Difference between revisions of "MLE Example OldKiwi" - Rhea

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-**Exponential Distribution**
-Let <img alt="tex:{X}_{1}, {X}_{2}, {X}_{3}.....{X}_{n}" /> be a random sample from the exponential distribution with p.d.f.
-<img alt="tex:f(x;\theta)=\frac{1}{\theta}{e}^{\frac{-x}{\theta}} 0<x<\infty, \theta\in\Omega=\{\theta|0<\theta<\infty\}" />
-The likelihood function is given by:
-<img alt="tex:L(\theta)=L\left(\theta;{x}_{1},{x}_{2}...{x}_{n} \right)=\left(\frac{1}{\theta}{e}^{\frac{{-x}_{1}}{\theta}}\right)\left(\frac{1}{\theta}{e}^{\frac{{-x}_{2}}{\theta}}\right)...\left(\frac{1}{\theta}{e}^{\frac{{-x}_{n}}{\theta}} \right)=\frac{1}{{\theta}^{n}}exp\left(\frac{-\sum_{1}^{n}{x}_{i}}{\theta} \right)" />
-Taking log, we get,
-<img alt="tex:lnL\left(\theta\right)=-\left(n \right)ln\left(\theta\right)
--\frac{1}{\theta}\sum_{1}^{n}{x}_{i}, 0<\theta<\infty" />
-Differentiating the above expression, and equating to zero, we get
-<img alt="tex:\frac{d\left[lnL\left(\theta\right) \right]}{d\theta}=\frac{-\left(n \right)}{\left(\theta\right)}
-+\frac{1}{{\theta}^{2}}\sum_{1}^{n}{x}_{i}=0" />
-The solution of equation for <img alt="tex:\theta" /> is:
-<img alt="tex:\theta=\frac{\sum_{1}^{n}{x}_{i}}{n}" />
-Thus, the maximum likelihood estimator of <img alt="tex:\Theta" /> is
-<img alt="tex:\Theta=\frac{\sum_{1}^{n}{X}_{i}}{n}" />
-**Geometric Distribution**
-Let <img alt="tex:{X}_{1}, {X}_{2}, {X}_{3}.....{X}_{n}" /> be a random sample from the geometric distribution with p.d.f.
-<img alt="tex:f\left(x;p \right)={\left(1-p \right)}^{x-1}p, x=1,2,3...." />
-The likelihood function is given by:
-<img alt="tex:L\left(p \right)={\left(1-p \right)}^{{x}_{1}-1}p
-{\left(1-p \right)}^{{x}_{2}-1}p...{\left(1-p \right)}^{{x}_{n}-1}p
-={p}^{n}{\left(1-p \right)}^{\sum_{1}^{n}{x}_{i}-n}" />
-Taking log,
-<img alt="tex:lnL\left(p \right)=
-nln{p}+\left(\sum_{1}^{n}{x}_{i}-n \right)ln{\left(1-p \right)} " />
-Differentiating and equating to zero, we get,
-<img alt="tex:\frac{d\left[lnL\left(p \right)\right]}{dp}=\frac{n}{p}
--\frac{\left(\sum_{1}^{n}{x}_{i}-n \right)}{\left(1-p \right)}=0" />
-Therefore,
-<img alt="tex:p=\frac{n}{\left(\sum_{1}^{n}{x}_{i} \right)}" />
-So, the maximum likelihood estimator of P is:
-<img alt="tex:P=\frac{n}{\left(\sum_{1}^{n}{X}_{i} \right)}=\frac{1}{X}" />
-This agrees with the intuition because, in n observations of a geometric random variable, there are n successes in the <img alt="tex:\sum_{1}^{n}{X}_{i}" /> trials. Thus the estimate of p is the number of successes divided by the total number of trials.

Difference between revisions of "MLE Example OldKiwi" - Rhea

Latest revision as of 13:40, 16 March 2008

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