y and how it affects the probability calculation. This simplifies our expression to:

G(u) = P(0 ≤ Y ≤ √u)

Calculating the Probability: Integrating the PDF of Y

Now we're getting somewhere! We've expressed the CDF of U in terms of the probability of Y falling within a specific interval. To actually calculate this probability, we'll use the PDF of Y, f(y), that was given to us initially. The probability that Y lies between two values is the integral of its PDF over that interval. Therefore:

G(u) = ∫[from 0 to √u] f(y) dy

Remember that f(y) = 2(1 - y) for 0 ≤ y ≤ 1. So, we can substitute that into our integral:

G(u) = ∫[from 0 to √u] 2(1 - y) dy

This is a standard integral that we can solve using basic calculus. Let's integrate! The antiderivative of 2(1 - y) is 2y - y². So, we evaluate this at the limits of integration, √u and 0:

G(u) = [2y - y²] evaluated from 0 to √u

G(u) = (2√u - (√u)²) - (2(0) - 0²)

G(u) = 2√u - u

So, we've found the CDF of U, G(u) = 2√u - u. This is a significant milestone. We now have a function that tells us the probability that U is less than or equal to any value u. But remember, our ultimate goal is to find the PDF of U. And the good news is, we're just one step away!

Finding the Probability Density Function (PDF) of U

Alright, guys, we're in the home stretch! We've successfully navigated the tricky terrain of finding the CDF of U. Now, for the grand finale: determining the probability density function (PDF) of U, which we'll denote as g(u). The fundamental relationship between the CDF and the PDF is that the PDF is the derivative of the CDF. Think of it this way: the CDF tells us the accumulated probability, while the PDF tells us the rate at which that probability is accumulating. Therefore, to find g(u), we simply need to differentiate G(u) with respect to u. Mathematically:

g(u) = dG(u)/du

We found earlier that G(u) = 2√u - u. So, let's differentiate this with respect to u. Remember that √u can be written as u^(1/2). So, we have:

g(u) = d/du (2u^(1/2) - u)

Using the power rule for differentiation (d/dx x^n = nx^(n-1)), we get:

g(u) = 2 * (1/2) * u^(-1/2) - 1

g(u) = u^(-1/2) - 1

We can rewrite u^(-1/2) as 1/√u, so our PDF becomes:

g(u) = 1/√u - 1

Defining the PDF: Considering the Range of U

We've found the functional form of the PDF, but we need to be precise about its definition. Remember, the PDF needs to specify the density for all possible values of U. We know that U = Y², and Y is between 0 and 1. Therefore, U will also be between 0 and 1. The range of U is crucial for defining the PDF completely. Outside this range, the probability density is zero. So, the complete definition of the PDF of U is:

g(u) = egin{cases} 1/√u - 1 & \text{if } 0 	ext{ ≤ } u 	ext{ ≤ } 1 \\ 0 & \text{otherwise} 
\end{cases}

This is our final answer! The PDF of U, g(u), is given by 1/√u - 1 for 0 ≤ u ≤ 1, and 0 otherwise. This function tells us the probability density for different values of U. Notice that as u approaches 0, the density increases, indicating that values of U near 0 are more likely than values near 1. This makes sense because squaring a number between 0 and 1 will result in a number closer to 0 than to 1, on average.

Conclusion: A Journey Through Probability Transformations

Woohoo! We made it! Finding the PDF of a transformed random variable can seem daunting at first, but by breaking it down into steps, it becomes manageable. We started with the PDF of Y, f(y), transformed it to the CDF of U, G(u), and finally differentiated to get the PDF of U, g(u). This process is a powerful tool in probability and statistics. It allows us to understand how transformations affect the distribution of random variables. Remember, the key steps are:

  1. Find the CDF of the transformed variable.
  2. Differentiate the CDF to get the PDF.
  3. Carefully consider the range of the transformed variable.

This problem highlights the importance of understanding the relationship between PDFs and CDFs, as well as the techniques of variable transformation. It's a fundamental concept that pops up in various applications, so mastering it is well worth the effort. So, keep practicing, and you'll become a pro at handling probability transformations in no time! Keep rocking, guys! Understanding PDF transformations is very essential in probability and statistics. Remember, practice makes perfect, so keep tackling those problems! You've got this!