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Best rejection region

StatisticsHard

A coin is tossed $5$ times, independently. You test $H 0 : P (H) = \frac{1}{2}$ against $H 1 : P (H) = \frac{3}{4}$ . The rejection region may be any set of the $32$ possible head-tail sequences, but the probability of rejecting under $H 0$ must be at most $\frac{1}{8}$ . What is the largest possible probability of rejecting under $H 1$ ?

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Hint

Rank the sequences by how much more likely they are under the alternative.

Solution

Under $H 0$ each sequence has probability $\frac{1}{32}$ , so the region holds at most $4$ sequences.
Under $H 1$ a sequence with $k$ heads has probability $\frac{3 ^{k}}{1024}$ , so sequences with more heads are the best ones to include.
Take the sequence with $5$ heads and any $3$ of the sequences with $4$ heads.
Their total probability is $\frac{243 + 3 \times 81}{1024} = \frac{486}{1024}$ , which reduces to $\frac{243}{512}$ .

Answer: $\frac{243}{512}$

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