
Advanced Statistics Advanced Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
May 10th, 2017, 12:19 PM  #1 
Senior Member Joined: Feb 2015 From: london Posts: 121 Thanks: 0  Likelihood function
Group A  women with no children Group B  women with children, where number of children is distributed with poisson with mean u Probabilities of being in group A and B P(A) = p P(B) = 1p $\displaystyle \phi =(p,u)$ $\displaystyle L(\phi;y) = [p + (1p)e^{u}]^{y_0} [\frac{(1p)e^{u} u^1}{1!}]^{y_1} [\frac{(1p)e^{u} u^2}{2!}]^{y_2} ...$ where $\displaystyle y_k$ denotes the number of women with k children Please can someone explain how this likelihood function is constructed. In the notes it says we can think of this problem as a mixture of a Bionomial and a poisson distribution. The thing that is really confusing me, is why is each section to the power of $\displaystyle y_k$ 
May 11th, 2017, 01:37 PM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,602 Thanks: 816 
I assume you got as far as $p_k = P[\text{a woman has k children}] = \begin{cases} p + e^{u} &k=0 \\ (1p)\dfrac{u^k e^{u}}{k!} &0 < k \end{cases}$ We sample a bunch of independently selected women and note the number of children they have. The joint probability of the number of children of all these women is just the product of all their individual distributions above. Now you coalesce these observations into a count of how many women have $k$ children. To try and save me some typing imagine that we have a product of different $k$s $p = k_1 k_0 k_1 k_2 k_3 k_0 k_1 k_4 k_3 \dots$ we can rewrite this product as $p = (k_0 k_0)(k_1 k_1 k_1)(k_2)(k_3 k_3)(k_4) \dots$ $p = \prod \limits_{j=0}^\infty k_j^{\text{# of women w/k children}}= \prod \limits_{j=0}^\infty k_j^{y_k}$ Now just replace $k_j$ with $p_k$ above $p = (p + e^{u})^{y_0} + \prod \limits_{k=1}^\infty ~\left((1p)\dfrac{u^k e^{u}}{k!}\right)^{y_k}$ and this is the formula you're given. 
May 12th, 2017, 11:05 AM  #3 
Senior Member Joined: Feb 2015 From: london Posts: 121 Thanks: 0 
Thanks romsek, that is a very clear explanation.


Tags 
function, likelihood, liklihood 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Loglikelihood function  bleh  Advanced Statistics  0  February 3rd, 2012 07:06 PM 
Likelihood calculation  AndiZ  Advanced Statistics  3  February 2nd, 2011 01:47 PM 
Log Likelihood~ how to get the value of ?i and ?j ?  ryusukekenji  Advanced Statistics  1  April 20th, 2009 04:08 PM 
Likelihood and Probability  a1d0ru  Algebra  3  October 15th, 2008 09:39 AM 
Marginalize likelihood  John_Smith  Advanced Statistics  0  February 6th, 2008 04:28 AM 