# Deriving likelihood function of binomial distribution, confusion over exponents

This question focuses on a specific aspect of this one: How to derive the likelihood function for binomial distribution for parameter estimation?

In my own derivation, I start with: $$f(x\mid p) = mC_x~p^x(1-p)^{m-x}$$

Ignoring $$mC_x$$, the likelihood function is then given by:

$$L(p) = \prod_{i=1}^np^{x_i}(1-p)^{m-x_i} = p^{\sum_1^n x_i}(1-p)^{\sum_1^n m-x_i} = p^{x}(1-p)^{nm-x}$$

However, in the question I referenced, they have this instead: $$\prod_{i=1}^np^{x_i}(1-p)^{1-x_i} = p^{\sum_1^n x_i}(1-p)^{\sum_1^n1-x_i} = p^{x}(1-p)^{n-x}$$

My question is, are both approaches correct? If so, why does the referenced question use $$1$$ in place of $$m$$ on the exponents?

It looks as if you intended $$X_1,\ldots,X_n \sim \operatorname{i{.}i{.}d{.}} \operatorname{Binomial}(m,p).$$ Then you have $$L(p) \propto \prod_{i=1}^n p^{x_i} (1-p)^{m-x_i} = p^{\sum_{i=1}^n x_i} (1-p)^{nm - \sum_{i=1}^n x_i} = p^x (1-p)^{nm-x}.$$ It appears that in the question that you "referenced" (I don't know what "referenced" means in this context, but that doesn't appear to matter.) one has $$X_1,\ldots,X_n \sim \operatorname{i{.}i{.}d{.}} \operatorname{Bernoulli}(p)$$, so that $$X_1+\cdots+X_n \sim \operatorname{Binomial} (n,p).$$ That leads to $$L(p) \propto p^{\sum_{i=1}^n x_i} (1-p)^{n - \sum_{i=1}^n x_i} = p^x(1-p)^{n-x}.$$ Therefore both are right, but they're answers to different questions.

The question you are referencing is starting with a Bernoulli distribution. To be sure, $$x$$ only takes on 0 or 1 in that question. In your work, you are starting with a Binomial distribution. To be sure, your values of $$x=0, 1, 2, ... m$$.

Remember that the sum of $$n$$ independent $$Bernoulli(p)$$ variables is a $$Binomial(n, p)$$ distribution. This should account for the differences you are seeing.

Your derivation for the likelihood of a binomial is just fine, ignoring the $$mCx$$ term, but you shouldn't ignore it. You can treat it as ignorable for the purposes of calculating the likelihood function since the likelihood function is a function only of the parameter $$p$$ and $$p$$ does not show up in $$mCx$$.

• Are you using "To be sure, [...]" to mean "You can be sure of this because [...]"? (I ask because "To be sure, [...]" means something quite different in English, which makes your usage very confusing. If you didn't intend the idiomatic meaning, I recommend finding a different phrase to express what you meant.) – ruakh Dec 24 '18 at 4:07
• @ruakh, indeed the intention was idiomatic. It is correctly used to mean "certainly, undoubtedly, admittedly." I will keep it. – StatsStudent Dec 26 '18 at 17:27
• But "admittedly" makes no sense in this context. You must be misunderstanding the idiom. :-/ – ruakh Dec 26 '18 at 18:02
• No, I have a perfect understanding. I'm using the idiomatic meaning of "certainly" or "undoubtedly" - not "admittedly." – StatsStudent Dec 26 '18 at 18:03
• That's not a separate meaning. You can use "certainly" to mean "to be sure" (for example, "To be sure, not everyone does it; but most people do" can be rephrased as "Certainly, not everyone does it; but most people do"); but you can't use "to be sure" in all cases where you can use "certainly" (for example, "It's certainly possible; would you like us to do it now?" can't be rephrased as "To be sure, it's possible; would you like us to do it now?"). If the word "admittedly" doesn't work, then the phrase "to be sure" doesn't work, either. – ruakh Dec 26 '18 at 18:38

The PMF in the question, $$f(x_i)=p^{x_i}(1-p)^{1-x_i}$$ belongs to Bernoulli distribution, where $$x_i$$ is a binary variable. Yours is the PMF of Binomial distribution, and $$x_i$$'s are Binomial RVs ($$n$$ of them actually) with parameters $$(m,p)$$.