I am currently trying to figure out some strangeness about using the Binomial distribution in Bayesian modeling to define the likelihood.

To make an example assume I have two conditions, and in each condition I do five repeated measurements, each of which can be defined as a single Bernoulli trial. So let's just say I get the results $Y_{1,i}=(1,1,1,0,0)$ for the first condition and $Y_{2,i}=(1,0,0,0,0)$. I want to compare the hypotheses that both $Y_{1}$ and $Y_{2}$ have the same probability of producing a $1$ (H1) vs. that they have a different probability (H2). For simplicity, I assume equal prior probability of both hypotheses. Also, assume flat prior on all parameters.

So the first hypothesis can be parametrized by two probabilities $\theta_1$ and $\theta_2$. So for $Y_1$ using a Binomial distribution I get three $1$s out of five and therefore $P(N=3|\theta_1)={5 \choose 3}\theta_1^3(1-\theta_1)^2$. Similarly, for $Y_2$ I get one $1$ out of five and therefore $P(N=1|\theta_2)={5 \choose 1}\theta_2^1(1-\theta_2)^4$. Now to get the total probability of H1 independently of $\theta_1$ and $\theta_2$ I need to multiply the two and marginalize out the parameters (i.e. integrate over the prior). Since I can split the multidimensional integral, I can just integrate each probability separately and then integrate:

$\int_0^1 {5 \choose 3}\theta_1^3(1-\theta_1)^2\;d\theta_1= {5 \choose 3}B(4,3)={5 \choose 3}\frac{\Gamma(4)\Gamma(3)}{\Gamma(7)}= \frac{5!}{(3!)(2!)}\frac{(3!)(2!)}{6!}=1/6$ for $Y_1$


$\int_0^1 {5 \choose 1}\theta_2^1(1-\theta_1)^4\;d\theta_2= {5 \choose 1}B(2,5)={5 \choose 1}\frac{\Gamma(2)\Gamma(5)}{\Gamma(7)}= \frac{5!}{(1!)(4!)}\frac{(1!)(4!)}{6!}=1/6$ for $Y_2$

and therefore $P(H1)=1/36$.

For the second hypothesis I only need a single parameter $\theta_1$, and thus I get four $1$s out of 10 and therefore $P(N=4|\theta_1)={10\choose 4}\theta_1^4(1-\theta_1)^6$. Now again I marginalize out $\theta_1$ and thus I get

$P(H2)=\int_0^1{10\choose 4}\theta_1^4(1-\theta_1)^6\;d\theta_1= \frac{10!}{(4!)(6!)}\frac{(4!)(6!)}{11!}=1/11$

So hypothesis H2 seems more likely. But looking at the formulas, I find that I will get $P(H1)=1/36$ and $P(H2)=1/11$ independently of the observation, because all values determined by the numbers of $1$s completely cancel out.

If I instead use a Bernoulli likelihood I get (derivation only for H2)

$P(H2)=\int_0^1 \theta_1^4(1-\theta_1)^6=\frac{(4!)(6!)}{11!}$



Which is actually dependent on the observation and therefore seems more correct. Now I have seen people using Binomial distributions as the final step in the likelihood definition in Bayesian samplers. So the question is, when would this work, and when would it fail?

I can see, that this might work (however I am not sure) when one is trying to estimate the parameters of each of the two models for H1 and H2. However, I have also seen this in tutorials about Baysian model selection, where a discrete random variable is used to switch between the two models. As far as I understood this method, the discrete variable just compares the integrals for each of the models (i.e. the probabilities after marginalizing out the parameters). So in that case, I assume that I would just get results independent of the observation?

So when is summarizing the data and then using a Binomial distribution safe, and when will it fail?


In this computation, you are not comparing the same events:

  1. Under the Binomial model and $H_1$, $P(H_1)=1/36$ is actually $P(N_1=3,N_2=1|H_1)$
  2. Under the Binomial model and $H_2$, $P(H_2)=1/11$ is actually $P(N_1+N_2=4|H_2)$
  3. Under the Bernoulli model and $H_2$, $P(H_2)=\frac{(4!)(6!)}{11!}$ is actually $P(Y_1=(1,1,1,0,0),Y_2=(1,0,0,0,0,0)|H_2)$
  4. Under the Bernoulli model and $H_1$, $P(H_1)=\frac{(3!)(2!)(1!)(4!)}{6!}$ is actually $P(Y_1=(1,1,1,0,0),Y_2=(1,0,0,0,0,0)|H_1)$

While the last two probabilities are about the same event, this is not true for the first two probabilities. One should compare $P(N_1=3,N_2=1|H_1)$ and $P(N_1=3,N_2=1|H_2)$ to make them commensurable, in which case one recovers the same probabilities as in 3. and 4.

  • $\begingroup$ Ok, I figured something like that should be at play here, but I am not sure I understand your answer completely. The comparison of 1 vs 2 uses the assumption that $ P(H_1|N_1=3,N_2=1)= P(N_1=3,N_2=1|H_1)\cdot C$ and $P(H_2|N_1+N_2=4|H_2)=P(H_2|N_1+N_2 = 4)\cdot C$ for the same $C$. But it seems this assumption is incorrect? Because obviously $P(Y_1=(1,1,1,0,0),Y=(1,0,0,0,0)|H_1)\neq P(N_1=3,N_2=1|H_1)$. So one does not recover the same probabilities in (correct versions of) 1 & 2 as in 3 & 4, but only up to a multiplicative constant? $\endgroup$ – LiKao Dec 14 '18 at 10:09
  • $\begingroup$ Never mind, I think I have figured out the answer to my commet on my own. So $C$ in the first equation should be $C_1=P(H_1)/P(N_1=3,N_2=1)$ and $C_2=P(H_2)/P(N_1+N_2=4)$. Now if $P(H_1)=P(H_2)$ (posteriori), then obviously $C_1\neq C_2$, because $N_1+N_2=4$ covers the (atomic) event $Y_1=(1,1,1,1,0)$ and $Y_2=(0,0,0,0,0)$, which is not covered by $N_1=3$ and $N_2=1$. Correct? $\endgroup$ – LiKao Dec 14 '18 at 10:38

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