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Let \(X\) be a continuous random variable with probability density function $$ f_{X}(x)= \begin{cases}\frac{3}{4} x(2-x) & \text { for } 0 \leq x \leq 2 \\\ 0 & \text { elsewhere }\end{cases} $$ a. Determine the distribution function \(F_{X}\). b. Let \(Y=\sqrt{X}\). Determine the distribution function \(F_{Y}\). c. Determine the probability density of \(Y\).

Short Answer

Expert verified
a. \(F_X(x) = \frac{3}{4} x^2 - \frac{x^3}{4}\); b. \(F_Y(y) = \frac{3}{4} y^4 - \frac{y^6}{4}\); c. \(f_Y(y) = 3y^3 - \frac{3}{2}y^5\).

Step by step solution

01

Define Cumulative Distribution Function (CDF)

The cumulative distribution function (CDF) of a continuous random variable is given by\[ F_X(x) = \int_{-\infty}^{x} f_X(t) \, dt \]where \( f_X(t) \) is the probability density function (pdf). Since \( f_X(x) \) is non-zero only between 0 and 2, we can write:\[ F_X(x) = \int_{0}^{x} \frac{3}{4} t(2-t) \, dt \text{ for } 0 \leq x \leq 2 \]
02

Calculate the Integral

Evaluate the integral to determine the CDF:\[ F_X(x) = \int_{0}^{x} \frac{3}{4} (2t - t^2) \, dt = \frac{3}{4} \left[ \int_{0}^{x} 2t \, dt - \int_{0}^{x} t^2 \, dt \right] \]This can be further split and solved as two simple integrals.
03

Solve the First Integral

Solve \( \int_{0}^{x} 2t \, dt \):\[ \int_{0}^{x} 2t \, dt = \left[ t^2 \right]_{0}^{x} = x^2 \]
04

Solve the Second Integral

Solve \( \int_{0}^{x} t^2 \, dt \):\[ \int_{0}^{x} t^2 \, dt = \left[ \frac{t^3}{3} \right]_{0}^{x} = \frac{x^3}{3} \]
05

Simplify the CDF Expression

Substitute the expressions from Steps 3 and 4 back into the integral solved in Step 2:\[ F_X(x) = \frac{3}{4} \left[ x^2 - \frac{x^3}{3} \right] = \frac{3}{4} x^2 - \frac{x^3}{4} \]So,\[ F_X(x) = \frac{3}{4} x^2 - \frac{x^3}{4} \text{ for } 0 \leq x \leq 2 \]
06

Define the Relationship between X and Y

Given \( Y = \sqrt{X} \), express \( X \) in terms of \( Y \):\[ X = Y^2 \]Since \( X \) ranges from 0 to 2, \( Y \) will range from 0 to \( \sqrt{2} \) (i.e., \( 0 \leq Y \leq \sqrt{2} \)).
07

Determine the CDF of Y

The CDF of \( Y \) is related to the CDF of \( X \):\[ F_Y(y) = P(Y \leq y) = P(\sqrt{X} \leq y) = P(X \leq y^2) = F_X(y^2) \]Using the expression from Step 5,\[ F_Y(y) = \frac{3}{4} y^4 - \frac{(y^2)^3}{4} = \frac{3}{4} y^4 - \frac{y^6}{4} \text{ for } 0 \leq y \leq \sqrt{2} \]
08

Find the Probability Density Function of Y

Differentiate the CDF of \( Y \) to find its probability density function:\[ f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{d}{dy} \left( \frac{3}{4} y^4 - \frac{y^6}{4} \right) \]Calculating the derivative gives:\[ f_Y(y) = 3y^3 - \frac{3}{2}y^5 \text{ for } 0 \leq y \leq \sqrt{2} \]

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Key Concepts

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

Cumulative Distribution Function
The cumulative distribution function (CDF) is a fundamental tool in probability and statistics used to describe the probability that a random variable takes on a value less than or equal to a certain threshold. For a continuous random variable, the CDF is the integral of the probability density function (PDF), which in this case we denote as \(f_X(x)\). The CDF is expressed mathematically as: \[ F_X(x) = \int_{-\infty}^{x} f_X(t) \, dt \] When dealing with a specific interval for a variable \(x\), only the section where the PDF is non-zero is considered. In our example, \(f_X(x)\) is non-zero from 0 to 2, so the CDF is calculated over this range. The CDF provides the total probability accumulated up to a given point \(x\) in the interval. This helps in understanding the distribution and likelihood of different outcomes for the variable. Importantly, the CDF always increases with \(x\) since probabilities cannot decrease, and it approaches 1 as \(x\) approaches infinity, covering the entire probability space.
Probability Density Function
The probability density function (PDF) is crucial in defining the likelihood of a continuous random variable. Unlike the discrete counterpart, for which probabilities of exact values are defined, a PDF gives a density over an interval and thus can yield probabilities only in the context of an interval. For a random variable \(X\) with PDF \(f_X(x)\), the function must satisfy the property that:
  • \(f_X(x) \geq 0\) for all \(x\)
  • \(\int_{-\infty}^{\infty} f_X(x) \, dx = 1\)
These conditions ensure that the total probability sums to 1 and no negative probabilities are assigned. Understanding a PDF allows us to determine CDF by integration. From the CDF, one can also derive back to the PDF by differentiation. For the given function, the specific form describes how probabilities distribute over the interval 0 to 2. Notably, outside this interval, the PDF is naturally addressed by being zero, implying no probability beyond these bounds.
Transformation of Variables
Transforming variables is a common technique used to study how one variable relates to another via a functional relationship. When a function of a random variable is involved, the distribution of this new variable can be derived through transformation rules. For a random variable \(X\) and a transformation \(Y = g(X)\), the new random variable's distribution needs to be evaluated. Critical steps include using the inverse transformation and correctly adjusting for changes in probability intervals. When we transform \(X\) to \(Y = \sqrt{X}\), observing the range of \(Y\) is essential, which in this context changes from 0 to \(\sqrt{2}\). To find the CDF of \(Y\), we look at probabilities up to a certain \(y\), expressed as: \[ F_Y(y) = P(Y \leq y) = P(\sqrt{X} \leq y) = P(X \leq y^2) = F_X(y^2) \] The PDF for \(Y\) can subsequently be found by differentiating the CDF. These steps are foundational in analyzing how transformations affect data and help reveal insights that may not be evident in the original variable's distribution.

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Most popular questions from this chapter

Let \(W\) have a \(U(\pi, 2 \pi)\) distribution. What is larger: \(\mathrm{E}[\sin (W)]\) or \(\sin (\mathrm{E}[W])\) ? Check your answer by computing these two numbers.

Let \(X\) be a continuous random variable with probability density \(f_{X}\) that takes only positive values and let \(Y=1 / X\). a. Determine \(F_{Y}(y)\) and show that $$ f_{Y}(y)=\frac{1}{y^{2}} f_{X}\left(\frac{1}{y}\right) \quad \text { for } y>0 . $$ b. Let \(Z=1 / Y\). Using a, determine the probability density \(f_{Z}\) of \(Z\), in terms of \(f_{X}\).

Transforming exponential distributions. a. Let \(X\) have an \(\operatorname{Exp}\left(\frac{1}{2}\right)\) distribution. Determine the distribution function of \(\frac{1}{2} X\). What kind of distribution does \(\frac{1}{2} X\) have? b. Let \(X\) have an \(\operatorname{Exp}(\lambda)\) distribution. Determine the distribution function of \(\lambda X\). What kind of distribution does \(\lambda X\) have?

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be independent random variables, all with a \(U(0,1)\) distribution. Let \(Z=\max \left\\{X_{1}, \ldots, X_{n}\right\\}\) and \(V=\min \left\\{X_{1}, \ldots, X_{n}\right\\}\). a. Compute \(\mathrm{E}\left[\max \left\\{X_{1}, X_{2}\right\\}\right]\) and \(\mathrm{E}\left[\min \left\\{X_{1}, X_{2}\right\\}\right]\). b. Compute \(\mathrm{E}[Z]\) and \(\mathrm{E}[V]\) for general \(n\). c. Can you argue directly (using the symmetry of the uniform distribution (see Exercise 6.3) and not the result of the computation in b) that \(1-\mathrm{E}\left[\max \left\\{X_{1}, \ldots, X_{n}\right\\}\right]=\mathrm{E}\left[\min \left\\{X_{1}, \ldots, X_{n}\right\\}\right] ?\)

Let \(X\) be a random variable with the following probability mass function: \begin{tabular}{ccccc} \(x\) & 0 & 1 & 100 & 10000 \\ \hline \(\mathrm{P}(X=x)\) & \(\frac{1}{4}\) & \(\frac{1}{4}\) & \(\frac{1}{4}\) & \(\frac{1}{4}\) \end{tabular} a. Determine the distribution of \(Y=\sqrt{X}\). b. Which is larger \(\mathrm{E}[\sqrt{X}]\) or \(\sqrt{\mathrm{E}[X]}\) ? c. Compute \(\sqrt{\mathrm{E}[X]}\) and \(\mathrm{E}[\sqrt{X}]\) to check your answer (and to see that it makes a big difference!).

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