Maximum likelihood estimator of binominal distribution

superwolf

[tex]
L(x_1,...,x_n;p)=\Pi_{i=1}^{n}(\stackrel{n}{x_i}) p^{x_i}(1-p)^{n-x_i}
[/tex]

Correct so far?

The solution tells me to skip the [tex]\Pi[/tex]:

[tex]
L(x_1,...,x_n;p)=(\stackrel{n}{x}) p^{x}(1-p)^{n-x}
[/tex]

This is contradictory to all the examples in my book. Why?

Defennder

I don't understand why you wrote L(x1...xn,p). I thought the purpose was to estimate p, the probability of a designated success outcome in a Bernoulli trial. So it should be L (p) as p is the only parameter.

I also don't see any sense in omitting the multiplicative pi symbol. What is x here, anyway? x_i all refer to the observed no. of succeses of each sample size n. So what is x?

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superwolf	Maximum likelihood estimator of binominal distribution Tweet [tex] L(x_1,...,x_n;p)=\Pi_{i=1}^{n}(\stackrel{n}{x_i}) p^{x_i}(1-p)^{n-x_i} [/tex] Correct so far? The solution tells me to skip the [tex]\Pi[/tex]: [tex] L(x_1,...,x_n;p)=(\stackrel{n}{x}) p^{x}(1-p)^{n-x} [/tex] This is contradictory to all the examples in my book. Why?

Defennder Recognitions: Homework Help	I don't understand why you wrote L(x1...xn,p). I thought the purpose was to estimate p, the probability of a designated success outcome in a Bernoulli trial. So it should be L (p) as p is the only parameter. I also don't see any sense in omitting the multiplicative pi symbol. What is x here, anyway? x_i all refer to the observed no. of succeses of each sample size n. So what is x?