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The following model is proposed for the distribution of family size in a large population:

P(k children in family;$\theta$) = $\theta^{k}$, for $k = 1, 2, ...$
P(0 children in family;$\theta$) = $\frac{1-2\theta}{1-\theta}$.

I tried to multiply them together and got the likelihood function
$L(\theta)=\frac{1-2\theta}{1-\theta}\prod_{k=1}^\infty \theta^{k}$

I don't think it is the right way to approach. Any help is appreciated.

Lawrence
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  • That is correct. It doesnt matter if the given distribution is partial or not. You will just multiply them. Then take the logarithm of it for convenience. Then take the derivative with respect to the parameter and in your case it is $\theta$, and make it equal to $0$ and evetually you will get something like this $\theta=f(k)$, which will be your MLE. I am however not sure if a closed form solution can be obtained in your case. one needs to check. – Seyhmus Güngören Mar 26 '17 at 16:24
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    The previous comment by Seyhmus is incorrect. First of all, the likelihood function depends on your data; suppose you observe $X_1,\dots,X_n$, iid from this distribution. WLOG, assume the first $m$ random variables have realization 0. Then the likelihood is the product of the individual probabilities: $$L(\theta) = \left( \frac{1-2\theta}{1-\theta} \right)^m \theta^{X_{m+1}} \cdots \theta^{X_n}.$$ Now you can take the logarithm and perform the standard ritual. – user217285 Mar 26 '17 at 16:32

1 Answers1

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Likelihood function is a function of samples $x_1,\ldots,x_n$ and of $\theta$.

$$ f(x_1,\ldots,x_n|\theta)=f(x_1|\theta)\times\dots\times f(x_n|\theta). $$ Here p.m.f. $f(x_i|\theta)$ is equal to $$ f(x_i|\theta) = \left(\frac{1-2\theta}{1-\theta}\right)^{1_{\{x_i=0\}}}\times \theta^{x_i}, $$ where $1_{\{x_i=0\}}=1$ if ${x_i=0}$ and zero otherwise.

Next, multiplying p.m.f. for different samples get $$ f(x_1,\ldots,x_n|\theta)= \prod_{i=1}^n\left(\frac{1-2\theta}{1-\theta}\right)^{1_{\{x_i=0\}}}\times \theta^{x_i}=\left(\frac{1-2\theta}{1-\theta}\right)^{S_0}\times \theta^{\sum_{i=1}^n x_i}, $$ where $S_0=\sum_{i=1}^n 1_{\{x_i=0\}}$ is the number of zeros in the samples.

Next, calculate logarithm and MLE.

NCh
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  • Thank you for your explanation. The data I got is 17 families have 0 children, 22 have 1, 7 have 2, 3 have 3, and 1 has 4. So the likelihood function I get becomes $L(\theta)=(\frac{1-2\theta}{1-\theta})^{17} \times \theta^{22\times1+7\times2+3\times3+1\times4}=(\frac{1-2\theta}{1-\theta})^{17} \times \theta^{49}$. Is it correct? – Lawrence Mar 26 '17 at 16:56
  • Yes, exactly. Unfortunately, further finding MLE requires quadratic equation solving. MLE will be $\hat\theta = \dfrac{3+\frac{17}{49}- \sqrt{\left(3+\frac{17}{49}\right)^2-8}}{4}$. – NCh Mar 26 '17 at 17:04
  • Again, thank you so much for your answer. I really appreciate it. – Lawrence Mar 26 '17 at 18:01
  • If you get an acceptable answer, you can "accept" the answer by clicking the check mark ✓ next to it. You can find out more about accepting answers here: https://math.meta.stackexchange.com/q/3399/413376 and https://math.meta.stackexchange.com/q/3286/413376: – NCh Mar 26 '17 at 18:09