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Homelikelihood function

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# likelihood function

Let X=($X_{1},\ldots,X_{n}$) be a random vector and

$\{f_{{\mathbf{X}}}(\boldsymbol{x}\mid\boldsymbol{\theta}):\boldsymbol{\theta}% \in\Theta\}$ |

a statistical model parametrized by $\boldsymbol{\theta}=(\theta_{1},\ldots,\theta_{k})$, the parameter vector in the *parameter space* $\Theta$. The *likelihood function* is a map $L:\Theta\to\mathbb{R}$ given by

$L(\boldsymbol{\theta}\mid\boldsymbol{x})=f_{{\mathbf{X}}}(\boldsymbol{x}\mid% \boldsymbol{\theta}).$ |

In other words, the likelikhood function is functionally the same in form as a probability density function. However, the emphasis is changed from the $\boldsymbol{x}$ to the $\boldsymbol{\theta}$. The pdf is a function of the $x$’s while holding the parameters $\theta$’s constant, $L$ is a function of the parameters $\theta$’s, while holding the $x$’s constant.

When there is no confusion, $L(\boldsymbol{\theta}\mid\boldsymbol{x})$ is abbreviated to be $L(\boldsymbol{\theta})$.

The parameter vector $\hat{\boldsymbol{\theta}}$ such that $L(\hat{\boldsymbol{\theta}})\geq L(\boldsymbol{\theta})$ for all $\boldsymbol{\theta}\in\Theta$ is called a *maximum likelihood estimate*, or *MLE*, of $\boldsymbol{\theta}$.

Many of the density functions are exponential in nature, it is therefore easier to compute the MLE of a likelihood function $L$ by finding the maximum of the natural log of $L$, known as the log-likelihood function:

$\ell(\boldsymbol{\theta}\mid\boldsymbol{x})=\operatorname{ln}(L(\boldsymbol{% \theta}\mid\boldsymbol{x}))$ |

due to the monotonicity of the log function.

Examples:

1. A coin is tossed $n$ times and $m$ heads are observed. Assume that the probability of a head after one toss is $\pi$. What is the MLE of $\pi$?

*Solution*: Define the outcome of a toss be 0 if a tail is observed and 1 if a head is observed. Next, let $X_{i}$ be the outcome of the $i$th toss. For any single toss, the density function is $\pi^{x}(1-\pi)^{{1-x}}$ where $x\in\{0,1\}$. Assume that the tosses are independent events, then the joint probability density is$f_{{\mathbf{X}}}(\boldsymbol{x}\mid\pi)=\binom{n}{\Sigma x_{i}}\pi^{{\Sigma x_% {i}}}(1-\pi)^{{\Sigma(1-x_{i})}}=\binom{n}{m}\pi^{m}(1-\pi)^{{n-m}},$ which is also the likelihood function $L(\pi)$. Therefore, the log-likelihood function has the form

$\ell(\pi\mid\boldsymbol{x})=\ell(\pi)=\operatorname{ln}\binom{n}{m}+m% \operatorname{ln}(\pi)+(n-m)\operatorname{ln}(1-\pi).$ Using standard calculus, we get that the MLE of $\pi$ is

$\hat{\pi}=\frac{m}{n}=\overline{x}.$ 2. Suppose a sample of $n$ data points $X_{i}$ are collected. Assume that the $X_{i}\sim N(\mu,\sigma^{2})$ and the $X_{i}$’s are independent of each other. What is the MLE of the parameter vector $\boldsymbol{\theta}=(\mu,\sigma^{2})$?

*Solution*: The joint pdf of the $X_{i}$, and hence the likelihood function, is$L(\boldsymbol{\theta}\mid\boldsymbol{x})=\frac{1}{\sigma^{n}(2\pi)^{{n/2}}}% \operatorname{exp}(-\frac{\Sigma(x_{i}-\mu)^{2}}{2\sigma^{2}}).$ The log-likelihood function is

$\ell(\boldsymbol{\theta}\mid\boldsymbol{x})=-\frac{\Sigma(x_{i}-\mu)^{2}}{2% \sigma^{2}}-\frac{n}{2}\operatorname{ln}(\sigma^{2})-\frac{n}{2}\operatorname{% ln}(2\pi).$ Taking the first derivative (gradient), we get

$\frac{\partial\ell}{\partial\boldsymbol{\theta}}=(\frac{\Sigma(x_{i}-\mu)}{% \sigma^{2}},\frac{\Sigma(x_{i}-\mu)^{2}}{2\sigma^{4}}-\frac{n}{2\sigma^{2}}).$ Setting

$\frac{\partial\ell}{\partial\boldsymbol{\theta}}=\boldsymbol{0}\mbox{ See % score function}$ and solve for $\boldsymbol{\theta}=(\mu,\sigma^{2})$ we have

$\boldsymbol{\hat{\theta}}=(\hat{\mu},\hat{\sigma}^{2})=(\overline{x},\frac{n-1% }{n}s^{2}),$ where $\overline{x}=\Sigma x_{i}/n$ is the sample mean and $s^{2}=\Sigma(x_{i}-\overline{x})^{2}/(n-1)$ is the sample variance. Finally, we verify that $\hat{\boldsymbol{\theta}}$ is indeed the MLE of $\boldsymbol{\theta}$ by checking the negativity of the 2nd derivatives (for each parameter).

## Mathematics Subject Classification

62A01*no label found*

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