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The continuous random variable \(U\) is uniformly distributed over \([0,1]\). a. Determine the distribution function of \(V=2 U+7\). What kind of distribution does \(V\) have? b. Determine the distribution function of \(V=r U+s\) for all real numbers \(r>0\) and \(s\). See Exercise \(8.9\) for what happens for negative \(r\).

Short Answer

Expert verified
a. V is uniformly distributed over [7, 9]. b. V is uniformly distributed over [s, r+s].

Step by step solution

01

Understand Uniform Distribution

We start with the given continuous random variable \( U \) which is uniformly distributed over the interval \([0, 1]\). This means that the probability density function (PDF) of \( U \) is constant across this interval. The PDF is defined as \( f_U(u) = 1 \) for \( u \in [0, 1] \).
02

Transforming the Variable for Part (a)

Consider the transformation \( V = 2U + 7 \). The function shows a linear transformation of \( U \). We need to find the distribution of \( V \). Since \( U \) varies from 0 to 1, \( V \) will range from \( 2(0) + 7 = 7 \) to \( 2(1) + 7 = 9 \).
03

Determine the Distribution Function of V for Part (a)

The distribution of \( V \) will also be uniform but shifted and scaled. Specifically, \( V \) is uniformly distributed over the interval \([7, 9]\). Therefore, the probability density function of \( V \) is given by \( f_V(v) = \frac{1}{9-7} = \frac{1}{2} \) for \( v \in [7, 9] \).
04

General Transformation for Part (b)

Now consider \( V = rU + s \) where \( r > 0 \) and \( s \) are constants. This transformation linearly scales \( U \) by factor \( r \) and shifts it by \( s \). Since \( U \) ranges from 0 to 1, \( V \) will range from \( r \times 0 + s = s \) to \( r \times 1 + s = r + s \).
05

Distribution Function of V for Part (b)

With the linear transformation \( V = rU + s \), \( V \) is uniformly distributed over the interval \([s, r+s]\). The probability density function is given by \( f_V(v) = \frac{1}{r} \) for \( v \in [s, r+s] \). The distribution retains a uniform nature due to the linear transformation.

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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) is a vital concept in understanding continuous random variables. When dealing with continuous distributions, PDFs describe how probabilities are distributed over different values. For our uniform distribution of the variable \( U \) over the interval \([0, 1]\), the PDF is constant. Practically, this means each point in the interval is equally likely. Thus, the PDF of \( U \) is given as:
  • \( f_U(u) = 1 \) for \( u \in [0, 1] \).
This indicates the total probability across the entire interval sums up to 1, ensuring it's a valid probability measure. Remember, for any uniform distribution across any interval \([a, b]\), the PDF is defined as \( f(x) = \frac{1}{b-a} \). This formula ensures that the sum of probabilities over the interval remains equal to 1. Knowing how to calculate and interpret PDFs is crucial for understanding the behavior of continuous random variables.
Linear Transformation
Linear transformation is a fundamental operation used to modify random variables. In our context, we dealt with a transformation of the form \( V = rU + s \), where \( r \) is a positive constant scaling factor, and \( s \) is a shifting constant. This type of transformation adjusts the range of the original random variable \( U \) while preserving the uniform nature of its distribution.With a given transformation \( V = 2U + 7 \), the interval of \( U \) from \([0, 1]\) shifts to \([7, 9]\) for \( V \). This is calculated by substituting the bounds of \( U \) into the transformation equation:
  • When \( U = 0 \): \( V = 2 \times 0 + 7 = 7 \).
  • When \( U = 1 \): \( V = 2 \times 1 + 7 = 9 \).
The resulting distribution of \( V \) is uniformly spread across \([7, 9]\). Generally, for any linear transformation \( V = rU + s \), the transformation results in \( V \) being uniformly distributed over \([s, r+s]\). Such transformations are key in altering the scale and position, all while preserving the overall shape of the distribution.
Continuous Random Variable
Continuous random variables, like the one we are examining, differ from discrete ones as they can take an infinite number of possible values within a given range. Imagine a continuous random variable as capable of covering every point in an interval, providing a smooth range of outcomes. For instance, if \( U \) is uniformly distributed across \([0, 1]\), it includes every real number between 0 and 1.An essential property of continuous random variables is their use of probability density functions (PDFs) to describe distributions instead of sum probability for discrete variables. This is because these variables are not restricted to distinct points. For example, the continuous random variable \( U \) has a PDF \( f_U(u) = 1 \) over \([0, 1]\). This constant PDF indicates equal chances for any number in this interval.Understanding continuous random variables greatly aids in modeling real-world phenomena where outcomes are not discrete, like time, temperature, or stock prices. The concepts of linear transformation and probability density functions are crucial in manipulating and interpreting these types of variables.

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Most popular questions from this chapter

Let \(X\) be a continuous random variable. Express the distribution function and probability density of the random variable \(Y=-X\) in terms of those of \(X\).

Let \(X\) have an \(\operatorname{Exp}(1)\) distribution, and let \(\alpha\) and \(\lambda\) be positive numbers. Determine the distribution function of the random variable $$ W=\frac{X^{1 / \alpha}}{\lambda} $$ The distribution of the random variable \(W\) is called the Weibull distribution with parameters \(\alpha\) and \(\lambda\).

From the "north pole" \(N\) of a circle with diameter 1 , a point \(Q\) on the circle is mapped to a point \(t\) on the line by its projection from \(N\), as illustrated in Figure \(8.2\). Suppose that the point \(Q\) is uniformly chosen on the circle. This is the same as saying that the angle \(\varphi\) is uniformly chosen from the interval \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) (can you see this?). Let \(X\) be this angle, so that \(X\) is uniformly distributed over the interval \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). This means that \(\mathrm{P}(X \leq \varphi)=1 / 2+\varphi / \pi\) (cf. Quick exercise \(5.3)\). What will be the distribution of the projection of \(Q\) on the line? Let us call this random variable \(Z\). Then it is clear that the event \(\\{Z \leq t\\}\) is equal to the event \(\\{X \leq \varphi\\}\), where \(t\) and \(\varphi\) correspond to each other under the projection. This means that \(\tan (\varphi)=t\), which is the same as saying that \(\arctan (t)=\varphi .\) a. What part of the circle is mapped to the interval \([1, \infty)\) ? b. Compute the distribution function of \(Z\) using the correspondence between \(t\) and \(\varphi\). c. Compute the probability density function of \(Z\). The distribution of \(Z\) is called the Cauchy distribution (which will be discussed in Chapter 11).

Often one is interested in the distribution of the deviation of a random variable \(X\) from its mean \(\mu=\mathrm{E}[X]\). Let \(X\) take the values \(80,90,100,110\), and 120, all with probability \(0.2 ;\) then \(\mathrm{E}[X]=\mu=100\). Determine the distribution of \(Y=|X-\mu|\). That is, specify the values \(Y\) can take and give the corresponding probabilities.

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?

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