So say I'm dealing with M&M. I have 5 M&Ms and one of them is yellow. I want to know if the true proportion of M&Ms in the entire bag is 10% or 20%.

In the frequentist approach, my hypothesis would be: H0: P = 10% H1: P > 10%

However, in the Bayesian approach, I would say:
H0: P = 10%
H1: P = 20%

Philosophically, why is this?

Updated question

I'm going to attempt to solve this using the Bayesian approach. What am I missing?

Say our priors are:
P(H1) = 0.5
P(H2) = 0.5

$P(k = 1 | H1) = 0.33$ (I used binomial distribution)
$P(k = 1 | H2) = 0.41$ (I used binomial distribution)


P(H1 | k = 1) = $\frac{0.5 * 33}{0.5 * 33 + 0.5 * 0.41} = 0.45$
P(H2 | k = 1) = $0.55$

So we choose H2.

  • 1
    $\begingroup$ Just an observation: you can test $H_0 : p=10\%$ against $H_1 : p=20\%$ using classical/frequentist methods. In fact, when both the null and the alternative are simple, there is a most powerful test available (which would probably have great appeal to a classical/frequentist statistician): en.wikipedia.org/wiki/Neyman%E2%80%93Pearson_lemma $\endgroup$
    – Zen
    Commented May 26, 2017 at 0:20
  • 1
    $\begingroup$ This is not correct: a Bayesian analysis can evaluate any (point or compound) hypothesis against any (point or compound) hypothesis. $\endgroup$
    – Xi'an
    Commented May 29, 2017 at 7:21

1 Answer 1


It's not a matter of philosophy, it's wether we get where is bayesian inference being done.

Remember that under bayesian inference estimates come from the posterior distribution.

In your problem the bayesian approach does not limit you on what type of hypothesis test you want to make. What limits you is wether your hypothesis statements make sense in relation to the posterior distribution.

For instance if your posterior is absolutely continous, making statements suck as $H_0 = k$ does not make any sense. This is because to make bayesian hypothesis tests you simply have to compute the probability of the $H_0$ statement from the posterior and atoms under continous distributions have 0 probability.

As for what your particular problem concerns, you can still test the simple vs compound case.

From your question I can see $P(H_0) = .45$ this is $H _0: p = .1$, it is clear that $P(H _1) = .65$ where $H_1: p > .1$ (this a kolmogorov axiom of probability), then you reject $H _0$


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