Results 1 to 10 of 10

Math Help - MLEs

  1. #1
    Newbie
    Joined
    Jun 2009
    From
    san diego
    Posts
    7

    Thumbs up MLEs

    Does anyone know how to find maximum likelihood estimates? I have a sample with 15 values but am not understanding exactly what I am supposed to be looking for with the maximum likelihood estimates. Thanks for any potential help. I haven't come across an explanation that would help me tackle this problem.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie drichie's Avatar
    Joined
    Aug 2008
    From
    india
    Posts
    11
    Quote Originally Posted by jerryc View Post
    Does anyone know how to find maximum likelihood estimates? I have a sample with 15 values but am not understanding exactly what I am supposed to be looking for with the maximum likelihood estimates. Thanks for any potential help. I haven't come across an explanation that would help me tackle this problem.
    What is your objective?.
    to Find MLE.. We assume a distribution first and for estimation of parameters we use MLE
    See this for more info.
    Maximum likelihood - Wikipedia, the free encyclopedia
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    First you need the likelihood function, which in the continuous case id the product of the densities via independence.
    In the discrete case you take the product of the distribution function.
    Then, as it's name says, you maximize wrt the parameter.
    In many cases that's just calculus.
    But in a case where the parameter is a boundary in your support, then calculus fails, but common sense prevails.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2009
    From
    san diego
    Posts
    7
    I am to find the MLE estimates for \theta_{1}=\mu and \theta_{2}=\sigma^{2} if a random sample of 15 from N(\mu,\sigma^{2}) has the values 31.5, 36.9, 33.8, 30.1, 33.9, 35.2, 29.6, 34.4, 30.5, 34.2, 31.6, 36.7, 35.8, 34.5, 32.7

    All I am coming up with is, for the mean,
    f(x_{k},\mu)=1/\sqrt{2\pi\sigma^{2}}*e^{-\frac{x_{k}-\mu)^{2}}{2\sigma^{2}}}

    L=f(x_{1},\mu)... f(x_{n},\mu)=(2\pi\sigma^{2})^{-\frac{n}{2}}e^{-\Sigma\frac{(x_{k}-\mu)^{2}}{2\sigma^{2}}}

    lnL=-\frac{n}{2};n(2\pi\sigma^{2})-\frac{1}{2\sigma^{2}}\Sigma(x_{k}-\mu)^{2}

    Partil derivative= \frac{1}{\sigma^{2}}\Sigma(x_{k}-\mu)

    0= \frac{1}{\sigma^{2}}\Sigma(x_{k}-\mu)

    \mu=\frac{\Sigma x_{k}}{n}

    And similar equations for the variance leading me to \sigma^{2}=\frac{\Sigma(x_{k}-\mu)^{2}}{n}

    The mean calculated from the sample is 33.4267 and the variance is 5.0979, but all these equations are straight out of a book from the library, so I'm not sure if they had to be tweaked or are standard. Also, I could find what the k stood for. It wasn't explained anywhere near that section. Thanks
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    The MLE of \mu is \bar X

    and the MLE of \sigma^2 is {\sum_{k=1}^n (X_k-\bar X)^2\over n}.
    Last edited by matheagle; June 14th 2009 at 08:29 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jun 2009
    From
    san diego
    Posts
    7
    It is really that simple? The MLE for the mean IS the mean, which is 33.4267, and the MLE for the variance IS the variance, which is 5.0979? I though I needed a whole slew of equations. Thanks
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by jerryc View Post
    It is really that simple? The MLE for the mean IS the mean, which is 33.4267, and the MLE for the variance IS the variance, which is 5.0979? I though I needed a whole slew of equations. Thanks
    Not exactly... The MLE for the mean is the empirical mean... the MLE for the variance is the empirical variance
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Jun 2009
    From
    san diego
    Posts
    7
    so...
    empirical mean =\frac{\Sigma X_{i}}{15}=\frac{501.4}{15}=33.4267

    empirical variance =\frac{\Sigma(X_{i}-EmpiricalMean)^{2}}{15}=\frac{76.4673}{15}=5.0979?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by jerryc View Post
    so...
    empirical mean =\frac{\Sigma X_{i}}{15}=\frac{501.4}{15}=33.4267

    empirical variance =\frac{\Sigma(X_{i}-EmpiricalMean)^{2}}{15}=\frac{76.4673}{15}=5.0979?
    I don't have your data, but I guess you're correct

    I just wanted to underline that the MLEs are the EMPIRICAL mean and variance, mean and variance issued from the data
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Jun 2009
    From
    san diego
    Posts
    7
    Quote Originally Posted by jerryc View Post
    I am to find the MLE estimates for \theta_{1}=\mu and \theta_{2}=\sigma^{2} if a random sample of 15 from N(\mu,\sigma^{2}) has the values 31.5, 36.9, 33.8, 30.1, 33.9, 35.2, 29.6, 34.4, 30.5, 34.2, 31.6, 36.7, 35.8, 34.5, 32.7
    My data is above. It seems to be the same formula as finding the sample mean and variance. That's why I wasn't sure. All the answers seem to be the same.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Asymptotic distributions of MLEs
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: March 17th 2011, 10:04 PM
  2. MLEs for a Geometric distribution
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: March 12th 2011, 07:58 AM
  3. Combining MLEs question
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: March 7th 2011, 08:03 AM

Search Tags


/mathhelpforum @mathhelpforum