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Chapter 4: Problem 165

Let \(Y\) have density function $$ f(y)=\left\\{\begin{array}{ll} c y e^{-2 y}, & 0 \leq y \leq \infty \\ 0, & \text { elsewhere } \end{array}\right. $$ a. Find the value of \(c\) that makes \(f(y)\) a density function. b. Give the mean and variance for \(Y\). c. Give the moment-generating function for \(Y\).

Short Answer

Expert verified
a. \(c = 4\). b. Mean = 0.5, Variance = 0.25. c. \(M(t) = \frac{4}{(2-t)^2}\) for \(t < 2\).

Step by step solution

01

Check Normalization Condition

For a function to be a valid probability density function, the total area under the curve must be 1. That means the integral of the function from 0 to infinity must equal 1:\[\int_{0}^{\infty} c y e^{-2y} \, dy = 1.\]
02

Perform the Integration

Integrate the function using integration by parts. Let \( u = y \) and \( dv = c e^{-2y} \, dy \), then \( du = dy \) and \( v = -\frac{c}{2} e^{-2y} \):\[\int c y e^{-2y} \, dy = \left[-\frac{c}{2} y e^{-2y}\right]_0^\infty + \frac{c}{2} \int e^{-2y} \, dy.\]Solving the second integral and evaluating, we have:\[\int c y e^{-2y} \, dy = -\frac{c}{2} \cdot 0 + \frac{c}{2} \cdot \left[-\frac{1}{2} e^{-2y} \right]_0^\infty = \frac{c}{4}.\]
03

Solve for c

Set \( \frac{c}{4} = 1 \) and solve for \( c \):\[c = 4.\]
04

Determine Mean (Expectation)

The mean can be calculated by evaluating:\[E[Y] = \int_{0}^{\infty} y f(y) \, dy = \int_{0}^{\infty} 4y^2 e^{-2y} \, dy.\]Use integration by parts, choosing \( u = y^2 \) and \( dv = 4e^{-2y} \, dy \). After applying integration by parts, the mean is \(E[Y] = 0.5\).
05

Calculate Variance

Variance is given by \(Var(Y) = E[Y^2] - (E[Y])^2\). First, find \(E[Y^2]\):\[E[Y^2] = \int_{0}^{\infty} y^2 f(y) \, dy = \int_{0}^{\infty} 4y^3 e^{-2y} \, dy = 0.25.\]Therefore, \(Var(Y) = 0.25 - (0.5)^2 = 0.25 - 0.25 = 0.25.\)
06

Find Moment-Generating Function

The moment-generating function \(M(t)\) is calculated by:\[M(t) = \int_{0}^{\infty} e^{ty} f(y) \, dy = \int_{0}^{\infty} 4y e^{(t-2)y} \, dy.\] Using integration by parts or recognizing this as a modification of a known result, \(M(t) = \frac{4}{(2-t)^2}\) for \(t < 2\).

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Key Concepts

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

Normalization Condition
When dealing with probability density functions (PDFs), it's important to ensure they are properly normalized. This means the total probabilistic "mass" under the curve must equal one.
To check this, you integrate the PDF over all possible outcomes. For our function, the range is from 0 to infinity:
  • The integral of the PDF must be exactly one, which ensures it's a legitimate probability curve.
  • Mathematically, it means solving the equation: \[ \int_{0}^{\infty} c y e^{-2y} \, dy = 1. \]
By solving this integral, you identify the constant c that makes the function a proper PDF. This principle underpins much of probability theory since it assures all potential outcomes are accounted for.
Integration by Parts
Integration by parts is a mathematical technique used to solve complex integrals. It's particularly helpful for integrating products of functions.
The rule is based on the formula:\[\int u \, dv = uv - \int v \, du,\]where you choose parts of the integral to be u and dv. This simplifies the original integral into parts that are easier to tackle.
In our case:
  • u was chosen as y and dv as c e^{-2y} dy.
  • From here, du becomes dy and v solves to \(-\frac{c}{2} e^{-2y} \).
This transforms the problem, breaking it down step-by-step, to determine the normalization constant and calculate expectations.
Moment-Generating Function
The moment-generating function (MGF) is a powerful tool in probability that helps you understand the distribution of a random variable.
It is defined as:\[M(t) = E[e^{tY}] = \int_{0}^{\infty} e^{ty} f(y) \, dy,\]and encodes all moments (mean, variance, etc.) of a distribution within it. The MGF helps in determining and comparing different distributions.
  • The primary use is in finding moments like the mean by differentiating the MGF and evaluating at t = 0.
  • For our example, we obtained: \( M(t) = \frac{4}{(2-t)^2} \) for t less than 2, which simplifies understanding through polynomial expressions.
MGFs offer an elegant and efficient way to process probabilistic data.
Expectation and Variance
Expectation (or mean) and variance are critical statistical measures that describe a probability distribution.
Expectation gives the average outcome, calculated as:\[E[Y] = \int_{0}^{\infty} y f(y) \, dy.\]In our scenario, integration by parts was used to find this value, resulting in an expectation of 0.5.
Variance gives an idea of the distribution's spread or variability:\[Var(Y) = E[Y^2] - (E[Y])^2.\]Here, E[Y^2] was found to be 0.25 and since E[Y] = 0.5,
  • Var(Y) turned out to be 0.25, indicating moderate variability within this distribution.
Understanding both these measures helps a lot in characterizing the underlying randomness and behavior of variables.

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Most popular questions from this chapter

The Markov Inequality Let \(g(Y)\) be a function of the continuous random variable \(Y\), with \(E(l g(Y) |)<\infty .\) Show that, for every positive constant \(k.\) $$ P(|g(Y)| \leq k) \geq 1-\frac{E(|g(Y)|)}{k} $$ [Note: This inequality also holds for discrete random variables, with an obvious adaptation in the proof.]

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