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May 10th, 2017, 11:19 AM   #1
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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) = 1-p

$\displaystyle \phi =(p,u)$
$\displaystyle L(\phi;y) = [p + (1-p)e^{-u}]^{y_0} [\frac{(1-p)e^{-u} u^1}{1!}]^{y_1} [\frac{(1-p)e^{-u} u^2}{2!}]^{y_2} ...$

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 $\displaystyle y_k$
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May 11th, 2017, 12:37 PM   #2
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I assume you got as far as

$p_k = P[\text{a woman has k children}] = \begin{cases}

p + e^{-u} &k=0 \\

(1-p)\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

$p = (p + e^{-u})^{y_0} + \prod \limits_{k=1}^\infty ~\left((1-p)\dfrac{u^k e^{-u}}{k!}\right)^{y_k}$

and this is the formula you're given.
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May 12th, 2017, 10:05 AM   #3
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Thanks romsek, that is a very clear explanation.
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