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This is the third part of a question I am working on. So after solving for the parameter estimate $p$ by using the MLE method, I've been asked to plot the log likelihood function for X = 5, n = 10. To solve for the parameter estimate $p<hat>$. I had to find the likelihood function of the binomial distribution. I was under the impression that the likelihood function was defined as:

$$\Pi_{i = 1}^{k} f(X_{i}|p) = \Pi_{i = 1}^{k} \Bigg(\binom{n}{x}\Bigg)^k p^{x_i}(1-p)^{n-x_i} $$

But apparently all it is only: $$\binom{n}{x}p^{x_i}(1-p)^{n-x_i}$$

So assuming that it is the second form the solution I've found expresses the log liklihood function as:

$$ (p)^{\sum x_i} (1-p) ^{n - \sum x_i} = p ^{n} (1-p)^{n - n}$$

Two related questions:

i) Why is the second expression the "liklihood" function in this case compared to the usual definition?

ii) I'm not sure how the author was able to arrive at those expression in my third line. How did we get a summation of the $x_i$ if we are not using the original definition of the likelihood?

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If I am following correctly, I conclude that you have $N$ Binomial($n,p$) samples $x_i$ for given n and you want to estimate p by MLE. The likelihood will be of the form you wrote at first, but the binomial coefficients have no effect on the p dependence. So you can ignore that and focus on $p^{\sum_{i=1}^N x_i} (1-p)^{nN-\sum_{i=1}^N x_i}$. Or in other words $(p^{\overline{x}} (1-p)^{(n-\overline{x})})^N$. Then you can remove the power of $N$ leaving a routine problem.

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  • $\begingroup$ Everything you've explained to me makes sense in how I should be interpreting it. I have another question though...why is it $\sum_{i = 1}^{N} x_i$ ? I ask because each random variable is independent, but if $\sum_{i = 1}^{N} x_i = N$ That would mean that the mass functions are dependent on each other no? $\endgroup$ – dc3rd Jan 22 at 3:41
  • $\begingroup$ @dc3rd The summation just comes about from simplifying $\prod_{i=1}^n p^{x_i}$ and the analogous thing for $1-p$. And it could be any integer between $0$ and $nN$ (it's a property of the given sample). $\endgroup$ – Ian Jan 22 at 13:08

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