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Chapter 4: Problem 2

Find the likelihood for a random sample \(y_{1}, \ldots, y_{n}\) from the geometric density \(\operatorname{Pr}(Y=y)=\pi(1-\pi)^{y}, y=0,1, \ldots\), where \(0<\pi<1\)

Short Answer

Expert verified
The likelihood is \( L(\pi; y_1, \ldots, y_n) = \pi^n (1-\pi)^{\sum_{i=1}^{n} y_i} \).

Step by step solution

01

Understand the Geometric Distribution

The geometric distribution models the number of trials needed to get the first success in a series of independent Bernoulli trials. The probability mass function is \( \operatorname{Pr}(Y=y)=\pi(1-\pi)^{y} \), where \( \pi \) is the probability of success on each trial, and \( y \) is the number of failures before the first success.
02

Write the Joint Probability for the Sample

Given a random sample \( y_1, y_2, \ldots, y_n \) from a geometric distribution, the joint probability (likelihood) of the sample is the product of the individual probabilities: \( L(\pi; y_1, \ldots, y_n) = \prod_{i=1}^{n} \pi (1-\pi)^{y_i} \).
03

Simplify the Likelihood Expression

The likelihood function can be rewritten using properties of exponents: \[ L(\pi; y_1, \ldots, y_n) = \pi^n (1-\pi)^{\sum_{i=1}^{n} y_i} \]. Here, \( \pi^n \) comes from multiplying \( \pi \) by itself \( n \) times, and the exponent of \( 1-\pi \) is the sum of all \( y_i \).

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Key Concepts

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

Geometric Distribution
The geometric distribution is a fundamental concept in probability and statistics. It is used to model the number of trials required to achieve the first success in a sequence of independent and identically distributed Bernoulli trials. Imagine a game where you flip a coin until you get heads for the first time. The number of flips (trials) before you get a head is described by a geometric distribution.

Some key features of the geometric distribution include:
  • The trials are independent, meaning each flip does not affect the outcome of the others.
  • The probability of success, denoted by \( \pi \), remains constant for each trial.
  • The outcome of each trial is binary鈥攅ither a success (like getting heads on a coin flip) or a failure (like getting tails).
Overall, the geometric distribution is a pivotal model for understanding processes involving repeated trials until success.
Probability Mass Function
Each probability distribution has a function that describes the likelihood of different outcomes. For the geometric distribution, this is called the Probability Mass Function (PMF). The PMF gives the probability that the random variable \( Y \), representing the number of failures before the first success, is equal to a specific value \( y \).

The formula is:
  • \( \operatorname{Pr}(Y=y)=\pi(1-\pi)^{y} \)
This formula breaks down into:
  • \( \pi \) - the probability of success on an individual trial.
  • \((1-\pi)^{y} \) - the probability of failing \( y \) times before a success happens.
By using the PMF, one can calculate how likely specific sequences of failures and a first success are, giving detailed insight into the structure of the distribution.
Bernoulli Trials
A Bernoulli trial is a simple experiment with only two possible outcomes: success or failure. Each trial acts independently of the previous ones, which is crucial for accurately modeling processes through probability distributions like the geometric distribution.

Here's what makes Bernoulli trials important:
  • They provide a simple yet powerful way to model random processes with binary outcomes.
  • They allow for the derivation of various distributions, such as the geometric distribution and binomial distribution.
  • They enable the understanding of the success probability \( \pi \) across multiple experiments.
The focus in Bernoulli trials is maintaining independence and constant probability of success, which ensures that models based on these trials accurately reflect the random processes involved.
Likelihood Expression Simplification
When dealing with a sample from a geometric distribution, the likelihood function represents the joint probability of the observed data. By itself, this likelihood function can be quite complex. However, simplification using algebraic properties makes it more manageable.

Given the likelihood function for a geometric distribution:
  • \( L(\pi; y_1, \ldots, y_n) = \prod_{i=1}^{n} \pi (1-\pi)^{y_i} \)
This can be simplified into:
  • \( L(\pi; y_1, \ldots, y_n) = \pi^n (1-\pi)^{\sum_{i=1}^{n} y_i} \)
How this simplification works:
  • Multiply \( \pi \) for each sample, resulting in \( \pi^n \).
  • Sum up all the failures \( y_i \) to get the exponent of \( 1-\pi \).
Simplifying the likelihood expression in this manner makes it easier to work with through further statistical analysis and understanding the model parameters.

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Most popular questions from this chapter

Find maximum likelihood estimates for \(\theta\) based on a random sample of size \(n\) from the densities (i) \(\theta y^{\theta-1}, 00 ;\) (ii) \(\theta^{2} y e^{-\theta y}, y>0, \theta>0 ;\) and (iii) \((\theta+1) y^{-\theta-2}\), \(y>1, \theta>0\)

If \(Y_{1}, \ldots, Y_{n} \stackrel{\text { iid }}{\sim} N\left(\mu, c \mu^{2}\right)\), where \(c\) is a known constant, show that the minimal sufficient statistic for \(\mu\) is the same as for the \(N\left(\mu, \sigma^{2}\right)\) distribution. Find the maximum likelihood estimate of \(\mu\) and give its large-sample standard error. Show that the distribution of \(\bar{Y}^{2} / S^{2}\) does not depend on \(\mu\).

A location-scale model with parameters \(\mu\) and \(\sigma\) has density $$ f(y ; \mu, \sigma)=\frac{1}{\sigma} g\left(\frac{y-\mu}{\sigma}\right), \quad-\infty0 $$ (a) Show that the information in a single observation has form $$ i(\mu, \sigma)=\sigma^{-2}\left(\begin{array}{ll} a & b \\ b & c \end{array}\right) $$ and express \(a, b\), and \(c\) in terms of \(h(\cdot)=\log g(\cdot) .\) Show that \(b=0\) if \(g\) is symmetric about zero, and discuss the implications for the joint distribution of the maximum likelihood estimators \(\widehat{\mu}\) and \(\widehat{\sigma}\) when \(g\) is regular. (b) Find \(a, b\), and \(c\) for the normal density \((2 \pi)^{-1 / 2} e^{-u^{2} / 2}\) and the log-gamma density \(\exp \left(\kappa u-e^{u}\right) / \Gamma(\kappa)\), where \(\kappa>0\) is known.

Suppose a random sample \(Y_{1}, \ldots, Y_{n}\) from the exponential density is rounded down to the nearest \(\delta\), giving \(\delta Z_{j}\), where \(Z_{j}=\left\lfloor Y_{j} / \delta\right\rfloor .\) Show that the likelihood contribution from a rounded observation can be written \(\left(1-e^{-\lambda \delta}\right) e^{-Z \lambda \delta}\), and deduce that the expected information for \(\lambda\) based on the entire sample is \(n \delta^{2} \exp (-\lambda \delta)\\{1-\exp (-\lambda \delta)\\}^{-2}\). Show that this has limit \(n / \lambda^{2}\) as \(\delta \rightarrow 0\), and that if \(\lambda=1\), the loss of information when data are rounded down to the nearest integer rather than recorded exactly, is less than \(10 \%\). Find the loss of information when \(\delta=0.1\), and comment briefly.

Counts \(y_{1}, y_{2}, y_{3}\) are observed from a multinomial density $$ \operatorname{Pr}\left(Y_{1}=y_{1}, Y_{2}=y_{2}, Y_{3}=y_{3}\right)=\frac{m !}{y_{1} ! y_{2} ! y_{3} !} \pi_{1}^{y_{1}} \pi_{2}^{y_{2}} \pi_{3}^{y_{3}}, y_{r}=0, \ldots, m, \sum y_{r}=m $$ where \(0<\pi_{1}, \pi_{2}, \pi_{3}<1\) and \(\pi_{1}+\pi_{2}+\pi_{3}=1\). Show that the maximum likelihood estimate of \(\pi_{r}\) is \(y_{r} / m\). It is suspected that in fact \(\pi_{1}=\pi_{2}=\pi\), say, where \(0<\pi<1\). Show that the maximum likelihood estimate of \(\pi\) is then \(\frac{1}{2}\left(y_{1}+y_{2}\right) / m\). Give the likelihood ratio statistic for comparing the models, and state its asymptotic distribution.
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