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I am unsure why I got this question wrong:

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So if my posterior probability of a head-based coin is currently 60%, then that means I'll choose heads and so my EV of a loss is 4 (40% * 10). However, if I wait and flip another coin, I immediately incur a loss of 1 + whatever my new EV is. What's my new EV seems to be the heart of the question.

How do I know what my new posterior is after I flip another coin?

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    $\begingroup$ Your question is answered in the pop-up commentary: "Using Bayes rule." Each possible outcome can lead to a different "new posterior," of course--but you only need to consider two cases. $\endgroup$
    – whuber
    Commented Oct 2, 2017 at 15:31
  • $\begingroup$ can you flesh it out for me? $\endgroup$
    – Jwan622
    Commented Oct 2, 2017 at 16:06
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    $\begingroup$ Sure: the coin can land heads. Compute the posterior and make a decision accordingly. It can land tails. Compute the posterior and make a decision accordingly. Assess your prospects in light of these possible outcomes. $\endgroup$
    – whuber
    Commented Oct 2, 2017 at 16:53
  • $\begingroup$ With numbers? If you write the answer and why I can credit you? $\endgroup$
    – Jwan622
    Commented Oct 2, 2017 at 22:30
  • $\begingroup$ Does the question provide me with enough information? $\endgroup$
    – Jwan622
    Commented Oct 2, 2017 at 22:34

1 Answer 1

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As indicated in the hint, you can find the posterior using Bayes rule. For example, if $P\left( B_H \right)=0.6$ is your current posterior probability that the coin is head-biased, and $P\left( {H|B_H} \right)=0.75$ the likelihood of obtaining head in the next flip (under the hypothesis of a head-biased coin), then the posterior probability of a head-biased coin after having observed head in the next flip is

$$ P\left( {{B_H}|H} \right) = \frac{{P\left( {H|{B_H}} \right)P\left( {{B_H}} \right)}}{{P\left( {H|{B_H}} \right)P\left( {{B_H}} \right) + P\left( {H|{B_T}} \right)P\left( {{B_T}} \right)}} \approx0.82 $$

Hence the expected loss (after having observed head and decided that the coin was biased toward heads) would be $$L(B_H,H)=1 + \left[1-P\left( {{B_H}|H} \right) \right] \cdot 10 \approx 2.8$$

You can do the same calculation and compute the expected loss after having observed $T$ (tail). I find that $P\left( {{B_T}|T} \right) \approx 0.67$, and that the expected loss (after having observed tail and decided that the coin was biased toward tails) is $L(B_T,T)\approx 4.3$.

However I am not entirely sure of what they mean with minimum posterior expected loss in the question, in particular when they write that the "minimum expected loss of making the decision now is 4": I think this is confusing, and should be just the "expected loss of making the decision now". The minimum expected loss is instead associated with the other possible course of action, that is making the decision after another flip. You can compute the expected loss of flipping another time the coin by taking the average of the expected loss of each of the outcomes of the flip (which we computed above), weighted by their probability. The probability $p(H)$, for example, is given by the denominator of the equation above $$ p(H) = {P\left( {H|{B_H}} \right)P\left( {{B_H}} \right) + P\left( {H|{B_T}} \right)P\left( {{B_T}} \right)} = 0.55$$ The expected loss of flipping the coin another time is then $$ p(H)\cdot L(B_H,H) + p(T)\cdot L(B_T,T) = 3.5 $$ So the correct answer should be the third one.

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  • $\begingroup$ Is this the right way to interpret the question? What does the question mean when it says the p of the coin being tail biased is p(0.25) $\endgroup$
    – Jwan622
    Commented Oct 3, 2017 at 13:11
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    $\begingroup$ The outcome of the coin flip is a Bernoulli random variable, so the $p$ is the parameter that indicates the expected proportion of heads over a large number of coin flips. The question asks you to decide between two options: either the coin is head-biased ($p=0.75$, hence heads are obtained on 75% of the times on average) or it is tail-biased ($p=0.25$, meaning that tails are obtained 75% of the times). $\endgroup$
    – matteo
    Commented Oct 3, 2017 at 17:05
  • $\begingroup$ I don't see another way to interpret the question. However I am not entirely sure about what it means with "minimum expected posterior loss of making the decision now is 4". I think this should be called just the "expected posterior loss" of the first possible course of action (that is: deciding now). Bayes theorem allows then to compute the expected loss of the other possible course of action (flipping the coin and deciding after). Only then one could compare the two expected loss and know which one has the minimum expected loss $\endgroup$
    – matteo
    Commented Oct 3, 2017 at 17:18
  • $\begingroup$ I have edited the answer. let me know if the third is the correct one :-) as I explain in the edited answer I find a bit odd the way how the word minimum is used in the text of the question. $\endgroup$
    – matteo
    Commented Oct 3, 2017 at 18:32

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