Problem 77
Question
Calculate \(E(Y)\) for the following pdfs: (a) \(f_{Y}(y)=3(1-y)^{2}, 0 \leq y \leq 1\) (b) \(f_{Y}(y)=4 y e^{-2 y}, y \geq 0\) (c) \(f_{Y}(y)= \begin{cases}\frac{3}{4}, & 0 \leq y \leq 1 \\ \frac{1}{4}, & 2 \leq y \leq 3 \\ 0, & \text { elsewhere }\end{cases}\) (d) \(f_{Y}(y)=\sin y, \quad 0 \leq y \leq \frac{\pi}{2}\)
Step-by-Step Solution
Verified Answer
(a) \(E(Y)\) is equal to \(\frac{1}{3}\) (b) \(E(Y)\) is equal to \(\frac{1}{2}\) (c) \(E(Y)\) is equal to \(\frac{3}{4}\) (d) \(E(Y)\) is equal to \(1\).
1Step 1: Calculating \(E(Y)\) for \(f_{Y}(y)=3(1-y)^2\) between \(0 \leq y \leq 1\)
Set up the integral to find the expected value over the range [0, 1]. \(E(Y) = \int_0^1 y * 3(1-y)^2 dy\). Evaluate this integral.
2Step 2: Calculating \(E(Y)\) for \(f_{Y}(y)=4 y e^{-2 y}\) for \(y \geq 0\)
Set up the integral to find the expected value over the range [0, infinity). \(E(Y) = \int_0^\infty y * 4 y e^{-2 y} dy\). Evaluate this integral.
3Step 3: Calculating \(E(Y)\) for piecewise function between \([0, 1]\) and \([2, 3]\)
Set up the integral for each piece of the function, multiply by the pdf and sum them. \(E(Y) = \int_0^1 y * \frac{3}{4} dy + \int_2^3 y * \frac{1}{4} dy\). Evaluate these integrals.
4Step 4: Calculating \(E(Y)\) for \(f_{Y}(y)=\sin y\) between \(0 \leq y \leq \frac{\pi}{2}\)
Set up the integral to find the expected value over the range [0, \(\frac{\pi}{2}\)]. \(E(Y) = \int_0^\frac{\pi}{2} y * \sin y dy\). Evaluate this integral.
Key Concepts
Probability Density FunctionMathematical StatisticsIntegral Evaluation
Probability Density Function
Understanding the probability density function (pdf) is critical when it comes to analyzing continuous random variables. A pdf provides a function that describes the likelihood of a random variable taking on a particular value. In essence, the pdf, which is denoted as fY(y), helps us find probabilities for intervals rather than specific points.
For a continuous random variable, the probability that it takes on a specific value is always zero. Instead, we look at the probability that the variable falls within a certain range. This is where the integral comes into play. The area under the curve of the pdf between two values gives us the probability that the random variable falls within that range.
Moreover, it's important to remember that a valid pdf must satisfy two conditions: it must be non-negative (fY(y) ≥ 0 for all y), and the total area under the pdf curve must be equal to 1, indicating that the probability of all possible outcomes sums to 1. When calculating expected values, we use the pdf to weigh each possible outcome by its respective probability.
For a continuous random variable, the probability that it takes on a specific value is always zero. Instead, we look at the probability that the variable falls within a certain range. This is where the integral comes into play. The area under the curve of the pdf between two values gives us the probability that the random variable falls within that range.
Moreover, it's important to remember that a valid pdf must satisfy two conditions: it must be non-negative (fY(y) ≥ 0 for all y), and the total area under the pdf curve must be equal to 1, indicating that the probability of all possible outcomes sums to 1. When calculating expected values, we use the pdf to weigh each possible outcome by its respective probability.
Mathematical Statistics
Mathematical statistics involves the application of probability theory to analyze and interpret data. One fundamental concept within this field is the expected value, commonly denoted as E(Y) for a random variable Y. This value provides a measure of the central tendency, or the 'average' outcome we would expect if we could repeat an experiment an infinite number of times.
To calculate the expected value of a continuous random variable, we take the integral of the product of the variable itself (y) and its pdf (fY(y)) over all possible values. The integral, in simple terms, adds up an infinite number of infinitesimally small products of the variable's value and the probability of that value, as given by the pdf.
In the context of the given exercises, E(Y) refers to the mean of the probability distribution created by the pdf. The interpretation of E(Y) is that it represents the long-run average outcome if we observed the random variable Y many times. This concept is key in fields from finance to engineering, where decision-making often depends on the expected outcomes.
To calculate the expected value of a continuous random variable, we take the integral of the product of the variable itself (y) and its pdf (fY(y)) over all possible values. The integral, in simple terms, adds up an infinite number of infinitesimally small products of the variable's value and the probability of that value, as given by the pdf.
In the context of the given exercises, E(Y) refers to the mean of the probability distribution created by the pdf. The interpretation of E(Y) is that it represents the long-run average outcome if we observed the random variable Y many times. This concept is key in fields from finance to engineering, where decision-making often depends on the expected outcomes.
Integral Evaluation
Integral evaluation is a mathematical process used to calculate areas under curves, total quantities, and in the context of probability, the expected value of a random variable. When faced with pdfs, as in the exercises provided, evaluating an integral allows for the precise calculation of expected values.
The integral of a function can be viewed as the accumulation of the area under the curve from one point to another. In the exercises, we are asked to perform this operation over specific ranges to find the average value the random variable Y will take. This requires a good grasp of integral calculus, as the type of function determines the method of integration.
Some functions, like polynomial expressions or exponential functions, have straightforward antiderivatives, while others may require special techniques like integration by parts, substitution, or even numerical methods for more complex expressions. The skillful application of these techniques is what allows us to find the expected value for different pdfs, highlighting the importance of integral evaluation in solving statistical problems.
The integral of a function can be viewed as the accumulation of the area under the curve from one point to another. In the exercises, we are asked to perform this operation over specific ranges to find the average value the random variable Y will take. This requires a good grasp of integral calculus, as the type of function determines the method of integration.
Some functions, like polynomial expressions or exponential functions, have straightforward antiderivatives, while others may require special techniques like integration by parts, substitution, or even numerical methods for more complex expressions. The skillful application of these techniques is what allows us to find the expected value for different pdfs, highlighting the importance of integral evaluation in solving statistical problems.
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