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Let the density function of a random variable \(Y\) be given by $$ f(y)=\left\\{\begin{array}{ll} \frac{2}{\pi\left(1+y^{2}\right)}, & -1 \leq y \leq 1 \\ 0 & \text { elsewhere } \end{array}\right. $$ a. Find the distribution function. b. Find \(E(Y)\)

Short Answer

Expert verified
a. \(F(y) = \frac{\tan^{-1}(y)}{\pi} + \frac{1}{4}\); b. \(E(Y) = 0\).

Step by step solution

01

Understanding the Density Function

The density function given is defined for the interval \(-1 \leq y \leq 1\). Outside this range, the function is zero. We need to calculate the cumulative distribution function (CDF) and the expected value of this density function.
02

Determine the Cumulative Distribution Function (CDF)

The CDF, F(y), is given by the integral of the density function \(f(y)\) over the interval from the lower bound to \(y\). Thus, \[F(y) = \int_{-1}^{y} \frac{2}{\pi(1+t^2)} \, dt, \, -1 \leq y \leq 1.\]This integral is known as the arctangent function: \[F(y) = \frac{1}{\pi} [\tan^{-1}(t)]_{-1}^{y} = \frac{1}{\pi} [\tan^{-1}(y) - \tan^{-1}(-1)].\]Since \(\tan^{-1}(-1) = -\frac{\pi}{4}\), the CDF becomes\[F(y) = \frac{1}{\pi} [\tan^{-1}(y) + \frac{\pi}{4}].\]Therefore, \[F(y) = \frac{\tan^{-1}(y)}{\pi} + \frac{1}{4}.\]
03

Compute the Expected Value E(Y)

The expected value is calculated by integrating the product of \(y\) and the density function \(f(y)\) over its range:\[E(Y) = \int_{-1}^{1} y \cdot \frac{2}{\pi(1+y^2)} \, dy.\]By symmetry, the function is odd over a symmetric interval around 0; the positive and negative contributions will cancel out:\[E(Y) = 0.\]Thus, \(E(Y) = 0\) without explicit computation.

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Key Concepts

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

Density Function
The density function of a random variable is a crucial part of probability theory. It defines how the values of a random variable, say \(Y\), are distributed across different intervals. The density function you're dealing with here, \(f(y)\), is specifically defined for \(-1 \leq y \leq 1\), and it is zero elsewhere.
This means that \(Y\) can only take values between -1 and 1, and this is where we should focus our attention. The function you have is \(f(y) = \frac{2}{\pi(1+y^{2})}\), which indicates the likelihood of \(Y\) being around any point \(y\) in the interval. Since it's a density function, when you integrate it over the entire space, it should equal to 1. Thus:
  • The function maps each point \(y\) to its likelihood within the specified interval.
  • For points outside \( -1 \leq y \leq 1\), the likelihood is zero, given by \(f(y) = 0\).
Understanding the density function helps us to determine other aspects like cumulative distribution function and expected value, which give us better insights into the behavior of the random variable \(Y\).
Cumulative Distribution Function (CDF)
The Cumulative Distribution Function (CDF), denoted as \(F(y)\), is a function that describes the probability that your random variable \(Y\) will take a value less than or equal to \(y\).
As calculated in the solution, the CDF for the given density function is derived by integrating \(f(y)\) from the lower bound up to \(y\). This integration results in the function \(F(y) = \frac{\tan^{-1}(y)}{\pi} + \frac{1}{4}\), valid for \(-1 \leq y \leq 1\).
  • The CDF informs us how probabilities accumulate as \(y\) increases from the lower to the upper bound.
  • At \(y = -1\), the CDF value is 0, indicating no probability accumulated below \(y = -1\).
  • At \(y = 1\), the CDF equals 1, representing the total probability of 1 when all possibilities are considered.
Exponentially growing as you move from -1 to 1, the CDF is a helpful tool for understanding which values of \(Y\) are more probable as compared to others. It climbs smoothly as \(y\) increases, reflecting how the likelihood builds up.
Expected Value
The expected value, also known as the mean or average value, is a key concept in probability theory. It's like the center of gravity for the random variable representing its "average" outcome across many repetitions of a scenario.
For a continuous random variable like \(Y\), the expected value is computed by integrating the product of the variable and its density function over the range of \(Y\). In simpler terms, it is the sum of each possible value weighted by its probability.
In the exercise, the expected value \(E(Y)\) of the random variable \(Y\) is computed by:
  • integrating the expression \(y \cdot f(y)\) over its defined interval \([-1, 1]\).
  • Given the symmetry in the density function around zero, the expected value becomes 0 without detailed calculation.
This indicates that \(Y\), on average, balances out to zero, equally likely to be positive or negative within its given interval. The expected value provides insights into the distribution's "center point" and is useful for various statistical and engineering applications.

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Most popular questions from this chapter

The proportion of time per day that all checkout counters in a supermarket are busy is a random variable \(Y\) with density function $$f(y)=\left\\{\begin{array}{ll} c y^{2}(1-y)^{4}, & 0 \leq y \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Find the value of \(c\) that makes \(f(y)\) a probability density function. b. Find \(E(Y)\)

The duration \(Y\) of long-distance telephone calls (in minutes) monitored by a station is a random variable with the properties that $$P(Y=3)=.2 \quad \text { and } \quad P(Y=6)=.1$$ Otherwise, \(Y\) has a continuous density function given by $$f(y)=\left\\{\begin{array}{ll} (1 / 4) y e^{-y / 2}, & y>0 \\ 0, & \text { elsewhere } \end{array}\right.$$ The discrete points at 3 and 6 are due to the fact that the length of the call is announced to the caller in three-minute intervals and the caller must pay for three minutes even if he talks less than three minutes. Find the expected duration of a randomly selected long-distance call.

During an eight-hour shift, the proportion of time \(Y\) that a sheet-metal stamping machine is down for maintenance or repairs has a beta distribution with \(\alpha=1\) and \(\beta=2\). That is, $$f(y)=\left\\{\begin{array}{ll}2(1-y), & 0 \leq y \leq 1 \\\0, & \text { elsewhere }\end{array}\right.$$ The cost (in hundreds of dollars) of this downtime, due to lost production and cost of maintenance and repair, is given by \(C=10+20 Y+4 Y^{2} .\) Find the mean and variance of \(C\).

If a point is randomly located in an interval \((a, b)\) and if \(Y\) denotes the location of the point, then \(Y\) is assumed to have a uniform distribution over \((a, b) .\) A plant efficiency expert randomly selects a location along a 500 -foot assembly line from which to observe the work habits of the workers on the line. What is the probability that the point she selects is a. within 25 feet of the end of the line? b. within 25 feet of the beginning of the line? c. closer to the beginning of the line than to the end of the line?

A manufacturing plant uses a specific bulk product. The amount of product used in one day can be modeled by an exponential distribution with \(\beta=4\) (measurements in tons). Find the probability that the plant will use more than 4 tons on a given day.

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