# Estimating unknown parameter exercise.

• May 14th 2011, 05:14 AM
goroner
Estimating unknown parameter exercise.
Hi i started solving exercises for estimating parameters and as i started already i stuck my self with this task:

Estimate the unknonw parameter \theta from the sample : 3,3,3,3,3,7,7,7
drawn from discrete distribution with pmf: P(3) = /theta and P(7)=1-/theta

use two methods : a)moments estimator b) maximum likelihood

-i have some ideas but i kinda dont know how to really apply those, first i thought of taking the first sample moment m1=mean(sample) which is m1=4.2 and make the identity m1=EX (EX is the first moment of the population, the actual discrete distribution that the sample is drawn from) but the problem is i dont know how to express the first moment so i can solve equition for /theta and estimate /theta,
i dont know how the info about the pmf is going to be in use for the problem i dont have idea for that one hmm

for the second method i know that i need to find derivative of the pmf and apply Logarithm and make identity of that to 0 so i can find the extreme points from where ill end up finding the parameter in such manner that the parameter gives the maximum probability that the observed data comes in the same state if i make take sample again but this time with the estimated parameter..

Thanks for helping!
• May 14th 2011, 08:43 AM
CaptainBlack
Quote:

Originally Posted by goroner
Hi i started solving exercises for estimating parameters and as i started already i stuck my self with this task:

Estimate the unknonw parameter \theta from the sample : 3,3,3,3,3,7,7,7
drawn from discrete distribution with pmf: P(3) = /theta and P(7)=1-/theta

use two methods : a)moments estimator b) maximum likelihood

-i have some ideas but i kinda dont know how to really apply those, first i thought of taking the first sample moment m1=mean(sample) which is m1=4.2 and make the identity m1=EX (EX is the first moment of the population, the actual discrete distribution that the sample is drawn from) but the problem is i dont know how to express the first moment so i can solve equition for /theta and estimate /theta,
i dont know how the info about the pmf is going to be in use for the problem i dont have idea for that one hmm

for the second method i know that i need to find derivative of the pmf and apply Logarithm and make identity of that to 0 so i can find the extreme points from where ill end up finding the parameter in such manner that the parameter gives the maximum probability that the observed data comes in the same state if i make take sample again but this time with the estimated parameter..

Thanks for helping!

The first moment is the mean, so:

$\mu|_{\theta}=3\theta + 7 (1-\theta)$

and you set this equal to the sample mean

$\mu|_{\theta}=3\theta + 7 (1-\theta)=\frac{5\times3+3 \times 7}{8}$

The likelihood of the data is (assuming that your sample is not ordered):

$p(data|\theta)=b(5;8,\theta)$

where $b(n;8,\theta)$ is the binomial mass function for a sample of size 8 with probability of a 3 being $\theta$. Now find the value of $\theta$ that maximises the likelihood.

CB
• May 14th 2011, 09:00 AM
goroner
hmm, is not clear to me how come the first moment of the population is m0=3*theta + 7(1-theta), cause the distribution is unknown i dont understand how did you come up with the formula, if you could explain it a bit please, thanks! (Speechless)
• May 14th 2011, 09:07 AM
CaptainBlack
Quote:

Originally Posted by goroner
hmm, is not clear to me how come the first moment of the population is m0=3*theta + 7(1-theta), cause the distribution is unknown i dont understand how did you come up with the formula, if you could explain it a bit please, thanks! (Speechless)

The distribution is known, it is only the parameter that is unknown, the number of 3's is a binomial random variable with probability of a 3 on a single trial of $\theta$. The population mean is as given That is a 3 occurs with probability $\theta$ and a 7 with probability $1-\theta$.

(and both methods give the same estimate for $\theta=5/8$)

CB