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If there are four possible hypotheses and I calculate the likelihood of the data given each of these hypotheses, can I calculate the probability of one of the hypotheses as

P(H1|Data) = P(Data|H1) / (P(Data|H1)+P(Data|H2)+P(Data|H3)+P(Data|H4))

Sorry if it's too basic question Thanks!

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You're close, but not quite there yet.

Using Bayes' Theorem we know that: $$P(H_{1}|Data) = \frac{P(Data|H_{1}) P(H_1)}{P(Data)}$$.

However we cannot state that: $$P(Data) \neq \sum_{i}^{4} P(Data|H_{i})$$, from definition we know that $P(Data) = \sum_{i}^{4}P(Data \cap H_{i})$, which is equivalent to saying that $$P(Data) = \sum_{i}^{4}P(Data|H_{i})*P(H_{i})$$.

So in order to fix your solution you would only have to scale each of the terms in the denominator with the corresponding probability of $P(H_{i})$.

So to summarize using your original notation we can compute $P(H_{1}|Data) = \frac{P(Data|H_{1}) P(H_1)}{P(Data|H_{1})*P(H_{1}) + P(Data|H_{2})*P(H_{2}) + P(Data|H_{3})*P(H_{3}) + P(Data|H_{4})*P(H_{4})}$

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  • $\begingroup$ Thanks Felix! Do I need to multiply the numerator by P(H1)? If that's the case, when P(H1)=P(H2)=P(H3)=P(H4), it should be equivalent to the formula I presented. I do not know P(H1)... PH(4), but I think in my case is reasonable to assume 0.25 for all of them $\endgroup$
    – Fede
    Commented Nov 22, 2018 at 22:28
  • $\begingroup$ No, you should not multiply the numerator by $P(H_{1})$. Such an assumption seems pretty reasonable to start, but fortunately you can always plug in different numbers to this formula! $\endgroup$
    – Felix van Doorn
    Commented Nov 22, 2018 at 22:51
  • $\begingroup$ If you don't multiply the numerator you can easily get P > 1. Maybe I am missing something! $\endgroup$
    – Fede
    Commented Nov 22, 2018 at 23:09
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    $\begingroup$ It is worth noting that the rule that Felix is using here is called the law of total probability. $\endgroup$
    – Ben
    Commented Nov 22, 2018 at 23:13
  • $\begingroup$ Now I see, I had forgotten $P(H_{1})$ in the numerator (so you were right!), Ben has fortunately fixed this! Apologies for the confusion! $\endgroup$
    – Felix van Doorn
    Commented Nov 23, 2018 at 7:13

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