The relationship between the beta and binomial distributions is well known.

$$\frac{\pi^{\alpha - 1} (1 - \pi)^{\beta - 1}}{B(\alpha, \beta)} \leftrightarrow {{n}\choose{x}}\pi^{x} (1 - \pi)^{n-x}$$

By comparing the two, one can see:

  1. $\alpha - 1$ is analogous to the number of successes, $x$
  2. $\beta - 1$ is analogous to the number of failures, $n-x$, and thus
  3. $\alpha + \beta - 2$ is analogous to the number of trials, $n$.

I am faced with the problem of testing whether two independent binomial random variables of different size have the following relation:

First we observe a random binomial, $X_1 \sim Bin(n_1, p_1)$.

Later, we observe another random binomial, $X_2 \sim Bin(n_2, p_2)$.

  • $H_0: p2 = 0.65p_1$
  • $H_a: p2 \ge 0.65p_1$

My thought is to take a Bayesian approach.

If I assume an uniform prior for $\pi_1$, $\pi_1 \sim Beta(1,1)$, then the posterior distribution is $\pi_1 | x_1 \sim Beta(x_1 + 1, n_1 - x_1 + 1)$

Question: Is it reasonable to do the following?

Since we know the outcome of the first trial $\{x_1, n_1\}$, as well as the size of the second trial, $n_2$, I want to choose $\alpha_2$ and $\beta_2$ such that:

  1. $\frac{\alpha_2}{\alpha_2 + \beta_2} = \frac{13}{20} \frac{x_1 + 1}{n_1 + 2}$, and
  2. $\alpha_2 + \beta_2 = n_2 + 2$

This suggests choosing the following values of $\alpha_2$ and $\beta_2$:

  • $\alpha_2$

$$ \begin{align} \frac{\alpha_2}{\alpha_2 + \beta_2} = \frac{\alpha_2}{n_2 + 2} & = \frac{13}{20} \frac{x_1 + 1}{n_1 + 2} \\ \alpha_2 & = \frac{13}{20} \frac{x_1 + 1}{n_1 + 2} (n_2 + 2) \end{align} $$

This intuitively make sense, since we believe the number of successes, $x_2$, will be roughly $\frac{13}{20} p_1$ of the $n_2$ trials (i.e. $\alpha_2 + \beta_2 = n_2 + 2$).

  • $\beta_2$

$$ \begin{align} \beta_2 = n_2 + 2 - \alpha_2 & = n_2 + 2 - \frac{13}{20} \frac{x_1 + 1}{n_1 + 2} (n_2 + 2) \\ & = \left(1 - \frac{13}{20} \frac{x_1 + 1}{n_1 + 2}\right) (n_2 + 2) \end{align} $$

So, the prior would be $\pi_2 \sim Beta\left(\frac{13}{20} \frac{x_1 + 1}{n_1 + 2} (n_2 + 2), \left(1 - \frac{13}{20} \frac{x_1 + 1}{n_1 + 2}\right) (n_2 + 2)\right)$

Then the posterior distribution is $\pi_2 | x_2 \sim Beta\left(\frac{13}{20} \frac{x_1 + 1}{n_1 + 2} (n_2 + 2) + x_2, \left(1 - \frac{13}{20} \frac{x_1 + 1}{n_1 + 2}\right) (n_2 + 2) + n_2 - x_2\right)$

Is this reasonable? Thanks for any thoughts you have on the matter.


Based on Whuber's critique (and whuber is a stats.stackexchange God), perhaps I can replace the $n_2 + 2$ in my prior specification of $\pi_2$ with some other constant, $k$? This would maintain my assumption that $E(\Pi_2) = \frac{13}{20}E(\Pi_1)$, but I can adjust $k$ to modify the variance and thereby reflect my confidence in the chosen value?

So, the prior would be $\pi_2 \sim Beta\left(k * \frac{13}{20} \frac{x_1 + 1}{n_1 + 2}, k * \left(1 - \frac{13}{20} \frac{x_1 + 1}{n_1 + 2}\right) \right)$

Then the posterior distribution is $\pi_2 | x_2 \sim Beta\left(k * \frac{13}{20} \frac{x_1 + 1}{n_1 + 2} + x_2, k * \left(1 - \frac{13}{20} \frac{x_1 + 1}{n_1 + 2}\right) + n_2 - x_2\right)$

  • $\begingroup$ This is not a Bayesian approach, because you have let the data determine the prior. Just adopt reasonable priors for both outcomes, compute their posterior distributions, and evaluate the posterior probability that $p_2\ge 0.65p_1.$ If you also assume the two outcomes are independent, this is an easy integral. $\endgroup$
    – whuber
    Jul 6 at 20:01
  • $\begingroup$ Thanks for the quick answer, @whuber. However it's more accurate to say that I let the data from an earlier, independent trial determine the prior of a subsequent trial. I use the random data from the second trial only to update the prior (the sample size $n_2$ is not random). Is that distinction not relevant in this instance? $\endgroup$
    – Lewkrr
    Jul 6 at 20:11
  • $\begingroup$ Also, if including the $n_2$ value in my prior isn't Bayesian, how ought I construct a reasonable prior for $X_2$ if my assumption is that $p_2 = \frac{13}{20} p_1$? $\endgroup$
    – Lewkrr
    Jul 6 at 21:01
  • 1
    $\begingroup$ If you have two samples from the same distribution, you can either process them sequentially or simply combine them into one sample - the results will be the same either way. However, you still need a prior to start the whole process off. One way to check that what you are doing is valid (necessary, but not sufficient) is to do just that - combine the two samples and see if you get the same posterior with the same prior as when you do the exercise sequentially, first with one sample, then with the next. $\endgroup$
    – jbowman
    Jul 6 at 21:20
  • $\begingroup$ @jbowman: The problem with your suggestion is that I assume the two samples are NOT from the same distribution. From a frequentist perspective, the null hypothesis would be $H_0: p_2 = 0.65 p_1$, $H_a: p_2 < 0.65 p_1$. How would you approach such a problem? Thank you for the feedback. $\endgroup$
    – Lewkrr
    Jul 22 at 13:28


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