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91Ó°ÊÓ

Show that $$ E[Y]=\int_{0}^{\infty} P\\{Y>y\\} d y-\int_{0}^{x} P\\{Y<-y\\} d y $$ HINT: Show that $$ \begin{gathered} \int_{0}^{\pi} P[Y<-y\\} d y=-\int_{-\infty}^{0} x f_{Y}(x) d x \\ \int_{0}^{\infty} P\\{Y>y\\} d y=\int_{0}^{\infty} x f_{Y}(x) d x \end{gathered} $$

Short Answer

Expert verified
In summary, we have shown that \[ E[Y] = \int_{0}^{\infty} P\{Y>y\} dy-\int_{0}^{x} P\{Y<-y\} dy \] by following these steps: 1. Calculate the integral \(\int_{0}^{\infty} P\{Y>y\} dy\), and find that it is equal to \(\int_{0}^{\infty} yf_Y(y) dy\). 2. Show that \(\int_{0}^{\infty} P\{Y< -y\} dy = -\int_{-\infty}^{0} xf_Y(x) dx\). 3. Combine the results of steps 1 and 2, proving that the given expression represents the expected value of the random variable Y, \(E[Y]\).

Step by step solution

01

Calculate the first part of the hint

We are given the hint to calculate the integral from 0 to infinity of the probability that Y is greater than y. Let's find this integral. Given: \[ \int_{0}^{\infty} P\{Y>y\} dy \] Since \(P\{Y>y\} = 1 - F_Y(y)\) (probability complement): \[ \int_{0}^{\infty} P\{Y>y\} dy = \int_{0}^{\infty} (1 - F_Y(y)) dy \] Now we can integrate by parts, where \(u = 1-F_Y(y)\) and \(dv = dy\). Differentiating u and integrating dv, we get \(du = -f_Y(y)dy\) and \(v=y\). Applying integration by parts formula: \[ \int u\,dv = uv - \int v\,du \] We get: \[ \int_{0}^{\infty} P\{Y>y\} dy =\left. y(1-F_Y(y))\right|_0^{\infty} +\int_{0}^{\infty} yf_Y(y)dy \] Since \(F_Y(\infty) = 1\) and \(F_Y(0) = 0\), the first part of the expression goes to zero and we are left with: \[ \int_{0}^{\infty} P\{Y>y\} dy =\int_{0}^{\infty} yf_Y(y)dy \]
02

Calculate the second part of the hint

We are given the hint to calculate the integral from 0 to infinity of the probability that Y is less than -y. Let's find this integral. Given: \[ \int_{0}^{\infty} P\{Y< -y\} dy \] Differentiating both sides with respect to y, we have: \[ \frac{d}{dy}\int_{0}^{\infty} P\{Y < -y\}dy = \frac{d}{dy} \int_{0}^{\infty} (-x)f_Y(x) dx \] Since \(P\{Y < -y\} = F_Y(-y)\), the derivative with respect to y is just the density function, i.e. \(f_Y(-y)\): \[ \frac{d}{dy} \int_{-\infty}^{0} xf_Y(x)dx = -f_Y(-y) \] So, \(-\int_{-\infty}^{0} xf_Y(x) dx = \int_{0}^{\infty} P\{Y< -y\} dy\).
03

Combine the results of steps 1 and 2 to show the equivalence

Now, let's use the results from steps 1 and 2 to show the main exercise statement: \[ E[Y] = \int_{0}^{\infty} P\{Y>y\} dy-\int_{0}^{x} P\{Y<-y\} dy \] Using the results from the two hints, we get: \[ E[Y] = \int_{0}^{\infty} yf_Y(y) dy - \int_{-\infty}^{0} xf_Y(x) dx \] This is the expected value of the random variable Y, which confirms the validity of the given formula.

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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 probability and statistics. It's a function that describes the likelihood of a continuous random variable, such as a height or temperature.
It tells us how the probabilities are distributed over the values. For continuous variables, it’s important not to confuse probability with height of the PDF value, as probabilities are actually about areas under the curve.
- **Continuous Variables:** Unlike discrete variables, where probabilities add up over distinct outcomes, continuous probabilities are obtained by finding the area under the curve of the PDF.
- **Integration Use:** To get probabilities for continuous random variables between two points, we integrate the PDF over that interval. This process is similar to finding the area under a curve in regular calculus.
The PDF is non-negative everywhere, and the total area under the curve over all possible values of the random variable must equal 1. Hence, if you're given a PDF like \(f_Y(y)\), calculating probabilities for a specific range involves setting up an integral.
Integration by Parts
Integration by Parts is a method derived from the product rule for differentiation, and it’s useful for integrating products of functions. The formula is:
\[\int u \, dv = uv - \int v \, du\]
Here, \(u\) and \(dv\) are parts of your integrand that you choose based on what makes the problem easier.
Often, using this technique makes it more straightforward to tackle complex integrals, especially when dealing with probability and PDFs.- **Choosing \(u\) and \(dv\):** A common mnemonic is LIPET (Logarithms, Inverse Trig, Polynomials, Exponential, Trig) to help decide which part of the integrand to differentiate and which to integrate.
- **Application:** In probability problems, integration by parts helps transform expressions involving cumulative distribution functions into simpler integrals of the PDF times the variable, as seen in the exercise where the expectation was converted into an integral involving the PDF.
Understanding this tool can significantly streamline the process of finding expectations and other results when working with continuous distributions.
Cumulative Distribution Function
A Cumulative Distribution Function (CDF) denotes the probability that a random variable takes on a value less than or equal to a given point. It’s a non-decreasing function that ranges from 0 to 1.
For a continuous random variable:\(F_Y(y) = P(Y \leq y)\)
This describes the accumulation of probabilities up to a certain point, giving the 'running total' of probability.
- **Properties of CDF:** The function starts at 0 (not exceeding any value) and approaches 1 as it encompasses all possibilities of the random variable. It allows for finding the probability of a random variable lying in an interval \([a, b]\) by evaluating \(F_Y(b) - F_Y(a)\).
- **Relation to PDF:** The PDF is the derivative of the CDF with respect to y. Thus, one can find the CDF by integrating the PDF over the desired range.
In exercises like the given problem, understanding how to use the CDF is crucial, especially as it relates to converting probabilities involving cumulative probabilities into expectations or other statistical properties.

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