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I have the following problem:

Lets flip a coin where the probability of getting heads is $\theta$. Write down $\theta$'s likelihood function in the three following cases:

  1. Throw the coin $n$ times, getting a total of 6 heads.
  2. Throw the coin $n$ times, getting the result (heads, heads, heads, heads, heads, heads, tails, tails, ..., tails), i.e. the first 6 flips are heads and the rest $n-6$ are tails.
  3. Keep throwing the coin until you get 6 heads. It is noted that this required $n$ flips total.

This is my attempt to solve this:

  1. $L(\theta)= {{n}\choose{6}}\theta^6(1-\theta)^{n-6}\propto \theta^6(1-\theta)^{n-6}$
  2. $L(\theta)= \theta^6(1-\theta)^{n-6}$

  3. ?

My question is: Are my 1-2 solutions correct and what is the answer to part 3.? Is it the same as solution to part 2. or not?

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  • $\begingroup$ Your solution for parts $(1)$ and $(2)$ look good. $\endgroup$ – quasi Sep 9 '17 at 10:56
  • $\begingroup$ But let me ask, for part $(3)$, is the rule "stop when $6$ consecutive heads", or just "stop when $6$ heads so far, not necessarily consecutive"? $\endgroup$ – quasi Sep 9 '17 at 10:59
  • $\begingroup$ Hi @quasi I think in the problem it is "stop when $6$ heads so far, not necessarily consecutive" $\endgroup$ – jjepsuomi Sep 9 '17 at 12:09
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Your answers for parts $(1)$ and $(2)$ look good.

For part $(3)$, subtract the respective probabilities for $k\;$heads, where $0 \le k\le 5$, from $1$, hence $$L(\theta) = 1 - \sum_{k=0}^5{\small{\binom{n}{k}}}\theta^k(1-\theta)^{n-k}$$

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  • $\begingroup$ Thank you:) appreciate it very much. $\endgroup$ – jjepsuomi Sep 9 '17 at 17:17

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