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Suppose that \(f_{Y}(y)\) is a continuous and sym. pdf, where symmetry is the property that \(f_{Y}(y)=f\) for all \(y\). Show that \(P(-a \leq Y \leq a)=2 F_{Y}(a)-1\).

Short Answer

Expert verified
\(P(-a \leq Y \leq a) = 2 F_Y(a) - 1\)

Step by step solution

01

Understand the givens

We are given that \(f_{Y}(y)\) is a symmetric pdf. Symmetry in this context means that \(f_{Y}(y) = f_{Y}(-y)\) for all real numbers \(y\). The goal is to show that the probability \(P(-a \leq Y \leq a) = 2 F_{Y}(a) - 1\)
02

Writing the expressions

Firstly, the Probability \(P(-a \leq Y \leq a)\) can be expressed as the integral of the pdf over the interval \([-a, a]\) as follows: \(P(-a \leq Y \leq a) = \int_{-a}^{a} f_{Y}(y) dy\). Secondly, The CDF \(F_{Y}(a)\) is defined as: \(F_{Y}(a) = \int_{-\infty}^{a} f_{Y}(y) dy\)
03

Using symmetry

Now, let's break the integral from \(-a\) to \(a\) into two parts: \(\int_{-a}^{a} f_{Y}(y) dy = \int_{-a}^{0} f_{Y}(y) dy + \int_{0}^{a} f_{Y}(y) dy\). Due to the symmetry of \(f_Y(y)\), we know that the integral from \(-a\) to \(0\) is the same as the integral from \(0\) to \(a\) i.e.: \(\int_{-a}^{0} f_{Y}(y) dy = \int_{0}^{a} f_{Y}(y) dy\). By substitution, we get: \(2 \int_{0}^{a} f_{Y}(y) dy\)
04

Substituting CDF expression

The double integral from step three can be substituted by \(2F_{Y}(a)\), as \(F_{Y}(a) = \int_{-\infty}^{a} f_{Y}(y) dy\), thus, \(2F_{Y}(a) = \int_{-\infty}^{a} f_{Y}(y) dy + \int_{-\infty}^{a} f_{Y}(y) dy = 2F_{Y}(a)\). So, we have: \(2 \int_{0}^{a} f_{Y}(y) dy = 2 F_Y(a)\)
05

Subtracting 1

Finally, subtract 1 from both sides to get the desired result: \(2F_Y(a) - 1 = P(-a \leq Y \leq a) - 1\). Thus, we have shown: \(P(-a \leq Y \leq a) = 2 F_Y(a) - 1\)

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Key Concepts

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

Continuous Probability Distributions
A continuous probability distribution is a way to describe outcomes for random variables that can take on a continuum of values. Unlike discrete probability distributions, where outcomes can be counted, continuous distributions involve any possible value within a certain range.
The probability of observing any single, exact value in a continuous distribution is zero. Instead, we look at the probability of the variable falling within a range of values.
To model these probabilities, we use a function called a probability density function (PDF), denoted usually by \( f_{Y}(y) \). The function describes how the probability is distributed over the possible values of the random variable. For example, a symmetric PDF means \( f_{Y}(y) = f_{Y}(-y) \) for all \( y \). This symmetry indicates that the distribution looks the same on both sides of its center.
Understanding continuous probability distributions is crucial because many real-world phenomena can be modeled this way, such as measurements involving height, temperature, or stock returns.
Cumulative Distribution Function
The cumulative distribution function, or CDF, is a key concept in understanding continuous probability distributions. It offers a way to calculate the probability that a random variable, say \( Y \), will take on a value less than or equal to a specific point.
Mathematically, the CDF is defined as \( F_{Y}(a) = \int_{-\infty}^{a} f_{Y}(y) \, dy \), where this integral sums up the area under the PDF curve from negative infinity to \( a \).
The importance of the CDF lies in its ability to give complete information about the probability distribution of \( Y \). Where the PDF might only tell you how dense the probabilities are at each point, the CDF accumulates these probabilities, providing a clearer picture on how likely it is for \( Y \) to be less than \( a \).
By using the property of a symmetric PDF, we can simplify calculations. For instance, to find the probability that \( Y \) lies between \( -a \) and \( a \), you determine it using the CDF as \( P(-a \leq Y \leq a) = 2F_{Y}(a) - 1 \). This formula emerges naturally from the symmetry of the distribution and the properties of the CDF.
Integral Calculus in Probability
Integral calculus is an essential tool in probability, especially when working with continuous distributions. Integrals help compute probabilities and other statistical measures, such as expected values.
In the world of continuous distributions, integration allows us to sum up probabilities over a range of values. It effectively replaces the summation seen in discrete probability. With continuous variables, a probability density function (PDF), such as \( f_{Y}(y) \), becomes the key element, where the integral of the PDF over an interval gives the probability of the variable falling within that interval.
For our scenario, we explore the integral \( \int_{-a}^{a} f_{Y}(y) \, dy \), which calculates the probability that \( Y \) lies between \( -a \) and \( a \). Thanks to the properties of symmetry, this probability simplifies to \( 2\int_{0}^{a} f_{Y}(y) \, dy \), emphasizing the role of symmetry and integrals working together.
Integral calculus doesn't stop there, as it also helps in constructing and understanding the cumulative distribution function (CDF). By taking the integral of a PDF, we construct the CDF, providing insights into the distribution's behavior over its range. All of this makes integral calculus indispensable for anyone tackling continuous probability distributions.

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