What is Maximum likelihood: Definition and 64 Discussions
In statistics, maximum likelihood estimation (MLE) is a method of estimating the parameters of a probability distribution by maximizing a likelihood function, so that under the assumed statistical model the observed data is most probable. The point in the parameter space that maximizes the likelihood function is called the maximum likelihood estimate. The logic of maximum likelihood is both intuitive and flexible, and as such the method has become a dominant means of statistical inference.If the likelihood function is differentiable, the derivative test for determining maxima can be applied. In some cases, the first-order conditions of the likelihood function can be solved explicitly; for instance, the ordinary least squares estimator maximizes the likelihood of the linear regression model. Under most circumstances, however, numerical methods will be necessary to find the maximum of the likelihood function.
From the vantage point of Bayesian inference, MLE is a special case of maximum a posteriori estimation (MAP) that assumes a uniform prior distribution of the parameters. In frequentist inference, MLE is a special case of an extremum estimator, with the objective function being the likelihood.
Homework Statement
Let Y1<Y2<...<Yn be the order statistics of a random sample from a distribution with pdf f(x; \theta) = 1, \theta - 0.5 < x < \theta + 0.5. Show that every statistic u(X1,X2,...,Xn) such that Y_n - 0.5<u(X_1,X_2,...,X_n)<Y_1 + 0.5 is a mle of theta. In particular (4Y_1 +...
Homework Statement
Suppose X1...Xn are iid and have PDF f(x; \theta) = \frac{1}{\theta} e^{\frac{-x}{\theta}} \ \ \ 0<x<\infty
Find the MLE of P(X<2).
Homework Equations
The Attempt at a Solution
I know the MLE of theta is \overline{X}
so would P(X<2) = 1 -...
L(x_1,...,x_n;p)=\Pi_{i=1}^{n}(\stackrel{n}{x_i}) p^{x_i}(1-p)^{n-x_i}
Correct so far?
The solution tells me to skip the \Pi:
L(x_1,...,x_n;p)=(\stackrel{n}{x}) p^{x}(1-p)^{n-x}
This is contradictory to all the examples in my book. Why?
Homework Statement
pdf: f(x)=ax^(a-1) ; 0<x<1, a>0
estimate a by maximum likelihood
Homework Equations
let L be maximum likelihood
L=(a(x[1])^(a-1))(a(x[2])^(a-1))...(a(x[n])^(a-1))
The Attempt at a Solution
Im trying to make this into a nicer expression:
L=a^n... (now I am...
Hi,
I'm posting this in this particular forum because, though this's a statistics question, my application is in high energy.
My question is regarding a problem in Bevington's book (Data Reduction and Error Analysis..., Page 193, Ex. 10.1), but I'll give a general description here...
Say...
Homework Statement
Let's have random value X defined by its density function:
f(x; \beta) = \beta^2x \mbox{e}^{-\beta x}
where \beta > 0 for x > 0 and f(x) = 0 otherwise.
Expected value of X is EX = \frac{2}{\beta} and variance is \mbox{var } X = \frac{2}{\beta^2}.
Next...
Homework Statement
Suppose X has a Poisson distribution with parameter lambda. Given a random sample of n observations,
Find the MLE of lambda, and hat lambda.
Find the expected value and variance of hat lambda.
Show that hat lambda is a consistent estimator of lambda.
Homework...
in need of help for how to do this question
given probability mass function:
x 1 2 3 4
p(x) 1/4(θ+2) 1/4(θ) 1/4(1-θ) 1/4(1-θ)
Marbles
1=green
2=blue
3=red
4=white
For 3839 randomly picked marbles
green=1997
blue=32
red=906...
http://en.wikipedia.org/wiki/Maximum_likelihood
What exactly does the "arg" here mean? It seems to be an unnecessary - the max L(\theta) seems to be sufficient enough. Or am I missing something?
\widehat{\theta} = \underset{\theta}{\operatorname{arg\ max}}\ \mathcal{L}(\theta).
Hi,
I'm taking a basic course in statistical methods, and we recently learned of maximum likelihood estimation. We defined the likelihood as a function of some parameter 'a', and found the estimator of 'a' by requiring a maximum likelihood with respect to it.
As an example, we took the...
How do I estimate the standart deviation for the mean average of an poisson-distribution ?
The mean average was estimated with the maximum-likelihood method by graphing the likelihood in dependence of the mean average, then just reading off the value for which the likelihood became maximal.
Up...
Maximum likelihood estimator...
ok, I'm stil a bit lost...so tell me if this is right:
f_y(y;\theta) = \frac{2y}{\theta^2}, for 0 < y < \theta
find the MLE estimator for theta.
L(\theta) = 2yn\theta^{-2 \sum_1^n y_i .
is this even right to begin with?
then take the natural...