/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 Let \(Y_{1}, Y_{2}, \ldots, Y_{n... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) be independent random variables such that each \(Y_{i}\) has a gamma distribution with parameters \(\alpha_{i}\) and \(\beta\). That is, the distributions of the \(Y\) 's might have different \(\alpha\) 's, but all have the same value for \(\beta\). Prove that \(U=Y_{1}+Y_{2}+\cdots+Y_{n}\) has a gamma distribution with parameters \(\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}\) and \(\beta\).

Short Answer

Expert verified
The sum \(U = Y_1 + Y_2 + \cdots + Y_n\) is gamma-distributed with parameters \(\alpha_1 + \alpha_2 + \cdots + \alpha_n\) and \(\beta\).

Step by step solution

01

Understanding the Problem

We have a series of independent random variables \(Y_1, Y_2, \ldots, Y_n\), each distributed according to a gamma distribution with parameters \(\alpha_i\) and \(\beta\). We need to show that their sum \(U = Y_1 + Y_2 + \cdots + Y_n\) is also a gamma-distributed random variable but with updated parameters.
02

Properties of Gamma Distribution

Recall that if a random variable \(Y_i\) follows a gamma distribution with parameters \(\alpha_i\) and \(\beta\), then its probability density function (pdf) is given by \( f_{Y_i}(y) = \frac{1}{\Gamma(\alpha_i) \beta^{\alpha_i}} y^{\alpha_i - 1} e^{-y/\beta} \), for \( y > 0 \).
03

Sum of Independent Gamma Variables

A key property of gamma distributions is that if \(Y_1, Y_2, \ldots, Y_n\) are independent gamma variables with the same rate parameter \(\beta\), their sum \(U = Y_1 + Y_2 + \cdots + Y_n\) is also a gamma-distributed variable. The shape parameter of \(U\) becomes \(\alpha_1 + \alpha_2 + \cdots + \alpha_n\), and the rate parameter remains the same \(\beta\).
04

Justification Using Moment Generating Functions

The moment-generating function (MGF) for a random variable with a gamma distribution \(\text{Gamma}(\alpha, \beta)\) is \( M_Y(t) = (1 - \beta t)^{-\alpha} \). The MGF of the sum \(U = Y_1 + Y_2 + \cdots + Y_n\) is the product of the individual MGFs: \(M_U(t) = (1 - \beta t)^{-(\alpha_1 + \alpha_2 + \cdots + \alpha_n)}\), which corresponds to a gamma distribution with parameters \(\alpha_1 + \alpha_2 + \cdots + \alpha_n\) and \(\beta\).
05

Conclusion

Since the MGF of the sum \(U\) matches that of a gamma distribution with parameters \(\alpha_1 + \alpha_2 + \cdots + \alpha_n\) and \(\beta\), we have shown that \(U\) is indeed gamma-distributed with these parameters.

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Key Concepts

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

Random Variables
In probability and statistics, the concept of random variables is fundamental to understanding distribution and stochastic processes. A random variable is essentially a variable whose values depend on the outcomes of a random phenomenon.

This can be confusing at first, but think of a random variable as "capturing" the randomness in a mathematical way. When you roll a die, the result - whether it's a 1 or a 6 - is random. If we set a random variable, say \(X\), to represent the roll of the die, then \(X\) could be any number from 1 to 6.

Random variables can be classified mainly into two types:
  • Discrete Random Variables: These assume distinct, separate values. For example, the roll of a die is discrete, as it can only be a whole number between 1 and 6.
  • Continuous Random Variables: These can assume any value within a given range or interval. For example, the exact height of a person can be considered continuous, as it is not limited to any specific set of numbers.
The gamma distribution, a type of continuous distribution, deals with continuous random variables, and is particularly useful in a variety of fields, including actuarial science and engineering.
Moment Generating Function
Moment Generating Functions (MGFs) are a powerful tool in probability theory, used to summarize all the moments of a random variable. Moments are quantitative measures related to the shape of the variable's distribution, and include things like the mean and variance.

The MGF of a random variable \(Y\), denoted as \(M_Y(t)\), encompasses these measures by means of a function. If a distribution has an MGF, it uniquely determines the distribution.

The formula for an MGF might be intimidating at first, but it is just a neat way to capture future information about the random variable. It's defined as:\[M_Y(t) = \mathbb{E}[e^{tY}]\]where \(\mathbb{E}\) denotes the expected value.
  • The MGF can be especially useful for finding the sum of independent random variables. If \(Y_1, Y_2, \ldots, Y_n\) are independent, then the MGF for their sum \(U = Y_1 + Y_2 + \cdots + Y_n\) is simply the product of their individual MGFs.
  • In the case of the gamma distribution with parameters \(\alpha\) and \(\beta\), the MGF is \((1 - \beta t)^{-\alpha}\). Applying this to a sum of gamma-distributed variables gives insight into how the distribution of their total, \(U\), behaves.
In essence, understanding MGFs allows us to prove properties of random variables, like why the sum of independent gamma variables is itself gamma-distributed.
Probability Density Function
A Probability Density Function (PDF) is vital for understanding the behavior of continuous random variables. The PDF describes the likelihood of a random variable falling within a particular range of values.

It gives us the "density" of the probability at any point on the distribution, though not the probability of any specific point itself. Instead, the probability of the variable falling within a certain interval is found by integrating the PDF over that interval.
  • The key property of the PDF is that it integrates to 1 over its entire range, ensuring it represents a valid probability distribution.
  • For a gamma-distributed random variable with parameters \(\alpha\) and \(\beta\), the PDF is given by:\[f_{Y}(y) = \frac{1}{\Gamma(\alpha) \beta^{\alpha}} y^{\alpha - 1} e^{-y/\beta},\quad y > 0\]where \(\Gamma(\alpha)\) is the gamma function. This function generalizes factorials to non-integer values of \(\alpha\).
The PDF helps in comprehending the properties of gamma distributions and how changing the parameters affects the shape and scale of the distribution. Deep diving into PDFs can unveil the magic behind predicting patterns and behaviors in varied scientific and engineering contexts.

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Most popular questions from this chapter

A member of the power family of distributions has a distribution function given by $$F(y)=\left\\{\begin{array}{ll} 0, & y<0 \\ \left(\frac{y}{\theta}\right)^{\alpha}, & 0 \leq y \leq \theta \\ 1, & y>\theta \end{array}\right.$$ where \(\alpha, \theta>0\) a. Find the density function. b. For fixed values of \(\alpha\) and \(\theta\), find a transformation \(G(U)\) so that \(G(U)\) has a distribution function of \(F\) when \(U\) possesses a uniform ( 0,1 ) distribution. c. Given that a random sample of size 5 from a uniform distribution on the interval (0,1) yielded the values \(.2700, .6901, .1413, .1523,\) and \(.3609,\) use the transformation derived in part (b) to give values associated with a random variable with a power family distribution with \(\alpha=2, \theta=4\)

Let \(Y_{1}\) and \(Y_{2}\) be independent, standard normal random variables. Find the probability density $$\text { function of } U=Y_{1} / Y_{2}$$

Let \(Y_{1}\) and \(Y_{2}\) be independent and uniformly distributed over the interval (0,1) . Find a. the probability density function of \(U_{1}=\min \left(Y_{1}, Y_{2}\right)\) b. \(E\left(U_{1}\right)\) and \(V\left(U_{1}\right)\)

Suppose that the number of occurrences of a certain event in time interval \((0, t)\) has a Poisson distribution. If we know that \(n\) such events have occurred in \((0, t),\) then the actual times, measured from \(0,\) for the occurrences of the event in question form an ordered set of random variables, which we denote by \(W_{(1)} \leq W_{(2)} \leq \cdots \leq W_{(n)} .\left[W_{(i)} \text { actually is the waiting time from } 0\) until the \right. occurrence of the \(i \text { th event. }]\) It can be shown that the joint density function for \(W_{(1)}, W_{(2)}, \ldots, W_{(n)}\) is given by $$f\left(w_{1}, w_{2}, \ldots, w_{n}\right)=\left\\{\begin{array}{ll} \frac{n !}{t^{n}}, & w_{1} \leq w_{2} \leq \cdots \leq w_{n} \\ 0, & \text { elsewhere } \end{array}\right.$$ [This is the density function for an ordered sample of size \(n\) from a uniform distribution on the interval (0,t).] Suppose that telephone calls coming into a switchboard follow a Poisson distribution with a mean of ten calls per minute. A slow period of two minutes' duration had only four calls. Find the a. probability that all four calls came in during the first minute; that is, find \(P(W_{(4)} \leq 1\) ). b. expected waiting time from the start of the two-minute period until the fourth call.

Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) be independent, uniformly distributed random variables on the interval \([0, \theta] .\) Find the a. probability distribution function of \(Y_{(n)}=\max \left(Y_{1}, Y_{2}, \ldots, Y_{n}\right)\) b. density function of \(Y_{(n)}\) c. mean and variance of \(Y_{(n)}\)

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