a ) \"Using an unfair coin as a fair die\") Suppose you have an unfair coin with

Question

a ) "Using an unfair coin as a fair die") Suppose you have an unfair coin with a probability p   of coming up heads and

probability (1p) of coming up tails on any given flip.

Define a procedure for picking a number from 1 to 6 as follows:
- First, flip the coin 4 times.
- ifIf the total number of heads out of the 4 flips is anything other than 2, start over from the previous step.

- Assuming the number of heads is 2, report a number from 1 to 6 according to the table below:

If the sequence is,,,,Report the numbers

HHTT.                                                                                                     1
HTHT.                                                                                                     2
HTTH                                                                                                    3

THHT                                                                                                     4

THTH                                                                                                     5

TTHH                                                                                                     6

Show that this procedure generates the numbers 1 to 6 with equal probability.
What if anything does this have to do with sufficient statistics?

b) Suppose a certain unknown proportion F   of voters support a particular political candidate in the

upcoming election, and you are interested in making inferences about F . You ask a pollster to perform a

survey of n   people at random from the population, ask each person whether they support the candidate,

and report the sequence of their responses. Instead, the pollster just records the total number of people

out of n n who said that they support the candidate. Do you need to redo the survey? Why or why not?

c) Is it important to have a sufficient statistic for a parameter in order to do Bayesian inference? Why or why not?

Explanation / Answer

<The 3 questions are completey unrelated. Hence as per answering guidelines answering the first question in que>

The important point to note is that the first part of the repitition of the experiment in case there are anything other than 2 heads is not needed to determine the probability.

Out of 4 tosses there can be 2 heads and 2 tails in 4C2 ways i.e. 6. We have assigned a number to be chosen for each of these 6 cases. Probability of HHTT = p2(1-p)2.

Since the order do not matter of H and T hence for each of the 6 combinations, there are always 2 heads hence contributes a prob of p2 and there are 2 tails ans hence contributes a probability of (1-p)2. So the probability of each of these 6 events is always p2(1-p)2.

Hence the numbers are chosen with equal probability.

There is no link of sufficient statistic with this probability problem.

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