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Chapter 8: Problem 13

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.

Short Answer

Expert verified
\(\mathrm{E}[\sin(W)]\) is larger than \(\sin(\mathrm{E}[W])\).

Step by step solution

01

Understanding the Distribution

We are given that the random variable \(W\) follows a uniform distribution on the interval \([\pi, 2\pi]\). This means that any value within this interval is equally likely to occur.
02

Compute the Expected Value of W

For a uniform distribution \(U(a, b)\), the expected value \(\mathrm{E}[W]\) is calculated as the midpoint of the interval. Thus, \(\mathrm{E}[W] = \frac{\pi + 2\pi}{2} = \frac{3\pi}{2}\).
03

Compute \(\sin(\mathrm{E}[W])\)

Substitute \(\mathrm{E}[W] = \frac{3\pi}{2}\) into the sine function: \(\sin(\mathrm{E}[W]) = \sin\left(\frac{3\pi}{2}\right) = -1\).
04

Compute \(\mathrm{E}[\sin(W)]\)

The expectation \(\mathrm{E}[\sin(W)]\) for a uniform distribution can be calculated by integrating the sine function over the interval \([\pi, 2\pi]\). We find:\[\mathrm{E}[\sin(W)] = \frac{1}{2\pi - \pi} \int_{\pi}^{2\pi} \sin(x) \, dx\]Evaluate the integral:\[ = \frac{1}{\pi} \left[-\cos(x) \right]_\pi^{2\pi}\]\[ = \frac{1}{\pi} \left[ -\cos(2\pi) + \cos(\pi) \right]\]\[ = \frac{1}{\pi} \left[ -(1) + (-1) \right] = -\frac{2}{\pi}\]
05

Compare the Two Values

We found that \(\mathrm{E}[\sin(W)] = -\frac{2}{\pi}\) and \(\sin(\mathrm{E}[W]) = -1\). Since \(-\frac{2}{\pi} \approx -0.6366\), which is greater than \(-1\), we conclude that \(\mathrm{E}[\sin(W)]\) is larger than \(\sin(\mathrm{E}[W])\).

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Key Concepts

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

Expected Value
When dealing with probability distributions, the Expected Value is a key concept. It represents the average or mean of a random variable over a vast number of trials. For a uniform distribution, this is calculated as the midpoint of the interval. This is because each outcome within the interval is equally likely. So, if a random variable, say \(W\), ranges from \(\pi\) to \(2\pi\), its Expected Value would be:
  • The midpoint: \(\frac{\pi + 2\pi}{2} = \frac{3\pi}{2}\)
In this scenario, the Expected Value tells us the central tendency of \(W\) when observed over many occurrences.

The Expected Value is useful because it simplifies comparisons. For example, when analyzing transformations of variables, like applying a sine function to our uniform distribution, knowing the expected value beforehand makes the computation easier.
Sine Function
The Sine Function is a trigonometric function that can be represented as \(\sin(x)\), where \(x\) is an angle measured in radians. The sine function has an output range from -1 to 1, making it periodic with a cycle repeating every \(2\pi\). Knowing how it behaves is crucial when dealing with angles that return values outside the basic sine cycle.

For example, \(\sin\left(\frac{3\pi}{2}\right) = -1\). This can be visualized on the sine curve where \(\frac{3\pi}{2}\) corresponds to the lowest point on the curve, giving the minimum value of -1. The sinusoidal nature of this function often makes it challenging yet interesting to interact with in probability and calculus, especially when integrated.
Integration
Integration is a fundamental concept in calculus used for finding areas under curves, among other applications. When working with random variables, integration helps determine the expected values of more complex transformations.
  • In the given problem, integrating the sine function from \(\pi\) to \(2\pi\) helps compute the expectation: \(\mathrm{E}[\sin(W)]\).
This integral is formulated as:
  • \( \int_{\pi}^{2\pi} \sin(x) \, dx \)
Evaluating this integral leads to the average value of \(\sin(W)\) within the given interval, resulting in \(-\frac{2}{\pi}\). This result provides insight into how the sine function behaves across the interval from \(\pi\) to \(2\pi\) and assists in comparing it with other transformations of \(W\). Integration thus serves as a crucial tool in the determination of expected values, particularly when dealing with non-linear transformations like the sine function.

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Most popular questions from this chapter

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.

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).

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!).

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 \(f_{X}\).

Suppose \(X\) has a uniform distribution over the points \(\\{1,2,3,4,5,6\\}\) and that \(g(x)=\sin \left(\frac{\pi}{2} x\right)\). a. Determine the distribution of \(Y=g(X)=\sin \left(\frac{\pi}{2} X\right)\), that is, specify the values \(Y\) can take and give the corresponding probabilities. b. Let \(Z=\cos \left(\frac{\pi}{2} X\right)\). Determine the distribution of \(Z\). c. Determine the distribution of \(W=Y^{2}+Z^{2}\). Warning: in this example there is a very special dependency between \(Y\) and \(Z\), and in general it is much harder to determine the distribution of a random variable that is a function of two other random variables. This is the subject of Chapter 11 .
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